UP Board Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

प्रश्नावली 5.1

 

Question 1. प्रश्न 1 से 10 तक की सम्मिश्र संख्याओं में प्रत्येक को a + ib के रूप में व्यक्त कीजिए। \( (5i) (\frac{-3}{5} i) \)
Answer:\[ (5i) (\frac{-3}{5} i) \] \[ = -5 \times \frac{3}{5} \times i \times i \] \[ = -3i^2 \] \[ = -3(-1) \] \[ = 3 \]
इस प्रकार \( 3 = 3 + 0i \)
In simple words: This question simplifies a product of two complex numbers to the standard \( a+ib \) form, which results in a real number 3. This is achieved by multiplying the numerical parts and the imaginary units, remembering that \( i^2 = -1 \).

🎯 Exam Tip: Remember the fundamental property \( i^2 = -1 \) when simplifying expressions involving imaginary numbers, as this is crucial for converting to the standard \( a+ib \) form.

 

Question 2. \( i^9 + i^{19} \)
Answer:\[ i^9 + i^{19} \] \[ = i^{8} \cdot i + i^{18} \cdot i \] \[ = (i^2)^4 \cdot i + (i^2)^9 \cdot i \] \[ = (-1)^4 \cdot i + (-1)^9 \cdot i \] \[ = (1) \cdot i + (-1) \cdot i \] \[ = i - i \] \[ = 0 \]
इस प्रकार \( 0 = 0 + 0i \)
In simple words: The expression simplifies to 0 because powers of \( i \) cycle every four terms. \( i^9 \) becomes \( i \) and \( i^{19} \) becomes \( -i \), leading to their sum being zero.

🎯 Exam Tip: To simplify higher powers of \( i \), divide the exponent by 4 and use the remainder to determine the equivalent power of \( i \) (\( i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1 \)).

 

Question 3. \( i^{-39} \)
Answer:\[ i^{-39} \] \[ = \frac{1}{i^{39}} \] \[ = \frac{1}{i^{38} \cdot i} \] \[ = \frac{1}{(i^2)^{19} \cdot i} \] \[ = \frac{1}{(-1)^{19} \cdot i} \] \[ = \frac{1}{-1 \cdot i} \] \[ = \frac{1}{-i} \] \[ = \frac{1}{-i} \times \frac{i}{i} \] \[ = \frac{i}{-i^2} \] \[ = \frac{i}{-(-1)} \] \[ = \frac{i}{1} \] \[ = i \]
इस प्रकार \( i = 0 + 1i \)
In simple words: To simplify negative powers of \( i \), convert them to a fraction with a positive power. Then, rationalize the denominator by multiplying the numerator and denominator by \( i \) to eliminate \( i \) from the denominator, using the fact that \( i^2 = -1 \).

🎯 Exam Tip: When dealing with \( i \) in the denominator, always rationalize it by multiplying by \( \frac{i}{i} \) or the conjugate to express the complex number in the standard \( a+ib \) form.

 

Question 4. \( 3(7 + i7) + i(7 + i7) \)
Answer:\( 3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i^2 \) \( = 21 + 28i + 7(-1) \) \( [\because i^2 = -1] \) \( = 21 - 7 + 28i \) \( = 14 + 28i \)
In simple words: This question involves distributing and combining like terms. Multiply the real and imaginary parts separately, then substitute \( i^2 = -1 \) to simplify the expression into the standard \( a+ib \) format.

🎯 Exam Tip: Always group real and imaginary terms separately after expanding and substituting \( i^2 = -1 \) to clearly represent the complex number in the \( a+ib \) form.

 

Question 5. \( (1-i) - (-1+i6) \)
Answer:\( (1-i) - (-1+i6) = (1-i) + (1-6i) \) \( = 1-i+1-6i \) \( = 2-7i \)
In simple words: This problem asks for the subtraction of two complex numbers. Distribute the negative sign, then combine the real parts and the imaginary parts separately to get the final \( a+ib \) form.

🎯 Exam Tip: Be careful with signs when subtracting complex numbers; distributing the negative sign to all terms in the second complex number is a common point of error.

 

Question 6. \( (\frac{1}{5} + i\frac{2}{5}) - (4 + i\frac{5}{2}) \)
Answer:\[ (\frac{1}{5} + i\frac{2}{5}) - (4 + i\frac{5}{2}) \] \[ = \frac{1}{5} + i\frac{2}{5} - 4 - i\frac{5}{2} \] \[ = (\frac{1}{5} - 4) + i(\frac{2}{5} - \frac{5}{2}) \] \[ = (\frac{1-20}{5}) + i(\frac{4-25}{10}) \] \[ = -\frac{19}{5} + i(-\frac{21}{10}) \] \[ = -\frac{19}{5} - i\frac{21}{10} \]
In simple words: To subtract these complex numbers, first remove the parentheses, then group the real components and the imaginary components, and simplify each group by finding a common denominator.

🎯 Exam Tip: When combining fractions in complex numbers, ensure you find a common denominator for both the real and imaginary parts to avoid calculation errors.

 

Question 7. \( [(\frac{1}{3} + i\frac{7}{3}) + (4 + i\frac{1}{3})] - (-\frac{4}{3} + i) \)
Answer:\[ [(\frac{1}{3} + i\frac{7}{3}) + (4 + i\frac{1}{3})] - (-\frac{4}{3} + i) \] \[ = (\frac{1}{3} + 4 + i\frac{7}{3} + i\frac{1}{3}) - (-\frac{4}{3} + i) \] \[ = (\frac{1+12}{3} + i\frac{7+1}{3}) - (-\frac{4}{3} + i) \] \[ = (\frac{13}{3} + i\frac{8}{3}) - (-\frac{4}{3} + i) \] \[ = \frac{13}{3} + i\frac{8}{3} + \frac{4}{3} - i \] \[ = (\frac{13}{3} + \frac{4}{3}) + i(\frac{8}{3} - 1) \] \[ = (\frac{13+4}{3}) + i(\frac{8-3}{3}) \] \[ = \frac{17}{3} + i\frac{5}{3} \]
In simple words: Simplify the terms inside the square brackets first by combining real and imaginary parts, then subtract the final complex number by again combining the real and imaginary components.

🎯 Exam Tip: Work step-by-step when simplifying complex expressions with multiple operations, always grouping real parts with real parts and imaginary parts with imaginary parts.

 

Question 8. \( (1-i)^4 \)
Answer:\[ (1-i)^4 \] \[ = [(1-i)^2]^2 \] \[ = [1 - 2i + i^2]^2 \] \[ = [1 - 2i - 1]^2 \] \[ = [-2i]^2 \] \[ = (-2i) \times (-2i) \] \[ = 4i^2 \] \[ = 4(-1) \] \[ = -4 \]
इस प्रकार \( -4 = -4 + 0i \)
In simple words: To calculate \( (1-i)^4 \), first square \( (1-i) \), which simplifies to \( -2i \). Then, square the result, \( (-2i)^2 \), leading to \( 4i^2 \), which finally becomes \( -4 \) because \( i^2 = -1 \).

🎯 Exam Tip: For even powers of binomials like \( (a+bi)^n \), it's often easier to first calculate \( (a+bi)^2 \) and then raise that result to the remaining power, as it can significantly simplify intermediate steps.

 

Question 9. \( (\frac{1}{3} + 3i)^3 \)
Answer:\[ (\frac{1}{3} + 3i)^3 \] Applying \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \): \[ = (\frac{1}{3})^3 + 3(\frac{1}{3})^2(3i) + 3(\frac{1}{3})(3i)^2 + (3i)^3 \] \[ = \frac{1}{27} + 3(\frac{1}{9})(3i) + 3(\frac{1}{3})(9i^2) + 27i^3 \] \[ = \frac{1}{27} + 9 \times \frac{1}{9}i + 9(-1) + 27(-i) \] \[ = \frac{1}{27} + i - 9 - 27i \] \[ = (\frac{1}{27} - 9) + (1 - 27)i \] \[ = (\frac{1 - 9 \times 27}{27}) - 26i \] \[ = (\frac{1 - 243}{27}) - 26i \] \[ = -\frac{242}{27} - 26i \]
In simple words: Expand the given complex number using the binomial cube formula \( (a+b)^3 \). Substitute \( i^2 = -1 \) and \( i^3 = -i \) to simplify the terms, then combine the real and imaginary parts to express the result in the standard \( a+ib \) form.

🎯 Exam Tip: When using binomial expansion for complex numbers, remember to correctly evaluate powers of \( i \) (especially \( i^2, i^3 \)) and combine real and imaginary terms meticulously to avoid errors.

 

Question 10. \( (-2 - \frac{1}{3}i)^3 \)
Answer:\[ (-2 - \frac{1}{3}i)^3 \] \[ = (-1)^3 (2 + \frac{1}{3}i)^3 \] \[ = -(2 + \frac{1}{3}i)^3 \] Applying \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \): \[ = -[ (2)^3 + 3(2)^2(\frac{1}{3}i) + 3(2)(\frac{1}{3}i)^2 + (\frac{1}{3}i)^3 ] \] \[ = -[ 8 + 3(4)(\frac{1}{3}i) + 6(\frac{1}{9}i^2) + \frac{1}{27}i^3 ] \] \[ = -[ 8 + 4i + \frac{2}{3}i^2 + \frac{1}{27}i^3 ] \] Substitute \( i^2 = -1 \) and \( i^3 = -i \): \[ = -[ 8 + 4i + \frac{2}{3}(-1) + \frac{1}{27}(-i) ] \] \[ = -[ 8 + 4i - \frac{2}{3} - \frac{1}{27}i ] \] Group real and imaginary parts: \[ = -[ (8 - \frac{2}{3}) + (4 - \frac{1}{27})i ] \] \[ = -[ (\frac{24-2}{3}) + (\frac{108-1}{27})i ] \] \[ = -[ \frac{22}{3} + \frac{107}{27}i ] \] \[ = -\frac{22}{3} - \frac{107}{27}i \]
In simple words: Factor out \( -1 \) from the expression, then apply the binomial cube formula. Substitute \( i^2 = -1 \) and \( i^3 = -i \), combine the real and imaginary terms, and finally apply the initial negative sign to get the complex number in the \( a+ib \) form.

