GSEB Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations Misc. Questions

Question 1. Evaluate \( [i^{18} + (\frac{1}{i})^{25}]^3 \).
Answer: First, simplify the term \( \frac{1}{i} \):
\( \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i \)
Now substitute this back into the expression:
\( [i^{18} + (\frac{1}{i})^{25}]^3 = [i^{18} + (-i)^{25}]^3 \)
To simplify \( i^{18} \) and \( (-i)^{25} \), divide the exponent by 4 and use the remainder:
\( i^{18} = (i^4)^4 \cdot i^2 = (1)^4 \cdot (-1) = -1 \)
\( (-i)^{25} = (-1)^{25} \cdot i^{25} = -1 \cdot (i^4)^6 \cdot i^1 = -1 \cdot (1)^6 \cdot i = -i \)
Substitute these simplified terms back:
\( [-1 + (-i)]^3 = [-1-i]^3 \)
Use the binomial expansion formula \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \), with \( a = -1 \) and \( b = -i \):
\( (-1)^3 + 3(-1)^2(-i) + 3(-1)(-i)^2 + (-i)^3 \)
\( = -1 + 3(1)(-i) + 3(-1)(-1) + (-i^3) \)
\( = -1 - 3i + 3 + i \) (since \( i^2 = -1 \) and \( i^3 = -i \), so \( -i^3 = -(-i) = i \))
\( = (-1+3) + (-3i+i) \)
\( = 2 - 2i \)
In simple words: We first changed \( \frac{1}{i} \) to \( -i \). Then we simplified the powers of \( i \) and \( -i \) inside the brackets. After that, we used the cubic formula to expand the expression \( (-1-i)^3 \) and collected the real and imaginary parts to get the final answer.

Exam Tip: Remember to simplify negative exponents and complex fractions like \( \frac{1}{i} \) early on. For higher powers of \( i \), divide the exponent by 4 and use the remainder to find the equivalent power (e.g., \( i^{18} = i^2 = -1 \)).

Question 2. For complex numbers \( z_1 \) and \( z_2 \), prove that \( \text{Re}(z_1z_2) = \text{Re}(z_1)\text{Re}(z_2) - \text{Im}(z_1)\text{Im}(z_2) \).
Answer: Let \( z_1 = a + ib \), where \( a = \text{Re}(z_1) \) and \( b = \text{Im}(z_1) \).
Let \( z_2 = c + id \), where \( c = \text{Re}(z_2) \) and \( d = \text{Im}(z_2) \).
First, find the product \( z_1z_2 \):
\( z_1z_2 = (a + ib)(c + id) \)
\( = ac + aid + ibc + i^2bd \)
\( = ac + aid + ibc - bd \) (since \( i^2 = -1 \))
Rearrange the terms to separate the real and imaginary parts:
\( z_1z_2 = (ac - bd) + i(ad + bc) \)
From this, the real part of \( z_1z_2 \) is:
\( \text{Re}(z_1z_2) = ac - bd \)
Now, substitute the definitions of \( a, b, c, d \):
\( \text{Re}(z_1z_2) = \text{Re}(z_1)\text{Re}(z_2) - \text{Im}(z_1)\text{Im}(z_2) \)
Thus, the identity is proved.
In simple words: We started by writing our two complex numbers as \( a+ib \) and \( c+id \). Then we multiplied them together. After getting the product, we picked out the real part. Finally, we showed that this real part was the same as multiplying the real parts of the original numbers and subtracting the product of their imaginary parts.

Exam Tip: To prove identities involving complex numbers, always start by defining the complex numbers in their general form (e.g., \( a+ib \) or \( r(\cos \theta + i \sin \theta) \)) and then apply the given operations step-by-step, carefully separating real and imaginary parts.