🎯 Exam Tip: Factoring out a negative sign before expanding can simplify calculations, especially with odd powers. Be meticulous with fraction arithmetic and powers of \( i \).

 

Question 11. प्रश्न 11 से 13 तक की सम्मिश्र संख्याओं में प्रत्येक का गुणात्मक प्रतिलोम ज्ञात कीजिए। \( 4-3i. \)
Answer:\( 4-3i \) का गुणात्मक प्रतिलोम \( = \frac{1}{4-3i} \) \[ = \frac{1}{4-3i} \times \frac{4+3i}{4+3i} \] \[ = \frac{4+3i}{4^2 - (3i)^2} \] \[ = \frac{4+3i}{16 - 9i^2} \] \[ = \frac{4+3i}{16 - 9(-1)} \] \[ = \frac{4+3i}{16+9} \] \[ = \frac{4+3i}{25} \] \[ = \frac{4}{25} + \frac{3}{25}i \]
In simple words: To find the multiplicative inverse of a complex number, express it as \( \frac{1}{z} \) and then rationalize the denominator by multiplying the numerator and denominator by the conjugate of \( z \). Simplify the expression using \( i^2 = -1 \) to get the result in \( a+ib \) form.

🎯 Exam Tip: The multiplicative inverse of a complex number \( z = a+ib \) is \( z^{-1} = \frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2} \). Using this formula directly can save time in objective questions.

 

Question 12. \( \sqrt{5} + 3i. \)
Answer:\( \sqrt{5}+3i \) का गुणात्मक प्रतिलोम \[ = \frac{1}{\sqrt{5}+3i} \] \[ = \frac{1}{\sqrt{5}+3i} \times \frac{\sqrt{5}-3i}{\sqrt{5}-3i} \] \[ = \frac{\sqrt{5}-3i}{(\sqrt{5})^2 - (3i)^2} \] \[ = \frac{\sqrt{5}-3i}{5 - 9i^2} \] \[ = \frac{\sqrt{5}-3i}{5 - 9(-1)} \] \[ = \frac{\sqrt{5}-3i}{5+9} \] \[ = \frac{\sqrt{5}-3i}{14} \] \[ = \frac{\sqrt{5}}{14} - \frac{3}{14}i \]
In simple words: Find the multiplicative inverse by taking the reciprocal of the complex number. Then, multiply both the numerator and denominator by its conjugate to remove the imaginary part from the denominator, simplifying it to the standard \( a+ib \) form.

🎯 Exam Tip: Always remember that the product of a complex number and its conjugate \( (a+ib)(a-ib) \) is \( a^2+b^2 \), which is a real number, simplifying the rationalization process.

 

Question 13. \( -i. \)
Answer:\( -i \) का गुणात्मक प्रतिलोम \[ = \frac{1}{-i} \] \[ = \frac{1}{-i} \times \frac{i}{i} \] \[ = \frac{i}{-i^2} \] \[ = \frac{i}{-(-1)} \] \[ = \frac{i}{1} \] \[ = i \]
इस प्रकार \( i = 0+1i \)
In simple words: The multiplicative inverse of \( -i \) is found by taking its reciprocal \( \frac{1}{-i} \) and then multiplying by \( \frac{i}{i} \) to rationalize the denominator, which simplifies to \( i \).

🎯 Exam Tip: The multiplicative inverse of \( i \) is \( -i \), and of \( -i \) is \( i \). Recognize these common inverses to quickly solve simple cases.

 

Question 14. निम्नलिखित व्यंजक को a + ib के रूप में व्यक्त कीजिए: \[ \frac{(3+\sqrt{5}i)(3-\sqrt{5}i)}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-\sqrt{2}i)} \]
Answer:\[ \frac{(3+\sqrt{5}i)(3-\sqrt{5}i)}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-\sqrt{2}i)} \] Numerator: \( (3+\sqrt{5}i)(3-\sqrt{5}i) = 3^2 - (\sqrt{5}i)^2 = 9 - 5i^2 = 9 - 5(-1) = 9+5 = 14 \) Denominator: \( (\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-\sqrt{2}i) = \sqrt{3}+\sqrt{2}i-\sqrt{3}+\sqrt{2}i = 2\sqrt{2}i \) So, the expression becomes: \[ = \frac{14}{2\sqrt{2}i} \] \[ = \frac{7}{\sqrt{2}i} \] Rationalize the denominator: \[ = \frac{7}{\sqrt{2}i} \times \frac{i}{i} \] \[ = \frac{7i}{\sqrt{2}i^2} \] \[ = \frac{7i}{\sqrt{2}(-1)} \] \[ = -\frac{7i}{\sqrt{2}} \] Rationalize \( \sqrt{2} \) in the denominator: \[ = -\frac{7i}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \] \[ = -\frac{7\sqrt{2}i}{2} \]
इस प्रकार \( -\frac{7\sqrt{2}}{2}i = 0 - \frac{7\sqrt{2}}{2}i \)
In simple words: Simplify the numerator using \( (a+b)(a-b) = a^2-b^2 \) and the denominator by distributing the negative sign. Then, combine the results and rationalize the denominator by multiplying by \( \frac{i}{i} \) and \( \frac{\sqrt{2}}{\sqrt{2}} \) to get the final \( a+ib \) form.

🎯 Exam Tip: Simplify the numerator and denominator separately before dividing. Remember to rationalize both the imaginary unit \( i \) and any irrational real numbers in the denominator to reach the standard form.

 

प्रश्नावली 5.2

 

Question 1. प्रश्न 1 से 2 तक सम्मिश्र संख्याओं में प्रत्येक का मापांक और कोणांक ज्ञात कीजिए: \( z = -1 - i\sqrt{3} \)
Answer:मान लीजिए \( z = -1 - i\sqrt{3} = r(\cos \theta + i \sin \theta) \) अर्थात् \( r\cos \theta = -1 \) और \( r\sin \theta = -\sqrt{3} \) वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (-1)^2 + (-\sqrt{3})^2 \) \( r^2\cos^2 \theta + r^2\sin^2 \theta = 1 + 3 \) \( r^2(\cos^2 \theta + \sin^2 \theta) = 4 \) \( r^2(1) = 4 \) या \( r^2 = 4 \implies r = 2 \) (चूंकि मापांक हमेशा धनात्मक होता है) \( z \) का मापांक \( = 2 \) अब, \( \cos \theta = \frac{-1}{r} = \frac{-1}{2} \) और \( \sin \theta = \frac{-\sqrt{3}}{r} = \frac{-\sqrt{3}}{2} \) यहां पर \( \sin \theta \) व \( \cos \theta \) दोनों ऋणात्मक हैं, इसलिए \( \theta \) तीसरे चतुर्थांश में है। मानक कोण \( \alpha \) के लिए \( \tan \alpha = |\frac{-\sqrt{3}/2}{-1/2}| = |\sqrt{3}| \implies \alpha = \frac{\pi}{3} \) चूंकि \( \theta \) तीसरे चतुर्थांश में है, \( \theta = \pi + \alpha \) \( \theta = \pi + \frac{\pi}{3} = \frac{3\pi+\pi}{3} = \frac{4\pi}{3} \) या कोणांक \( = \frac{4\pi}{3} \) अतः कोणांक \( = \frac{4\pi}{3} \) और मापांक \( = 2 \).
In simple words: To find the modulus (\( r \)), square and add the real and imaginary parts of the complex number. To find the argument (\( \theta \)), use \( \cos \theta = \frac{Re(z)}{r} \) and \( \sin \theta = \frac{Im(z)}{r} \), then determine the correct quadrant for \( \theta \) based on the signs of \( \cos \theta \) and \( \sin \theta \).

🎯 Exam Tip: Always check the signs of \( \cos \theta \) and \( \sin \theta \) to correctly identify the quadrant of the argument \( \theta \), which determines whether to add or subtract from \( \pi \) or \( 2\pi \). For arguments, use principal value \( -\pi < \theta \le \pi \) or \( 0 \le \theta < 2\pi \) consistently.

 

Question 2. \( -\sqrt{3} + i. \)
Answer:मान लीजिए \( z = -\sqrt{3} + i = r(\cos \theta + i \sin \theta) \) \( \implies r\cos \theta = -\sqrt{3} \) और \( r\sin \theta = 1 \) वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (-\sqrt{3})^2 + (1)^2 \) \( r^2(\cos^2 \theta + \sin^2 \theta) = 3 + 1 \) \( r^2(1) = 4 \) या \( r^2 = 4 \implies r = 2 \) अब, \( \cos \theta = \frac{-\sqrt{3}}{r} = \frac{-\sqrt{3}}{2} \) \( \sin \theta = \frac{1}{r} = \frac{1}{2} \) चूंकि \( \sin \theta \) धनात्मक और \( \cos \theta \) ऋणात्मक है। \( \therefore \) \( \theta \) दूसरे चतुर्थांश में स्थित है। मानक कोण \( \alpha \) के लिए \( \tan \alpha = |\frac{1}{-\sqrt{3}}| = |\frac{1}{\sqrt{3}}| \implies \alpha = \frac{\pi}{6} \) चूंकि \( \theta \) दूसरे चतुर्थांश में है, \( \theta = \pi - \alpha \) \( \theta = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6} \) अतः कोणांक \( = \frac{5\pi}{6} \), मापांक \( = 2 \).
In simple words: Determine the modulus by squaring and adding the real and imaginary components. Then, find the argument by calculating \( \cos \theta \) and \( \sin \theta \), using their signs to correctly place \( \theta \) in the second quadrant, which requires subtracting the reference angle from \( \pi \).