Question 3. Reduce \( (\frac{1}{1 - 4i} - \frac{2}{1 + i}) (\frac{3 - 4i}{5 + i}) \) to the standard form.
Answer: We will evaluate each complex fraction separately first.
For the first term, \( \frac{1}{1 - 4i} \):
\( \frac{1}{1 - 4i} = \frac{1}{1 - 4i} \times \frac{1 + 4i}{1 + 4i} = \frac{1 + 4i}{1^2 - (4i)^2} = \frac{1 + 4i}{1 - 16i^2} = \frac{1 + 4i}{1 + 16} = \frac{1 + 4i}{17} = \frac{1}{17} + \frac{4}{17}i \)
For the second term, \( \frac{2}{1 + i} \):
\( \frac{2}{1 + i} = \frac{2}{1 + i} \times \frac{1 - i}{1 - i} = \frac{2(1 - i)}{1^2 - i^2} = \frac{2(1 - i)}{1 + 1} = \frac{2(1 - i)}{2} = 1 - i \)
Now, evaluate the first bracket: \( (\frac{1}{1 - 4i} - \frac{2}{1 + i}) \)
\( = (\frac{1}{17} + \frac{4}{17}i) - (1 - i) \)
\( = (\frac{1}{17} - 1) + (\frac{4}{17}i + i) \)
\( = (\frac{1 - 17}{17}) + (\frac{4 + 17}{17})i \)
\( = -\frac{16}{17} + \frac{21}{17}i \)
For the third term, \( \frac{3 - 4i}{5 + i} \):
\( \frac{3 - 4i}{5 + i} = \frac{3 - 4i}{5 + i} \times \frac{5 - i}{5 - i} = \frac{(3 - 4i)(5 - i)}{5^2 - i^2} = \frac{15 - 3i - 20i + 4i^2}{25 + 1} = \frac{15 - 23i - 4}{26} = \frac{11 - 23i}{26} \)
Finally, multiply the results of the two brackets:
\( (-\frac{16}{17} + \frac{21}{17}i) (\frac{11}{26} - \frac{23}{26}i) \)
\( = \frac{-16 \times 11}{17 \times 26} + \frac{-16 \times -23}{17 \times 26}i + \frac{21 \times 11}{17 \times 26}i + \frac{21 \times -23}{17 \times 26}i^2 \)
\( = \frac{-176}{442} + \frac{368}{442}i + \frac{231}{442}i - \frac{483}{442}(-1) \)
\( = \frac{-176}{442} + \frac{483}{442} + \frac{368 + 231}{442}i \)
\( = \frac{307}{442} + \frac{599}{442}i \)
In simple words: We broke the problem into three main parts, dealing with each complex fraction separately by multiplying the numerator and denominator by the conjugate. After simplifying each part, we performed the subtraction inside the first set of brackets. Finally, we multiplied the two resulting complex numbers to get the answer in the simplest \( a+ib \) form.

Exam Tip: When dealing with multiple complex fractions or operations, simplify each term separately before combining them. Remember to always multiply by the conjugate to rationalize the denominator and combine real and imaginary parts carefully.

Question 4. If \( x - iy = \sqrt{\frac{a-ib}{c-id}} \), prove that \( (x^2 + y^2)^2 = \frac{a^2+b^2}{c^2+d^2} \).
Answer: We are given the equation:
\( x - iy = \sqrt{\frac{a-ib}{c-id}} \) ... (1)
To get a term with \( x^2+y^2 \), we will use the conjugate. Replace \( i \) with \( -i \) in equation (1) to find its conjugate:
\( x + iy = \sqrt{\frac{a+ib}{c+id}} \) ... (2)
Now, multiply equation (1) by equation (2):
\( (x - iy)(x + iy) = \sqrt{\frac{a-ib}{c-id}} \times \sqrt{\frac{a+ib}{c+id}} \)
The left side simplifies using the difference of squares formula \( (A-B)(A+B) = A^2-B^2 \):
\( x^2 - (iy)^2 = \sqrt{\frac{(a-ib)(a+ib)}{(c-id)(c+id)}} \)
\( x^2 - i^2y^2 = \sqrt{\frac{a^2 - (ib)^2}{c^2 - (id)^2}} \)
Since \( i^2 = -1 \), we have:
\( x^2 + y^2 = \sqrt{\frac{a^2 + b^2}{c^2 + d^2}} \)
To remove the square root on the right side and obtain the desired expression, square both sides of this equation:
\( (x^2 + y^2)^2 = \left(\sqrt{\frac{a^2 + b^2}{c^2 + d^2}}\right)^2 \)
\( (x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2} \)
This proves the desired result.
In simple words: We started with the given equation. We then found its conjugate by changing all \( i \) to \( -i \). Multiplying the original equation by its conjugate simplified the left side to \( x^2+y^2 \). After that, we squared both sides to get rid of the square root on the right, which gave us the final expression we needed to prove.

Exam Tip: When asked to prove a relationship involving \( x^2+y^2 \) from an expression like \( x+iy \), always think of multiplying the given complex number by its conjugate. This will eliminate the imaginary parts and directly give terms like \( x^2+y^2 \).

Question 5. Convert the following in the polar form:
(i) \( \frac{1+7i}{(2-i)^2} \)
(ii) \( \frac{1 + 3i}{1 - 2i} \)
Answer:
(i) Convert \( \frac{1+7i}{(2-i)^2} \) to polar form:
First, simplify the denominator:
\( (2-i)^2 = 2^2 - 2(2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i \)
Now, the complex number becomes \( \frac{1+7i}{3-4i} \). Multiply the numerator and denominator by the conjugate of the denominator:
\( \frac{1+7i}{3-4i} = \frac{1+7i}{3-4i} \times \frac{3+4i}{3+4i} = \frac{(1+7i)(3+4i)}{3^2 - (4i)^2} = \frac{3 + 4i + 21i + 28i^2}{9 - 16i^2} \)
\( = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i \)
Now, convert \( -1 + i \) to polar form. Let \( z = -1 + i \).
The modulus \( r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \)
The argument \( \theta \) is found using \( \cos \theta = \frac{\text{Re}(z)}{r} = \frac{-1}{\sqrt{2}} \) and \( \sin \theta = \frac{\text{Im}(z)}{r} = \frac{1}{\sqrt{2}} \).
Since \( \cos \theta \) is negative and \( \sin \theta \) is positive, \( \theta \) lies in the second quadrant. The reference angle is \( \frac{\pi}{4} \).
So, \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)
The polar form is \( r(\cos \theta + i \sin \theta) = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \).

(ii) Convert \( \frac{1 + 3i}{1 - 2i} \) to polar form:
Multiply the numerator and denominator by the conjugate of the denominator:
\( \frac{1 + 3i}{1 - 2i} = \frac{1 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{(1 + 3i)(1 + 2i)}{1^2 - (2i)^2} = \frac{1 + 2i + 3i + 6i^2}{1 - 4i^2} \)
\( = \frac{1 + 5i - 6}{1 + 4} = \frac{-5 + 5i}{5} = -1 + i \)
This is the same complex number as in part (i), so its polar form will be identical.
The modulus \( r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2} \)
The argument \( \theta = \frac{3\pi}{4} \)
The polar form is \( \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \).
In simple words: For both parts, we first simplified the complex fraction into the basic \( a+ib \) form by multiplying the top and bottom by the complex conjugate of the denominator. Once we had \( -1+i \), we found its length (modulus) using Pythagoras' theorem, which was \( \sqrt{2} \). Then, we figured out its angle (argument) based on where it sits in the complex plane, which was \( \frac{3\pi}{4} \) radians.

Exam Tip: When converting complex numbers to polar form, always simplify the expression to \( x+iy \) form first. Then, calculate the modulus \( r = \sqrt{x^2+y^2} \) and determine the argument \( \theta \) using \( \tan \theta = \frac{y}{x} \) while paying careful attention to the quadrant of \( (x,y) \).

Solve each of the equations in questions 6 to 9:

Question 6. \( 3x^2 - 4x + \frac{20}{3} = 0 \).
Answer: First, eliminate the fraction by multiplying the entire equation by 3:
\( 3(3x^2 - 4x + \frac{20}{3}) = 3(0) \)
\( 9x^2 - 12x + 20 = 0 \)
This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a=9, b=-12, c=20 \).
Calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = (-12)^2 - 4(9)(20) \)
\( = 144 - 720 \)
\( = -576 \)
Since the discriminant is negative, the equation has complex solutions. Use the quadratic formula \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \):
\( x = \frac{-(-12) \pm \sqrt{-576}}{2(9)} \)
\( x = \frac{12 \pm \sqrt{576}i}{18} \)
\( x = \frac{12 \pm 24i}{18} \) (since \( \sqrt{576} = 24 \))
Divide both terms in the numerator by the denominator:
\( x = \frac{12}{18} \pm \frac{24i}{18} \)
\( x = \frac{2}{3} \pm \frac{4}{3}i \)
In simple words: We first got rid of the fraction by multiplying everything by 3. Then, we used the quadratic formula to solve for \( x \). Because the number under the square root was negative, our answers involved the imaginary unit \( i \), giving us two complex solutions.

Exam Tip: When a quadratic equation contains fractions, it's often easiest to clear the denominators first by multiplying by the Least Common Multiple (LCM) of the denominators. This prevents calculation errors with fractions in the quadratic formula.

Question 7. \( x^2 - 2x + \frac{3}{2} = 0 \).
Answer: First, eliminate the fraction by multiplying the entire equation by 2:
\( 2(x^2 - 2x + \frac{3}{2}) = 2(0) \)
\( 2x^2 - 4x + 3 = 0 \)
This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a=2, b=-4, c=3 \).
Calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = (-4)^2 - 4(2)(3) \)
\( = 16 - 24 \)
\( = -8 \)
Since the discriminant is negative, the equation has complex solutions. Use the quadratic formula \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \):
\( x = \frac{-(-4) \pm \sqrt{-8}}{2(2)} \)
\( x = \frac{4 \pm \sqrt{8}i}{4} \)
\( x = \frac{4 \pm 2\sqrt{2}i}{4} \) (since \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \))
Divide both terms in the numerator by the denominator:
\( x = \frac{4}{4} \pm \frac{2\sqrt{2}i}{4} \)
\( x = 1 \pm \frac{\sqrt{2}}{2}i \)
In simple words: We cleared the fraction by multiplying the whole equation by 2. Then, we applied the quadratic formula. Since the discriminant was negative, we knew the solutions would involve \( i \), the imaginary unit. We simplified the square root of \( -8 \) to \( 2\sqrt{2}i \) and then divided to get the final complex answers.