🎯 Exam Tip: When \( \cos \theta \) is negative and \( \sin \theta \) is positive, the complex number lies in the second quadrant, and the argument is \( \pi - \text{reference angle} \).

 

Question 3. प्रश्न 3 से 8 तक सम्मिश्र संख्याओं में प्रत्येक को ध्रुवीय रूप में रूपांतरित कीजिए: \( 1-i \)
Answer:मान लीजिए \( z = 1 - i = r(\cos \theta + i \sin \theta) \) \( \therefore r\cos \theta = 1 \) तथा \( r\sin \theta = -1 \) वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (1)^2 + (-1)^2 \) \( r^2(\cos^2 \theta + \sin^2 \theta) = 1 + 1 \) या \( r^2(1) = 2 \) या \( r^2 = 2 \implies r = \sqrt{2} \) अब \( \cos \theta = \frac{1}{\sqrt{2}} \) और \( \sin \theta = \frac{-1}{\sqrt{2}} \) यहां \( \cos \theta \) धनात्मक है और \( \sin \theta \) ऋणात्मक है। \( \therefore \theta \) चौथे चतुर्थांश में है। मानक कोण \( \alpha \) के लिए \( \tan \alpha = |\frac{-1}{1}| = 1 \implies \alpha = \frac{\pi}{4} \) चूंकि \( \theta \) चौथे चतुर्थांश में है, \( \theta = 2\pi - \alpha \) या \( \theta = -\alpha \) (प्रमुख मान के लिए) \( \theta = -\frac{\pi}{4} \) अतः \( z \) का ध्रुवीय रूप \( = \sqrt{2} (\cos (-\frac{\pi}{4}) + i \sin (-\frac{\pi}{4})) \).
In simple words: Convert the complex number to polar form by first calculating the modulus \( r \) using \( \sqrt{x^2+y^2} \). Then, find the argument \( \theta \) using \( \cos \theta = \frac{x}{r} \) and \( \sin \theta = \frac{y}{r} \), placing it in the correct quadrant (fourth, so \( -\frac{\pi}{4} \) for principal argument).

🎯 Exam Tip: When \( \cos \theta \) is positive and \( \sin \theta \) is negative, the complex number lies in the fourth quadrant. The principal argument can be expressed as \( -\alpha \) or \( 2\pi - \alpha \).

 

Question 4. \( -1 + i. \)
Answer:मान लीजिए \( z = -1 + i = r(\cos \theta + i \sin \theta) \) \( \implies r\cos \theta = -1 \) और \( r\sin \theta = 1 \) इनका वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (-1)^2 + (1)^2 \) \( r^2\cos^2 \theta + r^2\sin^2 \theta = 1 + 1 \) या \( r^2(\cos^2 \theta + \sin^2 \theta) = 2 \) \( r^2(1) = 2 \) या \( r^2 = 2 \implies r = \sqrt{2} \) यहाँ \( \cos \theta \) ऋणात्मक तथा \( \sin \theta \) धनात्मक है। \( \therefore \) \( \theta \) दूसरे चतुर्थांश में है। मानक कोण \( \alpha \) के लिए \( \tan \alpha = |\frac{1}{-1}| = 1 \implies \alpha = \frac{\pi}{4} \) चूंकि \( \theta \) दूसरे चतुर्थांश में है, \( \theta = \pi - \alpha \) \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \) अतः \( z \) का ध्रुवीय रूप \( = \sqrt{2} (\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \).
In simple words: To convert to polar form, first find the modulus \( r \) by taking the square root of the sum of the squares of the real and imaginary parts. Then, determine the argument \( \theta \) using the signs of the real and imaginary parts to place it in the second quadrant, so \( \theta = \pi - \text{reference angle} \).

🎯 Exam Tip: For a complex number in the second quadrant (negative real, positive imaginary), the argument \( \theta \) is \( \pi \) minus the reference angle, which is found using \( \tan^{-1}(|\frac{y}{x}|) \).

 

Question 5. \( -1-i. \)
Answer:मान लीजिए \( z = -1 - i = r(\cos \theta + i \sin \theta) \) \( \therefore r\cos \theta = -1, r\sin \theta = -1 \) इनका वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (-1)^2 + (-1)^2 \) \( r^2\cos^2 \theta + r^2\sin^2 \theta = 1 + 1 \) या \( r^2(\cos^2 \theta + \sin^2 \theta) = 2 \) \( r^2(1) = 2 \) या \( r^2 = 2 \implies r = \sqrt{2} \) यहाँ \( \cos \theta \) और \( \sin \theta \) दोनों ही ऋणात्मक हैं। \( \therefore \theta \) तीसरे चतुर्थांश में है। मानक कोण \( \alpha \) के लिए \( \tan \alpha = |\frac{-1}{-1}| = 1 \implies \alpha = \frac{\pi}{4} \) चूंकि \( \theta \) तीसरे चतुर्थांश में है, \( \theta = \pi + \alpha \) (अगर \( 0 \le \theta < 2\pi \) में) या \( \theta = -\pi + \alpha \) (प्रमुख मान के लिए \( -\pi < \theta \le \pi \)) \( \theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4} \) \( z \) का ध्रुवीय रूप \( = \sqrt{2} (\cos (-\frac{3\pi}{4}) + i \sin (-\frac{3\pi}{4})) \) या \( \sqrt{2} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) \).
In simple words: To convert \( -1-i \) to polar form, calculate the modulus \( r = \sqrt{2} \). Since both real and imaginary parts are negative, the argument \( \theta \) is in the third quadrant. Using the principal argument, \( \theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4} \).

🎯 Exam Tip: When both \( \cos \theta \) and \( \sin \theta \) are negative (third quadrant), the principal argument \( \theta \) is typically found as \( -\pi + \text{reference angle} \) to stay within \( (-\pi, \pi] \).

 

Question 6. \( -3. \)
Answer:मान लीजिए \( z = -3 = r(\cos \theta + i \sin \theta) \) \( \therefore r\cos \theta = -3, r\sin \theta = 0 \) इनका वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (-3)^2 + (0)^2 \) \( r^2\cos^2 \theta + r^2\sin^2 \theta = 9 \) \( r^2(\cos^2 \theta + \sin^2 \theta) = 9 \) \( r^2(1) = 9 \) या \( r^2 = 9 \implies r = 3 \) अब \( \cos \theta = \frac{-3}{3} = -1 \) और \( \sin \theta = \frac{0}{3} = 0 \) \( \cos \theta = -1 \) और \( \sin \theta = 0 \) से, \( \theta = \pi \) \( z \) का ध्रुवीय रूप \( = 3(\cos \pi + i \sin \pi) \).
In simple words: For a purely real negative number like \( -3 \), the modulus \( r \) is its absolute value (3). Since it lies on the negative real axis, its argument \( \theta \) is \( \pi \) (180 degrees).

🎯 Exam Tip: For complex numbers lying on the real axis: if \( z = x \) where \( x > 0 \), then \( \theta = 0 \); if \( z = x \) where \( x < 0 \), then \( \theta = \pi \). For numbers on the imaginary axis, use \( \pm \frac{\pi}{2} \).

 

Question 7. \( \sqrt{3} + i \)
Answer:मान लीजिए \( z = \sqrt{3} + i = r(\cos \theta + i \sin \theta) \) \( \therefore r\cos \theta = \sqrt{3}, r\sin \theta = 1 \) वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (\sqrt{3})^2 + (1)^2 \) \( r^2\cos^2 \theta + r^2\sin^2 \theta = 3 + 1 \) या \( r^2(\cos^2 \theta + \sin^2 \theta) = 4 \) \( r^2(1) = 4 \) या \( r^2 = 4 \implies r = 2 \) अब \( \cos \theta = \frac{\sqrt{3}}{2} \) और \( \sin \theta = \frac{1}{2} \) \( \sin \theta \) और \( \cos \theta \) दोनों ही धनात्मक है। \( \therefore \theta \) पहले चतुर्थांश में है। इससे \( \theta = \frac{\pi}{6} \) \( z \) का ध्रुवीय रूप \( = 2 (\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) \).
In simple words: To convert \( \sqrt{3}+i \) to polar form, calculate the modulus \( r=2 \). Since both real and imaginary parts are positive, the argument \( \theta \) is in the first quadrant, directly given by \( \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6} \).

🎯 Exam Tip: When both real and imaginary parts are positive, the argument \( \theta \) is simply the reference angle in the first quadrant, where \( \theta = \tan^{-1}(\frac{y}{x}) \).

 

Question 8. \( i. \)
Answer:मान लीजिए \( z = i = r(\cos \theta + i \sin \theta) \) \( \implies r\cos \theta = 0 \) और \( r\sin \theta = 1 \) वर्ग करके जोड़ने पर, \( (r\cos \theta)^2 + (r\sin \theta)^2 = (0)^2 + (1)^2 \) \( r^2\cos^2 \theta + r^2\sin^2 \theta = 0 + 1 \) या \( r^2(\sin^2 \theta + \cos^2 \theta) = 1 \) \( r^2(1) = 1 \) या \( r^2 = 1 \implies r = 1 \) तब \( \cos \theta = \frac{0}{1} = 0 \) और \( \sin \theta = \frac{1}{1} = 1 \) \( \cos \theta = 0, \sin \theta = 1 \implies \theta = \frac{\pi}{2} \) \( z \) का ध्रुवीय रूप \( = (\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) \).
In simple words: For a purely imaginary number like \( i \), the modulus \( r \) is its coefficient (1). Since it lies on the positive imaginary axis, its argument \( \theta \) is \( \frac{\pi}{2} \) (90 degrees).