Exam Tip: Always simplify square roots of negative numbers carefully, remembering that \( \sqrt{-N} = \sqrt{N}i \). Also, make sure to simplify the final fraction by dividing all terms in the numerator by the denominator.

Question 8. \( 27x^2 - 10x + 1 = 0 \).
Answer: This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a=27, b=-10, c=1 \).
Calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = (-10)^2 - 4(27)(1) \)
\( = 100 - 108 \)
\( = -8 \)
Since the discriminant is negative, the equation has complex solutions. Use the quadratic formula \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \):
\( x = \frac{-(-10) \pm \sqrt{-8}}{2(27)} \)
\( x = \frac{10 \pm \sqrt{8}i}{54} \)
\( x = \frac{10 \pm 2\sqrt{2}i}{54} \) (since \( \sqrt{8} = 2\sqrt{2} \))
Divide both terms in the numerator by the denominator:
\( x = \frac{10}{54} \pm \frac{2\sqrt{2}i}{54} \)
\( x = \frac{5}{27} \pm \frac{\sqrt{2}}{27}i \)
In simple words: We used the quadratic formula to find the values of \( x \). The discriminant was negative, indicating complex answers. We simplified the square root of the negative number and then divided by the denominator to get the two complex solutions.

Exam Tip: Even if the numbers are larger, the process for solving quadratic equations with complex roots remains the same. Calculate the discriminant, then apply the quadratic formula, and simplify the square root of the negative number using \( i \).

Question 9. \( 21x^2 - 28x + 10 = 0 \).
Answer: This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a=21, b=-28, c=10 \).
Calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = (-28)^2 - 4(21)(10) \)
\( = 784 - 840 \)
\( = -56 \)
Since the discriminant is negative, the equation has complex solutions. Use the quadratic formula \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \):
\( x = \frac{-(-28) \pm \sqrt{-56}}{2(21)} \)
\( x = \frac{28 \pm \sqrt{56}i}{42} \)
\( x = \frac{28 \pm \sqrt{4 \times 14}i}{42} \)
\( x = \frac{28 \pm 2\sqrt{14}i}{42} \)
Divide both terms in the numerator by the denominator:
\( x = \frac{28}{42} \pm \frac{2\sqrt{14}i}{42} \)
\( x = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i \)
In simple words: We used the quadratic formula to solve for \( x \). After computing the discriminant, which was negative, we knew our solutions would be complex. We simplified the square root of \( -56 \) and then divided both parts of the numerator by the denominator to get the final answers.

Exam Tip: Always remember to simplify radicals (like \( \sqrt{56} \)) to their simplest form before concluding the answer. This often involves factoring out perfect squares from the number under the radical.

Question 10. If \( z_1 = 2-i \) and \( z_2 = 1 + i \), find \( |\frac{z_1+z_2+1}{z_1-z_2+i}| \).
Answer: We are given \( z_1 = 2-i \) and \( z_2 = 1+i \). We need to find the modulus of the complex fraction.
First, calculate the numerator \( z_1+z_2+1 \):
\( z_1+z_2+1 = (2-i) + (1+i) + 1 \)
\( = 2 - i + 1 + i + 1 \)
\( = (2+1+1) + (-i+i) \)
\( = 4 \)
Next, calculate the denominator \( z_1-z_2+i \):
\( z_1-z_2+i = (2-i) - (1+i) + i \)
\( = 2 - i - 1 - i + i \)
\( = (2-1) + (-i-i+i) \)
\( = 1 - i \)
Now, substitute these into the expression:
\( \frac{z_1+z_2+1}{z_1-z_2+i} = \frac{4}{1-i} \)
To simplify this complex fraction, multiply the numerator and denominator by the conjugate of the denominator:
\( \frac{4}{1-i} = \frac{4}{1-i} \times \frac{1+i}{1+i} = \frac{4(1+i)}{1^2 - i^2} = \frac{4(1+i)}{1 - (-1)} = \frac{4(1+i)}{2} = 2(1+i) \)
So, the complex number is \( 2+2i \).
Finally, find the modulus of this complex number:
\( |2(1+i)| = |2+2i| \)
\( = \sqrt{2^2 + 2^2} \)
\( = \sqrt{4 + 4} \)
\( = \sqrt{8} \)
\( = 2\sqrt{2} \)
In simple words: We first calculated the top part and the bottom part of the fraction separately, substituting the given values of \( z_1 \) and \( z_2 \). Then we simplified the resulting fraction by multiplying by the conjugate of the denominator. Once we had a simple complex number, we found its modulus, which is its distance from the origin on the complex plane.

Exam Tip: When evaluating complex expressions, calculate the numerator and denominator separately before dividing. Remember that \( |zw| = |z||w| \) and \( |\frac{z}{w}| = \frac{|z|}{|w|} \), which can sometimes simplify modulus calculations.

Question 11. If \( a+ib = \frac{(x+i)^2}{2x^2+1} \), prove that \( a^2+b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \).
Answer: We are given the equation:
\( a+ib = \frac{(x+i)^2}{2x^2+1} \) ... (1)
To find \( a^2+b^2 \), we will use the property that \( |z|^2 = z \cdot \overline{z} \).
First, find the conjugate of the given complex number. Replace \( i \) with \( -i \) in equation (1):
\( a-ib = \frac{(x-i)^2}{2x^2+1} \) ... (2)
Now, multiply equation (1) by equation (2):
\( (a+ib)(a-ib) = \left(\frac{(x+i)^2}{2x^2+1}\right) \left(\frac{(x-i)^2}{2x^2+1}\right) \)
The left side simplifies to \( a^2 - (ib)^2 = a^2 + b^2 \).
The right side can be grouped as:
\( a^2+b^2 = \frac{(x+i)^2 (x-i)^2}{(2x^2+1)(2x^2+1)} \)
\( a^2+b^2 = \frac{[(x+i)(x-i)]^2}{(2x^2+1)^2} \)
Apply the difference of squares formula \( (A+B)(A-B) = A^2-B^2 \) to the term in the square bracket:
\( (x+i)(x-i) = x^2 - i^2 = x^2 - (-1) = x^2+1 \)
Substitute this back into the equation:
\( a^2+b^2 = \frac{[x^2+1]^2}{(2x^2+1)^2} \)
\( a^2+b^2 = \frac{(x^2+1)^2}{(2x^2+1)^2} \)
This proves the desired result.
In simple words: We started with the given equation relating \( a+ib \) to an expression with \( x \) and \( i \). To get \( a^2+b^2 \), we multiplied the original equation by its complex conjugate. This simplifies the left side directly to \( a^2+b^2 \). On the right side, we used the difference of squares formula to simplify the expression, leading to the final proven statement.

Exam Tip: When a problem asks to prove a relationship involving \( a^2+b^2 \) from \( a+ib = Z \), remember the identity \( a^2+b^2 = |a+ib|^2 = (a+ib)(a-ib) \). This method often significantly simplifies the proof compared to separating real and imaginary parts explicitly.

Question 12. Let \( z_1 = 2 - i \) and \( z_2 = -2 + i \). Find:
(i) \( \text{Re}(\frac{z_1}{z_2}) \)
(ii) \( \text{Im}(\frac{1}{z_1 z_2}) \)
Answer: We are given \( z_1 = 2-i \) and \( z_2 = -2+i \).

(i) Calculate \( \text{Re}(\frac{z_1}{z_2}) \):
First, find the fraction \( \frac{z_1}{z_2} \):
\( \frac{z_1}{z_2} = \frac{2-i}{-2+i} \)
To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( -2-i \):
\( \frac{2-i}{-2+i} = \frac{(2-i)(-2-i)}{(-2+i)(-2-i)} \)
\( = \frac{2(-2) + 2(-i) - i(-2) - i(-i)}{(-2)^2 - (i)^2} \)
\( = \frac{-4 - 2i + 2i + i^2}{4 - i^2} \)
\( = \frac{-4 - 1}{4 - (-1)} \) (since \( i^2 = -1 \))
\( = \frac{-5}{5} \)
\( = -1 \)
So, \( \frac{z_1}{z_2} = -1 \).
The real part of this is \( \text{Re}(\frac{z_1}{z_2}) = \text{Re}(-1) = -1 \).