🎯 Exam Tip: For complex numbers lying on the imaginary axis: if \( z = iy \) where \( y > 0 \), then \( \theta = \frac{\pi}{2} \); if \( z = iy \) where \( y < 0 \), then \( \theta = -\frac{\pi}{2} \).

 

प्रश्नावली 5.3

 

Question 1. निम्नलिखित समीकरणों में से प्रत्येक को हल कीजिए: \( x^2 + 3 = 0. \)
Answer:\( x^2 + 3 = 0 \) \( x^2 = -3 \) \( x = \pm\sqrt{-3} \) \( x = \pm\sqrt{3 \times -1} \) \( x = \pm\sqrt{3}i \)
In simple words: To solve \( x^2+3=0 \), isolate \( x^2 \) and take the square root of both sides. Since the square root of a negative number is imaginary, the solution involves \( i \).

🎯 Exam Tip: Remember that \( \sqrt{-k} = \sqrt{k}i \) for any positive real number \( k \). Always include both positive and negative roots when solving quadratic equations.

 

Question 2. \( 2x^2 + x + 1 = 0. \)
Answer:दिया गया है: \( 2x^2 + x + 1 = 0 \) समीकरण \( ax^2 + bx + c = 0 \) से तुलना करने पर, \( a = 2, b = 1, c = 1 \) द्विघाती सूत्र \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) का प्रयोग करने पर: \[ x = \frac{-(1) \pm \sqrt{(1)^2 - 4(2)(1)}}{2(2)} \] \[ x = \frac{-1 \pm \sqrt{1 - 8}}{4} \] \[ x = \frac{-1 \pm \sqrt{-7}}{4} \] \[ x = \frac{-1 \pm \sqrt{7}i}{4} \]
In simple words: This quadratic equation has complex roots because its discriminant (\( b^2-4ac \)) is negative. Apply the quadratic formula, substituting \( \sqrt{-7} \) with \( \sqrt{7}i \).

🎯 Exam Tip: When the discriminant \( b^2-4ac \) is negative, the quadratic equation will have two distinct complex conjugate roots. Be careful with the signs and the \( i \) term.

 

Question 3. \( x^2 + 3x + 9 = 0. \)
Answer:दिए गए समीकरण \( x^2 + 3x + 9 = 0 \) की \( ax^2 + bx + c = 0 \) से तुलना करने पर, \( a = 1, b = 3, c = 9 \) द्विघाती सूत्र \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) का प्रयोग करने पर: \[ x = \frac{-(3) \pm \sqrt{(3)^2 - 4(1)(9)}}{2(1)} \] \[ x = \frac{-3 \pm \sqrt{9 - 36}}{2} \] \[ x = \frac{-3 \pm \sqrt{-27}}{2} \] \[ x = \frac{-3 \pm \sqrt{9 \times -3}}{2} \] \[ x = \frac{-3 \pm 3\sqrt{3}i}{2} \]
In simple words: Use the quadratic formula to solve this equation. Calculate the discriminant, which will be negative, leading to complex roots. Express \( \sqrt{-27} \) as \( 3\sqrt{3}i \).

🎯 Exam Tip: Always simplify the radical \( \sqrt{-k} \) completely before writing the final complex root. For example, \( \sqrt{-27} = \sqrt{9 \times -3} = 3\sqrt{-3} = 3\sqrt{3}i \).

 

Question 4. \( -x^2 + x - 2 = 0. \)
Answer:दिया गया है: \( -x^2 + x - 2 = 0 \) \( -1 \) से दोनों पक्षों में गुणा करने पर \( x^2 - x + 2 = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर, \( a = 1, b = -1, c = 2 \) द्विघाती सूत्र \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) का प्रयोग करने पर: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(2)}}{2(1)} \] \[ x = \frac{1 \pm \sqrt{1 - 8}}{2} \] \[ x = \frac{1 \pm \sqrt{-7}}{2} \] \[ x = \frac{1 \pm \sqrt{7}i}{2} \]
In simple words: First, multiply the entire equation by -1 to make the \( x^2 \) coefficient positive. Then, apply the quadratic formula, recognizing that the negative discriminant yields complex solutions involving \( i \).

🎯 Exam Tip: It is often easier to solve quadratic equations when the leading coefficient (\( a \)) is positive. Multiplying by -1 changes the signs but not the roots.

 

Question 5. \( x^2 + 3x + 5 = 0. \)
Answer:दिया गया है: \( x^2 + 3x + 5 = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर, \( a = 1, b = 3, c = 5 \) द्विघाती सूत्र \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) का प्रयोग करने पर: \[ x = \frac{-(3) \pm \sqrt{(3)^2 - 4(1)(5)}}{2(1)} \] \[ x = \frac{-3 \pm \sqrt{9 - 20}}{2} \] \[ x = \frac{-3 \pm \sqrt{-11}}{2} \] \[ x = \frac{-3 \pm \sqrt{11}i}{2} \]
In simple words: Use the quadratic formula to find the roots of the equation. Since the discriminant is negative, the solutions will be complex conjugates, expressed using \( i \).

🎯 Exam Tip: Double-check the arithmetic, especially for \( b^2-4ac \), as a small calculation error can lead to an incorrect discriminant and thus incorrect roots.

 

Question 6. \( x^2 - x + 2 = 0. \)
Answer:दिया है: \( x^2 - x + 2 = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर, \( a = 1, b = -1, c = 2 \) द्विघाती सूत्र \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) का प्रयोग करने पर: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(2)}}{2(1)} \] \[ x = \frac{1 \pm \sqrt{1 - 8}}{2} \] \[ x = \frac{1 \pm \sqrt{-7}}{2} \] \[ x = \frac{1 \pm \sqrt{7}i}{2} \]
In simple words: Apply the quadratic formula to solve the equation. The discriminant will be negative, leading to complex roots, which are represented by using \( i \) for the square root of the negative number.

🎯 Exam Tip: Be careful with the negative sign of \( b \) in the quadratic formula. \( -(-1) \) simplifies to \( 1 \).

 

Question 7. \( \sqrt{2}x^2 + x + \sqrt{2} = 0. \)
Answer:दिया है: \( \sqrt{2}x^2 + x + \sqrt{2} = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a = \sqrt{2}, b = 1, c = \sqrt{2} \) द्विघाती सूत्र \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) का प्रयोग करने पर: \[ x = \frac{-(1) \pm \sqrt{(1)^2 - 4(\sqrt{2})(\sqrt{2})}}{2(\sqrt{2})} \] \[ x = \frac{-1 \pm \sqrt{1 - 4(2)}}{2\sqrt{2}} \] \[ x = \frac{-1 \pm \sqrt{1 - 8}}{2\sqrt{2}} \] \[ x = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}} \] \[ x = \frac{-1 \pm \sqrt{7}i}{2\sqrt{2}} \]
In simple words: Use the quadratic formula. The coefficients involve \( \sqrt{2} \), so carefully perform the multiplication \( 4(\sqrt{2})(\sqrt{2}) = 4 \times 2 = 8 \). The negative discriminant will result in complex roots.

🎯 Exam Tip: When coefficients involve irrational numbers, ensure that the products are simplified correctly. \( \sqrt{k} \times \sqrt{k} = k \).

 

Question 8. √3 x² - √2 x + 3√3 = 0.
Answer: हल : दिया है : \( \sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a = \sqrt{3}, b = -\sqrt{2}, c = 3\sqrt{3} \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2-4(\sqrt{3})(3\sqrt{3})}}{2\sqrt{3}} \)
\( = \frac{\sqrt{2} \pm \sqrt{2-36}}{2\sqrt{3}} \)
\( = \frac{\sqrt{2} \pm \sqrt{-34}}{2\sqrt{3}} \)
\( = \frac{\sqrt{2} \pm \sqrt{34}i}{2\sqrt{3}} \)In simple words: This question solves a quadratic equation with complex roots using the quadratic formula. The values of a, b, and c are identified from the given equation, and substituted into the formula to find the two complex solutions.

🎯 Exam Tip: Remember to correctly identify the coefficients a, b, and c, especially when they involve square roots, to avoid errors in calculations. Pay close attention to the sign of the discriminant (\( b^2-4ac \)) as it determines whether the roots are real or complex.

 

Question 9. x² + x + \( \frac { 1 }{ \sqrt{2} } \) = 0
Answer: हल : दिया है : \( x^2+x+\frac{1}{\sqrt{2}}=0 \) दोनों पक्षों में \( \sqrt{2} \) से गुणा करने पर, \( \sqrt{2}x^2 + \sqrt{2}x + 1 = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a = \sqrt{2}, b = \sqrt{2}, c = 1 \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{-\sqrt{2} \pm \sqrt{(\sqrt{2})^2-4(\sqrt{2})(1)}}{2\sqrt{2}} \)
\( = \frac{-\sqrt{2} \pm \sqrt{2-4\sqrt{2}}}{2\sqrt{2}} \)
\( = \frac{-\sqrt{2} \pm \sqrt{2(1-2\sqrt{2})}}{2\sqrt{2}} \)
\( = \frac{-\sqrt{2} \pm \sqrt{2} \sqrt{1-2\sqrt{2}}}{2\sqrt{2}} \)
\( = \frac{-1 \pm \sqrt{1-2\sqrt{2}}}{2} \)
\( = \frac{-1 \pm \sqrt{2\sqrt{2}-1} i}{2} \)In simple words: This problem solves a quadratic equation by first clearing the fraction using multiplication by \( \sqrt{2} \). Then, the coefficients are identified, and the quadratic formula is applied to find the complex roots, simplifying the expression involving square roots and imaginary unit.

🎯 Exam Tip: When dealing with fractional coefficients, multiplying by the common denominator can simplify the equation before applying the quadratic formula. Ensure the simplification of square roots and handling of imaginary numbers is precise.