(ii) Calculate \( \text{Im}(\frac{1}{z_1 z_2}) \):
First, find the product \( z_1 z_2 \):
\( z_1 z_2 = (2-i)(-2+i) \)
\( = 2(-2) + 2(i) - i(-2) - i(i) \)
\( = -4 + 2i + 2i - i^2 \)
\( = -4 + 4i - (-1) \)
\( = -4 + 4i + 1 \)
\( = -3 + 4i \)
Now, find the reciprocal \( \frac{1}{z_1 z_2} \):
\( \frac{1}{-3+4i} \)
To simplify, multiply the numerator and denominator by the conjugate of \( -3+4i \), which is \( -3-4i \):
\( \frac{1}{-3+4i} = \frac{1}{-3+4i} \times \frac{-3-4i}{-3-4i} \)
\( = \frac{-3-4i}{(-3)^2 - (4i)^2} \)
\( = \frac{-3-4i}{9 - 16i^2} \)
\( = \frac{-3-4i}{9 - 16(-1)} \)
\( = \frac{-3-4i}{9 + 16} \)
\( = \frac{-3-4i}{25} \)
So, \( \frac{1}{z_1 z_2} = -\frac{3}{25} - \frac{4}{25}i \).
The imaginary part of this is \( \text{Im}(\frac{1}{z_1 z_2}) = -\frac{4}{25} \).
In simple words: For the first part, we divided \( z_1 \) by \( z_2 \) and simplified the fraction. Then we took the real part of the result. For the second part, we multiplied \( z_1 \) and \( z_2 \), then found the reciprocal of that product, simplifying it to its \( a+ib \) form. Finally, we extracted the imaginary part of that simplified fraction.

Exam Tip: When dealing with complex numbers, always ensure you correctly identify the conjugate (change the sign of the imaginary part). Be meticulous with arithmetic, especially signs, when multiplying and dividing complex numbers.

Question 13. Find the modulus and argument of the complex number \( \frac{1 + 2i}{1 - 3i} \).
Answer: First, simplify the complex number to the standard form \( x+iy \). Let \( z = \frac{1 + 2i}{1 - 3i} \).
Multiply the numerator and denominator by the conjugate of the denominator, which is \( 1+3i \):
\( z = \frac{1 + 2i}{1 - 3i} \times \frac{1 + 3i}{1 + 3i} \)
\( = \frac{(1 + 2i)(1 + 3i)}{(1)^2 - (3i)^2} \)
\( = \frac{1(1) + 1(3i) + 2i(1) + 2i(3i)}{1 - 9i^2} \)
\( = \frac{1 + 3i + 2i + 6i^2}{1 - 9(-1)} \)
\( = \frac{1 + 5i - 6}{1 + 9} \)
\( = \frac{-5 + 5i}{10} \)
\( = -\frac{5}{10} + \frac{5}{10}i \)
\( z = -\frac{1}{2} + \frac{1}{2}i \)
Now, find the modulus \( r \) of \( z \):
\( r = |z| = \sqrt{(-\frac{1}{2})^2 + (\frac{1}{2})^2} \)
\( = \sqrt{\frac{1}{4} + \frac{1}{4}} \)
\( = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} \)
\( r = \frac{1}{\sqrt{2}} \)
Next, find the argument \( \theta \). We use \( \cos \theta = \frac{x}{r} \) and \( \sin \theta = \frac{y}{r} \).
\( \cos \theta = \frac{-\frac{1}{2}}{\frac{1}{\sqrt{2}}} = -\frac{1}{2} \times \sqrt{2} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}} \)
\( \sin \theta = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)
Since \( \cos \theta \) is negative and \( \sin \theta \) is positive, the angle \( \theta \) lies in the second quadrant. The reference angle is \( \frac{\pi}{4} \).
Therefore, \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
In simple words: First, we simplified the complex fraction into a standard \( x+iy \) form. Then, we calculated its length (modulus) using the formula \( \sqrt{x^2+y^2} \). After that, we found its angle (argument) by looking at its cosine and sine values, making sure to place it in the correct quadrant on the complex plane.

Exam Tip: When finding the argument, always ensure you correctly identify the quadrant of the complex number \( x+iy \) to select the proper angle for \( \theta \). The arctan function alone may not give the correct principal argument for numbers in the second or third quadrants.

Question 14. Find the real number \( x \) and \( y \), if \( (x - iy)(3 + 5i) \) is conjugate of \( - 6 - 24i \).
Answer: First, find the conjugate of \( -6 - 24i \). The conjugate is found by changing the sign of the imaginary part, so it is \( -6 + 24i \).
Next, expand the left side of the equation: \( (x - iy)(3 + 5i) \)
\( = x(3) + x(5i) - iy(3) - iy(5i) \)
\( = 3x + 5xi - 3yi - 5i^2y \)
\( = 3x + 5xi - 3yi + 5y \) (since \( i^2 = -1 \))
Group the real and imaginary parts:
\( = (3x + 5y) + (5x - 3y)i \)
Now, equate this expanded form with the conjugate of \( -6 - 24i \):
\( (3x + 5y) + (5x - 3y)i = -6 + 24i \)
For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. This gives us a system of two linear equations:
1. \( 3x + 5y = -6 \) (Equating real parts)
2. \( 5x - 3y = 24 \) (Equating imaginary parts)
To solve this system, we can use the elimination method. Multiply equation (1) by 3 and equation (2) by 5:
\( 3 \times (3x + 5y) = 3 \times (-6) \implies 9x + 15y = -18 \)
\( 5 \times (5x - 3y) = 5 \times (24) \implies 25x - 15y = 120 \)
Add the two new equations:
\( (9x + 15y) + (25x - 15y) = -18 + 120 \)
\( 34x = 102 \)
\( x = \frac{102}{34} \)
\( x = 3 \)
Substitute the value of \( x = 3 \) into equation (1):
\( 3(3) + 5y = -6 \)
\( 9 + 5y = -6 \)
\( 5y = -6 - 9 \)
\( 5y = -15 \)
\( y = \frac{-15}{5} \)
\( y = -3 \)
Thus, the real numbers are \( x=3 \) and \( y=-3 \).
In simple words: We first found the conjugate of the given complex number. Then, we multiplied out the left side of the equation and grouped the real and imaginary parts. By setting the real parts equal and the imaginary parts equal, we got two simple equations. Solving these equations gave us the values for \( x \) and \( y \).

Exam Tip: When a problem involves equality of complex numbers, always remember to equate their real parts and imaginary parts separately to form a system of linear equations. This is a common strategy for solving for unknown real variables in complex number problems.

Question 15. Find the modulus of \( \frac{1+i}{1-i} \), \( \frac{1-i}{1+i} \) and \( |\frac{1+i}{1-i} - \frac{1-i}{1+i}| \).
Answer: We will evaluate each part separately.
First, simplify \( \frac{1+i}{1-i} \):
\( \frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i \)
The modulus of \( i \) is \( |i| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1 \).

Next, simplify \( \frac{1-i}{1+i} \):
\( \frac{1-i}{1+i} = \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{(1-i)^2}{1^2 - i^2} = \frac{1 - 2i + i^2}{1 - (-1)} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i \)
The modulus of \( -i \) is \( |-i| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1 \).

Finally, find the modulus of the difference \( |\frac{1+i}{1-i} - \frac{1-i}{1+i}| \):
From the previous calculations, we know \( \frac{1+i}{1-i} = i \) and \( \frac{1-i}{1+i} = -i \).
So, the expression becomes \( |i - (-i)| \)
\( = |i + i| \)
\( = |2i| \)
The modulus of \( 2i \) is \( |2i| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2 \).
In simple words: We first simplified the first fraction to \( i \), and its length (modulus) is 1. Then we simplified the second fraction to \( -i \), and its length (modulus) is also 1. Lastly, we found the difference between the two simplified fractions, which was \( 2i \), and its length (modulus) is 2.

Exam Tip: Simplify complex fractions to their standard \( a+ib \) form before calculating the modulus. Remember that \( i^2 = -1 \) and \( |a+ib| = \sqrt{a^2+b^2} \). Also, be careful with signs when performing subtraction with complex numbers.

Question 16. If \( (x + iy)^3 = u + iv \), then show that \( \frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2) \).
Answer: We are given \( (x + iy)^3 = u + iv \).
Expand the left side using the binomial expansion formula \( (A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 \):
\( (x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 \)
\( = x^3 + 3ix^2y + 3xi^2y^2 + i^3y^3 \)
Substitute \( i^2 = -1 \) and \( i^3 = -i \):
\( = x^3 + 3ix^2y + 3x(-1)y^2 + (-i)y^3 \)
\( = x^3 + 3ix^2y - 3xy^2 - iy^3 \)
Group the real and imaginary parts:
\( = (x^3 - 3xy^2) + i(3x^2y - y^3) \)
Since \( (x + iy)^3 = u + iv \), by equating the real and imaginary parts, we get:
\( u = x^3 - 3xy^2 \) ... (1)
\( v = 3x^2y - y^3 \) ... (2)
Now, calculate \( \frac{u}{x} \) (assuming \( x \neq 0 \)):
\( \frac{u}{x} = \frac{x^3 - 3xy^2}{x} = x^2 - 3y^2 \)
Next, calculate \( \frac{v}{y} \) (assuming \( y \neq 0 \)):
\( \frac{v}{y} = \frac{3x^2y - y^3}{y} = 3x^2 - y^2 \)
Now, add \( \frac{u}{x} \) and \( \frac{v}{y} \):
\( \frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2) \)
\( = x^2 - 3y^2 + 3x^2 - y^2 \)
\( = (x^2 + 3x^2) + (-3y^2 - y^2) \)
\( = 4x^2 - 4y^2 \)
Factor out 4:
\( = 4(x^2 - y^2) \)
Thus, \( \frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2) \) is proved.
In simple words: We expanded \( (x+iy)^3 \) and separated it into its real part \( u \) and imaginary part \( v \). Then, we divided \( u \) by \( x \) and \( v \) by \( y \). Finally, adding these two results together and simplifying showed that they equaled \( 4(x^2-y^2) \), which was what we needed to prove.