 

Question 10. x² + \( \frac { x }{ \sqrt{2} } \) + 1 = 0
Answer: हल : दिया है : \( x^2+\frac{x}{\sqrt{2}}+1=0 \) दोनों पक्षों में \( \sqrt{2} \) से गुणा करने पर \( \sqrt{2}x^2 + x + \sqrt{2} = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a = \sqrt{2}, b = 1, c = \sqrt{2} \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{-1 \pm \sqrt{1^2-4(\sqrt{2})(\sqrt{2})}}{2\sqrt{2}} \)
\( = \frac{-1 \pm \sqrt{1-8}}{2\sqrt{2}} \)
\( = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}} \)
\( = \frac{-1 \pm \sqrt{7}i}{2\sqrt{2}} \)In simple words: The problem is to solve a quadratic equation with a fractional coefficient. First, multiply the entire equation by \( \sqrt{2} \) to eliminate the fraction. Then, identify the coefficients a, b, and c, and use the quadratic formula to find the complex solutions.

🎯 Exam Tip: Always clear fractions in quadratic equations to simplify calculations. Be cautious with the signs under the square root, as a negative discriminant indicates complex roots, requiring the imaginary unit 'i'.

अध्याय 5 पर विविध प्रश्नावली

 

Question 1. \( i^{18}+(\frac{1}{i})^{25}\)^{3} \) का मान ज्ञात कीजिए।
Answer: हल : \( [i^{18}+(\frac{1}{i})^{25}]^3 \)
\( = [(i^2)^9 + \frac{1}{(i^2)^{12} i}]^3 \)
\( = [(-1)^9 + \frac{1}{(-1)^{12} i}]^3 \)
\( = [-1+\frac{1}{i}]^3 \)
\( = [-1+\frac{i}{i^2}]^3 \)
\( = [-1-i]^3 = -(1+i)^3 \) [अब \( (a + b)^3 = [a^3+3a^2b+3ab^2 + b^3] \)]
\( = -[1^3 + 3(1)^2(i) + 3(1)(i)^2 + (i)^3] \)
\( = -[1 + 3i + 3i^2 + i^3] \)
\( = -[1 + 3i - 3 - i] \)
\( = -[-2 + 2i] \)
\( = 2 - 2i \)In simple words: This problem simplifies a complex number expression involving powers of 'i' raised to a cube. First, powers of 'i' are reduced to \( \pm 1 \) or \( \pm i \), then the expression is simplified to \( -(1+i)^3 \). Finally, the binomial expansion for \( (a+b)^3 \) is used, and powers of 'i' are simplified again to get the final result in \( a+ib \) form.

🎯 Exam Tip: Remember the cyclical nature of powers of 'i' (\( i^1=i, i^2=-1, i^3=-i, i^4=1 \)). For \( \frac{1}{i} \), multiply by \( \frac{i}{i} \) to get \( -i \). Apply binomial expansion carefully and simplify at each step.

 

Question 2. किन्हीं दो सम्मिश्र संख्याओं \( z_1 \) और \( z_2 \) के लिए सिद्ध कीजिए :**Re(\( z_1 z_2 \)) = Re\( z_1 \) Re\( z_2 \) - Im \( z_1 \) Im \( z_2 \)**
Answer: हल : मान लीजिए \( z_1 = a + ib, z_2 = c + id \)
\( z_1 z_2 = (a + ib)(c + id) \)
\( = ac + adi + bci + i^2bd \)
\( = (ac-bd) + (ad+bc) i \) Re \( (z_1z_2) \) का वास्तविक भाग \( = ac - bd \) यहाँ पर Re\( z_1 \) का वास्तविक भाग \( = a \), Im \( z_1 \) का काल्पनिक भाग \( = b \) इसी प्रकार Re \( z_2 = c \), Im \( z_2 = d \)
\( = \text{Re}z_1 \text{Re}z_2 - \text{Im}z_1 \text{Im}z_2 \)In simple words: This problem proves a property of complex numbers. By defining two complex numbers, \( z_1 = a+ib \) and \( z_2 = c+id \), their product \( z_1z_2 \) is computed. Then, the real part of this product is compared with the product of their real parts minus the product of their imaginary parts, showing they are equal.

🎯 Exam Tip: To prove properties involving complex numbers, always start by expressing the complex numbers in their general form, \( a+ib \). Carefully perform the operations (multiplication, addition, etc.) and then identify the real and imaginary parts to compare with the desired result.

 

Question 3. \( (\frac{1}{1-4i} - \frac{2}{1+i}) (\frac{3-4i}{5+i}) \) को मानक रूप में परिवर्तित कीजिए।
Answer: हल : \( (\frac{1}{1-4i} - \frac{2}{1+i}) (\frac{3-4i}{5+i}) \)
\( = (\frac{1(1+i) - 2(1-4i)}{(1-4i)(1+i)}) (\frac{3-4i}{5+i}) \)
\( = (\frac{1+i - 2+8i}{1+i-4i-4i^2}) (\frac{3-4i}{5+i}) \)
\( = (\frac{-1+9i}{1-3i+4}) (\frac{3-4i}{5+i}) \)
\( = (\frac{-1+9i}{5-3i}) (\frac{3-4i}{5+i}) \) अब, पहले पद को परिमेयकृत करें: \( \frac{-1+9i}{5-3i} = \frac{-1+9i}{5-3i} \times \frac{5+3i}{5+3i} = \frac{-5-3i+45i+27i^2}{25-9i^2} = \frac{-5+42i-27}{25+9} = \frac{-32+42i}{34} = \frac{-16+21i}{17} \) दूसरे पद को परिमेयकृत करें: \( \frac{3-4i}{5+i} = \frac{3-4i}{5+i} \times \frac{5-i}{5-i} = \frac{15-3i-20i+4i^2}{25-i^2} = \frac{15-23i-4}{25+1} = \frac{11-23i}{26} \) अब दोनों को गुणा करें: \( (\frac{-16+21i}{17}) (\frac{11-23i}{26}) \)
\( = \frac{(-16)(11) + (-16)(-23i) + (21i)(11) + (21i)(-23i)}{17 \times 26} \)
\( = \frac{-176 + 368i + 231i - 483i^2}{442} \)
\( = \frac{-176 + 599i + 483}{442} \)
\( = \frac{307 + 599i}{442} \)
\( = \frac{307}{442} + \frac{599}{442}i \)In simple words: This problem requires simplifying a complex expression involving subtraction and multiplication of fractions with complex numbers. Each fractional part is first rationalized by multiplying the numerator and denominator by the conjugate. Then, the resulting complex numbers are multiplied to obtain the final answer in the standard \( a+ib \) form.

🎯 Exam Tip: Rationalizing complex fractions by multiplying by the conjugate is a crucial step. Ensure accurate distribution when multiplying complex numbers and remember that \( i^2 = -1 \) to correctly combine real and imaginary parts.

 

Question 4. यदि \( x - iy = \sqrt{\frac{a-ib}{c-id}} \), तो सिद्ध कीजिए \( x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}} \).
Answer: हल : दिया है: \( x-iy = \sqrt{\frac{a-ib}{c-id}} \) ...(1) \( i \) के स्थान पर \( -i \) लिखने पर (Replacing \( i \) with \( -i \)) \( x+iy = \sqrt{\frac{a-(-i)b}{c-(-i)d}} \) \( x+iy = \sqrt{\frac{a+ib}{c+id}} \) ...(2) समी. (1) और (2) का गुणा करने पर, \( (x-iy)(x+iy) = \sqrt{\frac{a-ib}{c-id}} \times \sqrt{\frac{a+ib}{c+id}} \)
\( (x^2 - (iy)^2) = \sqrt{\frac{(a-ib)(a+ib)}{(c-id)(c+id)}} \)
\( x^2 - i^2y^2 = \sqrt{\frac{a^2-i^2b^2}{c^2-i^2d^2}} \)
\( x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}} \) दोनों पक्षों का वर्ग करने पर, \( (x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2} \) [नोट: पुस्तक में प्रश्न में गलती है।]In simple words: The problem asks to prove a relationship between \( x^2+y^2 \) and given complex numbers. By taking the conjugate of the given equation, a second equation is formed. Multiplying these two equations simplifies the left side to \( x^2+y^2 \) and the right side to the square root of the product of the original and conjugate fractions, leading to the desired result. The note indicates a potential error in the textbook's problem statement.

🎯 Exam Tip: When proving identities involving complex numbers and their moduli, taking the conjugate and multiplying the original equation by its conjugate is a common and effective technique. Remember that \( |z|^2 = z \bar{z} \) and \( \sqrt{AB} = \sqrt{A}\sqrt{B} \).

 

Question 5. निम्नलिखित को ध्रुवीय रूप में परिवर्तित कीजिए :
**(i) \( \frac{1+7i}{(2-i)^2} \)**
**(ii) \( \frac{1+3i}{1-2i} \)**
Answer: हल : (i) माना \( z = \frac{1+7i}{(2-i)^2} \)
\( = \frac{1+7i}{4-4i+i^2} \)
\( = \frac{1+7i}{4-4i-1} \)
\( = \frac{1+7i}{3-4i} \) अब इसे मानक रूप में बदलें: \( z = \frac{1+7i}{3-4i} \times \frac{3+4i}{3+4i} \)
\( = \frac{3+4i+21i+28i^2}{9-16i^2} \)
\( = \frac{3+25i-28}{9+16} \)
\( = \frac{-25+25i}{25} \)
\( = -1+i \) ध्रुवीय रूप के लिए, माना \( -1+i = r(\cos \theta + i \sin \theta) \) तो, \( r\cos \theta = -1 \) और \( r\sin \theta = 1 \) वर्ग करके जोड़ने पर \( r^2\cos^2 \theta + r^2\sin^2 \theta = (-1)^2 + (1)^2 \)
\( r^2(\cos^2 \theta + \sin^2 \theta) = 1+1 \)
\( r^2 = 2 \) या \( r = \sqrt{2} \) अब \( \cos \theta = \frac{-1}{\sqrt{2}} \) और \( \sin \theta = \frac{1}{\sqrt{2}} \) यहाँ \( \cos \theta \) ऋणात्मक, \( \sin \theta \) धनात्मक है। अतः \( \theta \) दूसरे चतुर्थांश में है। \( \tan \theta = \frac{r\sin \theta}{r\cos \theta} = \frac{1}{-1} = -1 \) \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \) अतः \( z \) का ध्रुवीय रूप \( = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \)
Answer: (ii) हल : मान लिया \( z = \frac{1+3i}{1-2i} \) इसे मानक रूप में बदलें: \( z = \frac{1+3i}{1-2i} \times \frac{1+2i}{1+2i} \)
\( = \frac{1+2i+3i+6i^2}{1-4i^2} \)
\( = \frac{1+5i-6}{1+4} \)
\( = \frac{-5+5i}{5} \)
\( = -1+i \) भाग (i) के अनुसार \( -1+i = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \) अतः \( \frac{1+3i}{1-2i} = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \)In simple words: This problem involves converting two complex number expressions into their polar form. First, each expression is simplified to the standard \( a+ib \) form by expanding the denominator and rationalizing. Then, the modulus \( r \) is found using \( \sqrt{a^2+b^2} \), and the argument \( \theta \) is found using \( \tan \theta = b/a \) and identifying the correct quadrant, to write the number as \( r(\cos \theta + i \sin \theta) \).

🎯 Exam Tip: Always convert complex numbers to the standard \( a+ib \) form before finding their polar form. Pay close attention to the signs of 'a' and 'b' to correctly determine the quadrant of the argument \( \theta \) (especially for \( \tan^{-1} \) results) to get the correct principal value.

प्रश्न 6 से 9 में दिए गए प्रत्येक समीकरण को हल कीजिए:

 

Question 6. 3x² - 4x + \( \frac { 20 }{ 3 } \) = 0.
Answer: हल : \( 3x^2 - 4x + \frac{20}{3} = 0 \) को 3 से गुणा करने पर \( 9x^2-12x + 20 = 0 \) इसकी \( ax^2 + bx + c = 0 \) से तुलना करने पर, \( a = 9, b = -12, c = 20 \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{-(-12) \pm \sqrt{(-12)^2-4(9)(20)}}{2(9)} \)
\( = \frac{12 \pm \sqrt{144-720}}{18} \)
\( = \frac{12 \pm \sqrt{-576}}{18} \)
\( = \frac{12 \pm 24i}{18} \)
\( = \frac{2 \pm 4i}{3} \)
\( = \frac{2}{3} \pm \frac{4}{3}i \)In simple words: This problem solves a quadratic equation with a fractional constant term. First, multiply the entire equation by 3 to remove the fraction. Then, identify the coefficients a, b, and c, and apply the quadratic formula to find the complex roots, simplifying the result.

🎯 Exam Tip: Clear fractions by multiplying the entire equation by the least common multiple of the denominators. Carefully calculate the discriminant; a negative value leads to complex conjugate roots, which should be expressed using the imaginary unit 'i'.

 

Question 7. x² - 2x + \( \frac { 3 }{ 2 } \) = 0.
Answer: हल: \( x^2 - 2x + \frac{3}{2} = 0 \), इसे 2 से गुणा करने पर \( 2x^2 - 4x + 3 = 0 \) \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a = 2, b = -4, c = 3 \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{-(-4) \pm \sqrt{(-4)^2-4(2)(3)}}{2(2)} \)
\( = \frac{4 \pm \sqrt{16-24}}{4} \)
\( = \frac{4 \pm \sqrt{-8}}{4} \)
\( = \frac{4 \pm 2\sqrt{2}i}{4} \)
\( = \frac{2 \pm \sqrt{2}i}{2} \)
\( = 1 \pm \frac{\sqrt{2}}{2}i \)In simple words: To solve this quadratic equation with a fractional constant, first multiply by 2 to eliminate the fraction. Then, identify the coefficients a, b, and c, and substitute them into the quadratic formula. Simplify the resulting expression to find the complex roots in standard form.

🎯 Exam Tip: Clearing fractions at the beginning simplifies calculations for the quadratic formula. Be careful when simplifying square roots of negative numbers, remembering that \( \sqrt{-N} = \sqrt{N}i \), and simplify the fraction by dividing all terms in the numerator by the denominator.

 

Question 8. 21x²-28x + 10 = 0.
Answer: हल : दिए गए समीकरण \( 21x^2 - 28x + 10 = 0 \) की \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a=21, b=-28, c=10 \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{-(-28) \pm \sqrt{(-28)^2-4(21)(10)}}{2(21)} \)
\( = \frac{28 \pm \sqrt{784-840}}{42} \)
\( = \frac{28 \pm \sqrt{-56}}{42} \)
\( = \frac{28 \pm \sqrt{4 \times 14}i}{42} \)
\( = \frac{28 \pm 2\sqrt{14}i}{42} \)
\( = \frac{14 \pm \sqrt{14}i}{21} \)
\( = \frac{14}{21} \pm \frac{\sqrt{14}}{21}i \)
\( = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i \)In simple words: This problem solves a quadratic equation by identifying its coefficients and directly applying the quadratic formula. The negative discriminant indicates complex roots, which are then expressed using the imaginary unit 'i' and simplified to their simplest form.

🎯 Exam Tip: Always simplify the square root of the discriminant, especially when it involves a negative number. Ensure all terms in the numerator are divided by the denominator when simplifying the final complex number expression.

 

Question 9. 21x² - 28x + 10 = 0.
Answer: हल : दिए गए समीकरण \( 21x^2 - 28x + 10 = 0 \) की \( ax^2 + bx + c = 0 \) से तुलना करने पर \( a=21, b=-28, c=10 \)
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( = \frac{-(-28) \pm \sqrt{(-28)^2-4(21)(10)}}{2(21)} \)
\( = \frac{28 \pm \sqrt{784-840}}{42} \)
\( = \frac{28 \pm \sqrt{-56}}{42} \)
\( = \frac{28 \pm \sqrt{4 \times 14}i}{42} \)
\( = \frac{28 \pm 2\sqrt{14}i}{42} \)
\( = \frac{14 \pm \sqrt{14}i}{21} \)
\( = \frac{14}{21} \pm \frac{\sqrt{14}}{21}i \)
\( = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i \)In simple words: This problem solves a quadratic equation by identifying its coefficients and directly applying the quadratic formula. The negative discriminant indicates complex roots, which are then expressed using the imaginary unit 'i' and simplified to their simplest form.

🎯 Exam Tip: Always simplify the square root of the discriminant, especially when it involves a negative number. Ensure all terms in the numerator are divided by the denominator when simplifying the final complex number expression. Note: This question is a duplicate of Question 8.

 

Question 10. यदि \( z_1 = 2-i, z_2 = 1+i \), \( \frac{|z_1+z_2+1|}{|z_1-z_2+i|} \) का मान ज्ञात कीजिए।
Answer: हल : दिया है: \( z_1 = 2-i, z_2 = 1+i \) पहले अंश को हल करें: \( z_1+z_2+1 = (2-i) + (1+i) + 1 = 2-i+1+i+1 = 4 \) तो, \( |z_1+z_2+1| = |4| = 4 \) अब हर को हल करें: \( z_1-z_2+i = (2-i) - (1+i) + i = 2-i-1-i+i = 1-i \) तो, \( |z_1-z_2+i| = |1-i| = \sqrt{1^2+(-1)^2} = \sqrt{1+1} = \sqrt{2} \) अब अनुपात ज्ञात करें: \( \frac{|z_1+z_2+1|}{|z_1-z_2+i|} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \)In simple words: This problem asks for the value of the ratio of moduli of complex expressions. First, substitute the given complex numbers \( z_1 \) and \( z_2 \) into the numerator and denominator expressions separately, simplifying them to standard \( a+ib \) form. Then, calculate the modulus of each simplified expression and finally, compute their ratio.

🎯 Exam Tip: When evaluating complex expressions, substitute the values carefully and simplify each part (numerator and denominator) to the \( a+ib \) form before calculating the modulus. Remember that \( |a+ib| = \sqrt{a^2+b^2} \) and rationalize denominators if necessary in the final answer.

 

Question 11. यदि \( a + ib = \frac{(x+i)^2}{2x^2+1} \), सिद्ध कीजिए कि \( a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \).
Answer: हल : दिया है: \( a+ib = \frac{(x+i)^2}{2x^2+1} \) ...(1) \( i \) के स्थान पर \( -i \) रखने से (Replacing \( i \) with \( -i \)) \( a-ib = \frac{(x-i)^2}{2x^2+1} \) ...(2) समी. (1) और (2) को गुणा करने पर \( (a+ib)(a-ib) = \frac{(x+i)^2}{2x^2+1} \times \frac{(x-i)^2}{2x^2+1} \)
\( a^2 - (ib)^2 = \frac{[(x+i)(x-i)]^2}{(2x^2+1)^2} \)
\( a^2 - i^2b^2 = \frac{[x^2-i^2]^2}{(2x^2+1)^2} \)
\( a^2 + b^2 = \frac{[x^2-(-1)]^2}{(2x^2+1)^2} \)
\( a^2 + b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \) इति सिद्धम् ।In simple words: This problem proves an identity involving complex numbers. Starting with the given equation \( a+ib \), its conjugate \( a-ib \) is found by replacing \( i \) with \( -i \). Multiplying these two equations simplifies the left side to \( a^2+b^2 \) and the right side uses the property \( (A+B)(A-B)=A^2-B^2 \) to reach the desired expression.

🎯 Exam Tip: When proving identities involving \( a^2+b^2 \) from a complex number \( a+ib \), consider using the property \( a^2+b^2 = (a+ib)(a-ib) \). This often simplifies the proof significantly. Remember \( i^2 = -1 \) for correct sign changes.

 

Question 12. यदि \( z_1 = 2-i, z_2 = -2+i \), निम्न का मान ज्ञात कीजिए :
**(i) Re \( (\frac{z_1 z_2}{\bar{z_1}}) \)**
**(ii) Im \( (\frac{1}{z_1 \bar{z_1}}) \)**
Answer: हल : (i) दिया है: \( z_1 = 2-i \), \( z_2 = -2+i \) सबसे पहले \( z_1 z_2 \) ज्ञात करें: \( z_1 z_2 = (2-i)(-2+i) = -4+2i+2i-i^2 = -4+4i+1 = -3+4i \) अब \( \bar{z_1} \) ज्ञात करें: \( \bar{z_1} = 2-(-i) = 2+i \) अब \( \frac{z_1 z_2}{\bar{z_1}} \) ज्ञात करें: \( \frac{z_1 z_2}{\bar{z_1}} = \frac{-3+4i}{2+i} \) इसे मानक रूप में बदलें: \( \frac{-3+4i}{2+i} \times \frac{2-i}{2-i} = \frac{(-3)(2) + (-3)(-i) + (4i)(2) + (4i)(-i)}{2^2-i^2} \)
\( = \frac{-6+3i+8i-4i^2}{4-(-1)} \)
\( = \frac{-6+11i+4}{4+1} \)
\( = \frac{-2+11i}{5} \)
\( = -\frac{2}{5} + \frac{11}{5}i \) अतः Re \( (\frac{z_1 z_2}{\bar{z_1}}) = -\frac{2}{5} \)
Answer: (ii) दिया है: \( z_1 = 2-i \) \( z_1 \bar{z_1} = (2-i)(2+i) = 2^2 - i^2 = 4-(-1) = 4+1 = 5 \) अब \( \frac{1}{z_1 \bar{z_1}} \) ज्ञात करें: \( \frac{1}{z_1 \bar{z_1}} = \frac{1}{5} \) इसे मानक रूप में लिखें: \( \frac{1}{5} + 0i \) अतः Im \( (\frac{1}{z_1 \bar{z_1}}) = 0 \)In simple words: This problem involves calculating the real and imaginary parts of complex expressions. For part (i), first find the product \( z_1z_2 \) and the conjugate \( \bar{z_1} \). Then, divide the product by the conjugate, rationalizing the denominator, and extract the real part. For part (ii), calculate \( z_1\bar{z_1} \), find its reciprocal, and extract the imaginary part.

🎯 Exam Tip: Remember that \( z\bar{z} = |z|^2 \) is always a real number. When performing division of complex numbers, always multiply the numerator and denominator by the conjugate of the denominator to simplify the expression into the \( a+ib \) form, making it easy to identify real and imaginary parts.

 

Question 13. सम्मिश्र संख्या \( \frac{1+2i}{1-3i} \) का मापांक और कोणांक ज्ञात कीजिए।
Answer: हल : माना \( z = \frac{1+2i}{1-3i} \) इसे मानक रूप \( a+ib \) में बदलें: \( z = \frac{1+2i}{1-3i} \times \frac{1+3i}{1+3i} \)
\( = \frac{1(1+3i) + 2i(1+3i)}{1^2 - (3i)^2} \)
\( = \frac{1+3i+2i+6i^2}{1-9i^2} \)
\( = \frac{1+5i-6}{1+9} \)
\( = \frac{-5+5i}{10} \)
\( = -\frac{5}{10} + \frac{5}{10}i \)
\( = -\frac{1}{2} + \frac{1}{2}i \) अब मापांक और कोणांक ज्ञात करें। माना \( z = r(\cos \theta + i \sin \theta) \) तो \( r\cos \theta = -\frac{1}{2} \) और \( r\sin \theta = \frac{1}{2} \) वर्ग करके जोड़ने पर, \( r^2\cos^2 \theta + r^2\sin^2 \theta = (-\frac{1}{2})^2 + (\frac{1}{2})^2 \)
\( r^2(\cos^2 \theta + \sin^2 \theta) = \frac{1}{4} + \frac{1}{4} \)
\( r^2 = \frac{2}{4} = \frac{1}{2} \)
\( r = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \) मापांक \( |z| = \frac{1}{\sqrt{2}} \) कोणांक के लिए: \( \tan \theta = \frac{r\sin \theta}{r\cos \theta} = \frac{1/2}{-1/2} = -1 \) चूँकि \( \cos \theta \) ऋणात्मक और \( \sin \theta \) धनात्मक है, \( \theta \) दूसरे चतुर्थांश में है। इसलिए, \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \) अतः कोणांक \( \text{arg}(z) = \frac{3\pi}{4} \)In simple words: This problem requires finding the modulus and argument of a given complex number. First, the complex number is simplified to the standard \( a+ib \) form by rationalizing the denominator. Then, the modulus \( r \) is calculated as \( \sqrt{a^2+b^2} \). The argument \( \theta \) is determined using \( \tan \theta = b/a \) and identifying the correct quadrant based on the signs of \( a \) and \( b \).

🎯 Exam Tip: Always convert the complex number to the \( a+ib \) form before calculating the modulus and argument. Correctly identifying the quadrant of the complex number is crucial for finding the principal value of the argument, which is typically in the range \( (-\pi, \pi] \).

 

Question 14. यदि \( (x - iy) (3 + 5i) \), \( 6-24i \) की संयुग्मी है तो वास्तविक संख्याएँ \( x \) और \( y \) ज्ञात कीजिए।
Answer: हल : दिया है कि \( (x-iy)(3+5i) \) यह \( 6-24i \) की संयुग्मी है। \( 6-24i \) की संयुग्मी \( \overline{6-24i} = 6+24i \) इसलिए, \( (x-iy)(3+5i) = 6+24i \) बायां पक्ष (LHS) गुणा करने पर: \( (x-iy)(3+5i) = 3x+5xi-3iy-5i^2y \)
\( = 3x+5xi-3iy+5y \)
\( = (3x+5y) + (5x-3y)i \) वास्तविक व काल्पनिक संख्याओं को समान लिखते हुए, \( (3x+5y) + (5x-3y)i = 6+24i \) वास्तविक भागों की तुलना करने पर: \( 3x+5y = 6 \) ...(1) काल्पनिक भागों की तुलना करने पर: \( 5x-3y = 24 \) ...(2) समीकरण (1) को 3 से और समीकरण (2) को 5 से गुणा करने पर: \( 3 \times (3x+5y) = 3 \times 6 \implies 9x+15y = 18 \) ...(3) \( 5 \times (5x-3y) = 5 \times 24 \implies 25x-15y = 120 \) ...(4) समीकरण (3) और (4) को जोड़ने पर, \( (9x+15y) + (25x-15y) = 18+120 \) \( 34x = 138 \) \( x = \frac{138}{34} = \frac{69}{17} \) x का मान समीकरण (1) में रखने पर, \( 3(\frac{69}{17}) + 5y = 6 \) \( \frac{207}{17} + 5y = 6 \) \( 5y = 6 - \frac{207}{17} \) \( 5y = \frac{102-207}{17} \) \( 5y = \frac{-105}{17} \) \( y = \frac{-105}{17 \times 5} = \frac{-21}{17} \) अतः \( x = \frac{69}{17}, y = \frac{-21}{17} \).In simple words: This problem states that the product of two complex numbers is the conjugate of another complex number. First, find the conjugate of \( 6-24i \), which is \( 6+24i \). Then, multiply out the left-hand side \( (x-iy)(3+5i) \) and equate its real and imaginary parts to those of \( 6+24i \). This generates a system of two linear equations in terms of \( x \) and \( y \), which can be solved simultaneously to find their values.

🎯 Exam Tip: Remember that the conjugate of \( a-ib \) is \( a+ib \). When equating two complex numbers, their real parts must be equal, and their imaginary parts must be equal. This forms a system of linear equations, which can be solved using substitution or elimination methods.

 

Question 15. \( |\frac{1+i}{1-i} - \frac{1-i}{1+i}| \) का मापांक ज्ञात कीजिए।
Answer: हल : पहले \( \frac{1+i}{1-i} \) को सरल करें: \( \frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2-i^2} = \frac{1+2i+i^2}{1-(-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i \) अब \( \frac{1-i}{1+i} \) को सरल करें: \( \frac{1-i}{1+i} = \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{(1-i)^2}{1^2-i^2} = \frac{1-2i+i^2}{1-(-1)} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i \) अब दी गई अभिव्यक्ति में मान सब्स्टिट्यूट करें: \( |\frac{1+i}{1-i} - \frac{1-i}{1+i}| = |i - (-i)| \)
\( = |i+i| \)
\( = |2i| \) मापांक ज्ञात करें: \( |2i| = \sqrt{0^2+2^2} = \sqrt{4} = 2 \)In simple words: This problem asks for the modulus of a difference between two complex fractions. Each fraction is first simplified by multiplying the numerator and denominator by the conjugate of the denominator. Once simplified, the difference is calculated, and the modulus of the resulting complex number is found using the formula \( |a+ib| = \sqrt{a^2+b^2} \).

🎯 Exam Tip: Always simplify complex fractions before performing arithmetic operations. Rationalizing the denominator by multiplying by the conjugate is a key step. Remember that \( |2i| \) represents the distance from the origin to \( 0+2i \) in the complex plane, which is 2.

 

Question 16. यदि \( (x + iy)^3 = u + iv \), तो दर्शाइए कि \( \frac{u}{x} + \frac{v}{y} = 4(x^2-y^2) \).
Answer: हल : दिया है: \( (x + iy)^3 = u + iv \) बाईं ओर का विस्तार करें: \( (x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 \)
\( = x^3 + 3x^2yi + 3x(i^2y^2) + i^3y^3 \)
\( = x^3 + 3x^2yi - 3xy^2 - iy^3 \) [: \( i^2 = -1 \), \( i^3 = -i \)] वास्तविक और काल्पनिक भागों को अलग करें: \( (x^3-3xy^2) + (3x^2y-y^3)i = u+iv \) वास्तविक और काल्पनिक भागों की तुलना करने पर: \( u = x^3-3xy^2 \)
\( u = x(x^2-3y^2) \) तो, \( \frac{u}{x} = x^2-3y^2 \) ...(1) और \( v = 3x^2y-y^3 \)
\( v = y(3x^2-y^2) \) तो, \( \frac{v}{y} = 3x^2-y^2 \) ...(2) समीकरण (1) और (2) को जोड़ने पर: \( \frac{u}{x} + \frac{v}{y} = (x^2-3y^2) + (3x^2-y^2) \)
\( = x^2-3y^2+3x^2-y^2 \)
\( = 4x^2-4y^2 \)
\( = 4(x^2-y^2) \) इति सिद्धम् ।In simple words: This problem proves an identity by expanding a complex number raised to the power of three. First, the given expression \( (x+iy)^3 \) is expanded using the binomial theorem, simplifying powers of \( i \). Then, the real and imaginary parts are equated to \( u \) and \( v \), respectively. From these, expressions for \( \frac{u}{x} \) and \( \frac{v}{y} \) are derived. Finally, these two expressions are added to show that their sum equals \( 4(x^2-y^2) \).

🎯 Exam Tip: Remember the binomial expansion formula for \( (a+b)^3 \) and the powers of \( i \). Carefully separate the real and imaginary parts after expansion. Ensure correct algebraic manipulation when deriving \( \frac{u}{x} \) and \( \frac{v}{y} \) and adding them together.

 

Question 17. यदि α और ẞभिन्न सम्मिश्र संख्याएँ हैं जहाँ |β| = 1, तब \(\left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|\) का मान ज्ञात कीजिए।
Answer:
हल : \[ \left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right| \] हमें ज्ञात है कि \(|z_1/z_2| = |z_1|/|z_2|\) तथा \(|z|^2 = z\bar{z}\)
अतः \( \left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2 = \frac{|\beta-\alpha|^2}{|1-\bar{\alpha}\beta|^2} = \frac{(\beta-\alpha)(\overline{\beta-\alpha})}{(1-\bar{\alpha}\beta)(\overline{1-\bar{\alpha}\beta})} \) \( = \frac{(\beta-\alpha)(\bar{\beta}-\bar{\alpha})}{(1-\bar{\alpha}\beta)(1-\alpha\bar{\beta})} \) \( = \frac{\beta\bar{\beta} - \beta\bar{\alpha} - \alpha\bar{\beta} + \alpha\bar{\alpha}}{1 - \alpha\bar{\beta} - \bar{\alpha}\beta + \bar{\alpha}\beta\alpha\bar{\beta}} \) \( = \frac{|\beta|^2 - \beta\bar{\alpha} - \alpha\bar{\beta} + |\alpha|^2}{1 - \alpha\bar{\beta} - \bar{\alpha}\beta + |\alpha|^2|\beta|^2} \) दिया है : \(|\beta| = 1\) हो, तब \(|\beta|^2 = 1\)
\( = \frac{1 - \beta\bar{\alpha} - \alpha\bar{\beta} + |\alpha|^2}{1 - \alpha\bar{\beta} - \bar{\alpha}\beta + |\alpha|^2 \cdot 1} \) \( = \frac{1 - \beta\bar{\alpha} - \alpha\bar{\beta} + |\alpha|^2}{1 - \alpha\bar{\beta} - \bar{\alpha}\beta + |\alpha|^2} \) \( = 1 \)
\( \implies \left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2 = 1 \)
\( \implies \left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right| = 1 \)
In simple words: This problem demonstrates how to use the properties of complex conjugates and the modulus of a complex number (\(|z|^2 = z\bar{z}\)) to simplify a complex expression. By substituting \(|\beta|^2 = 1\) (since \(|\beta| = 1\)), the numerator and denominator become identical, leading to a modulus of 1.

🎯 Exam Tip: Mastering the properties of complex conjugates and modulus, especially \(|z_1z_2| = |z_1||z_2|\) and \(|z|^2 = z\bar{z}\), is crucial for solving such identity proofs in complex numbers.

 

Question 18. समीकरण \(|1-i|^x = 2^x\) के शून्येत्तर पूर्णांक मूलों की संख्या ज्ञात कीजिए :
Answer:
हल : \( |1-i|^x = ( \sqrt{1^2 + (-1)^2} )^x = ( \sqrt{1+1} )^x = ( \sqrt{2} )^x = (2)^{x/2} \) अब \( (2)^{x/2} = 2^x \) घातांकों की तुलना से, \( \frac{x}{2} = x \)
\( \implies x = 2x \)
\( \implies x - 2x = 0 \)
\( \implies -x = 0 \)
\( \implies x = 0 \) इस समीकरण का 0 के अतिरिक्त और कोई हल नहीं हो सकता।
In simple words: We first simplify the modulus of the complex number \(|1-i|\) to \(\sqrt{2}\) or \(2^{1/2}\). Then, we equate \((2^{1/2})^x\) with \(2^x\), which simplifies to \(x/2 = x\), leading to \(x=0\) as the only integer solution. Since the question asks for non-zero integer roots, there are no such solutions.

🎯 Exam Tip: When dealing with equations involving the modulus of complex numbers and exponents, first simplify the modulus. Then, convert both sides to the same base to easily compare exponents and solve for the variable, paying close attention to any constraints on the solution (e.g., non-zero integer).

 

Question 19. यदि \((a + ib)(c + id)(e + if)(g + ih) = A + iB\) है. तो दर्शाइए कि \((a^2 + b^2) (c^2 + d^2)(e^2 + f^2) (g^2 + h^2) = A^2 + B^2\).
Answer:
हल : दिया है: \((a + ib)(c + id)(e + if)(g + ih) = A + iB\) ...(1)
i के स्थान पर -i रखने पर, हमें समीकरण का संयुग्मी मिलता है:
\((a - ib)(c - id)(e - if)(g - ih) = A - iB\) ...(2)
समीकरण (1) और (2) को गुणा करने पर,
\([ (a + ib)(a - ib) ][ (c + id)(c - id) ][ (e + if)(e - if) ][ (g + ih)(g - ih) ] = (A + iB)(A - iB)\)
हमें पता है कि \( (x+iy)(x-iy) = x^2 - (iy)^2 = x^2 + y^2 \).
इसलिए, \( (a^2 - (ib)^2)(c^2 - (id)^2)(e^2 - (if)^2)(g^2 - (ih)^2) = A^2 - (iB)^2 \)
\( (a^2 - i^2b^2)(c^2 - i^2d^2)(e^2 - i^2f^2)(g^2 - i^2h^2) = A^2 - i^2B^2 \)
जैसा कि \( i^2 = -1 \), हम इसे प्रतिस्थापित करते हैं:
\( (a^2 - (-1)b^2)(c^2 - (-1)d^2)(e^2 - (-1)f^2)(g^2 - (-1)h^2) = A^2 - (-1)B^2 \)
\( (a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2 \)
यह सिद्ध हुआ।
In simple words: This proof utilizes the property that the product of a complex number and its conjugate equals the square of its modulus (\(z\bar{z} = |z|^2 = x^2+y^2\)). By taking the conjugate of the given equation and multiplying it with the original equation, we effectively convert each complex factor \((x+iy)\) into \((x^2+y^2)\), thereby proving the identity.

🎯 Exam Tip: The key to solving such problems is to understand and apply the complex conjugate property \((z \cdot \bar{z} = |z|^2)\). This technique is very common for proving identities involving the squares of real and imaginary parts of complex products.

 

Question 20. यदि \(\left(\frac{1+i}{1-i}\right)^m = 1\), तो m का न्यूनतम पूर्णांक मान ज्ञात कीजिए।
Answer:
हल : पहले हम \(\frac{1+i}{1-i}\) को सरल करते हैं:
\( \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{(1+i)^2}{1^2 - i^2} \) \( = \frac{1^2 + i^2 + 2 \cdot 1 \cdot i}{1 - (-1)} \) \( = \frac{1 - 1 + 2i}{1 + 1} = \frac{2i}{2} = i \) अब, दिया गया समीकरण बन जाता है:
\( (i)^m = 1 \) हम जानते हैं कि \(i\) की घातें चक्र में चलती हैं: \(i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1\), और फिर यह दोहराता है।
इसलिए, \(i^m = 1\) के लिए, \(m\) का सबसे छोटा धनात्मक पूर्णांक मान 4 है।
\(\implies m\) संख्या 4 का गुणज है।
\(\implies m\) की कम से कम मूल्य = 4.
In simple words: First, simplify the complex fraction \(\frac{1+i}{1-i}\) by multiplying the numerator and denominator by the conjugate of the denominator, which results in \(i\). Then, we need to find the smallest positive integer \(m\) such that \(i^m = 1\). The powers of \(i\) cycle as \(i, -1, -i, 1\), so the minimum integer value for \(m\) is 4.

🎯 Exam Tip: Always simplify complex fractions before raising them to a power. Remember the cycle of powers of \(i\) (\(i^1=i, i^2=-1, i^3=-i, i^4=1\)) to quickly find the minimum integer exponent for equations like \(i^m=1\).