Exam Tip: Be careful with the powers of \( i \) ( \( i^2=-1 \), \( i^3=-i \), \( i^4=1 \)) when expanding binomials involving complex numbers. Once \( u \) and \( v \) are identified, ensure you divide by \( x \) and \( y \) correctly and combine like terms accurately.

Question 17. If \( \alpha \) and \( \beta \) are different complex numbers with \( |\beta| = 1 \), then find \( |\frac{\beta-\alpha}{1-\overline{\alpha}\beta}| \).
Answer: We are given that \( \alpha \) and \( \beta \) are different complex numbers and \( |\beta| = 1 \).
Since \( |\beta| = 1 \), we know that \( |\beta|^2 = 1 \), which implies \( \beta\overline{\beta} = 1 \).
We need to find \( |\frac{\beta-\alpha}{1-\overline{\alpha}\beta}| \). It is often easier to calculate the square of the modulus first:
\( |\frac{\beta-\alpha}{1-\overline{\alpha}\beta}|^2 = \left(\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right) \overline{\left(\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right)} \)
Recall that \( \overline{\left(\frac{Z_1}{Z_2}\right)} = \frac{\overline{Z_1}}{\overline{Z_2}} \) and \( \overline{Z_1 - Z_2} = \overline{Z_1} - \overline{Z_2} \), and \( \overline{Z_1 Z_2} = \overline{Z_1}\overline{Z_2} \).
\( = \frac{\beta-\alpha}{1-\overline{\alpha}\beta} \times \frac{\overline{\beta}-\overline{\alpha}}{\overline{1}-\overline{(\overline{\alpha}\beta)}} \)
\( = \frac{\beta-\alpha}{1-\overline{\alpha}\beta} \times \frac{\overline{\beta}-\overline{\alpha}}{1-\alpha\overline{\beta}} \)
Now, multiply the numerators and denominators:
\( = \frac{(\beta-\alpha)(\overline{\beta}-\overline{\alpha})}{(1-\overline{\alpha}\beta)(1-\alpha\overline{\beta})} \)
Numerator: \( \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha} \)
\( = |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2 \)
Denominator: \( 1(1) - 1(\alpha\overline{\beta}) - (\overline{\alpha}\beta)(1) + (\overline{\alpha}\beta)(\alpha\overline{\beta}) \)
\( = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \overline{\alpha}\alpha\beta\overline{\beta} \)
\( = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2|\beta|^2 \)
Since we are given \( |\beta|=1 \), then \( |\beta|^2 = 1 \). Substitute this into the numerator and denominator:
Numerator: \( 1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2 \)
Denominator: \( 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2(1) \)
\( = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2 \)
We can observe that the numerator and the denominator are identical.
Therefore,
\( |\frac{\beta-\alpha}{1-\overline{\alpha}\beta}|^2 = \frac{1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2}{1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2} = 1 \)
Taking the square root of both sides (since modulus must be non-negative):
\( |\frac{\beta-\alpha}{1-\overline{\alpha}\beta}| = \sqrt{1} = 1 \)
In simple words: We wanted to find the length (modulus) of a complex fraction. We used the fact that the square of the modulus is the number times its conjugate. By expanding this and using the given condition that \( |\beta| \) is 1, we found that the numerator and denominator became exactly the same, simplifying the whole expression to 1. Taking the square root, the modulus is 1.

Exam Tip: For problems involving the modulus of a complex fraction, it is often more efficient to compute the square of the modulus first, using the property \( |Z|^2 = Z\overline{Z} \). Remember that \( \overline{Z_1+Z_2} = \overline{Z_1}+\overline{Z_2} \) and \( \overline{Z_1Z_2} = \overline{Z_1}\overline{Z_2} \), and \( |Z|^2 = Z\overline{Z} \).

Question 18. Find the number of non-zero integral solutions of the equation \( |1 - i|^x = 2^x \).
Answer: First, calculate the modulus of the complex number \( 1 - i \):
\( |1 - i| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \)
We can write \( \sqrt{2} \) as a power of 2:
\( \sqrt{2} = 2^{1/2} \)
Now, substitute this into the given equation:
\( (|1 - i|)^x = (2^{1/2})^x \)
\( = 2^{x/2} \)
The original equation is \( |1 - i|^x = 2^x \). So, we have:
\( 2^{x/2} = 2^x \)
For this equality to hold, the exponents must be equal (since the bases are the same and greater than 1):
\( \frac{x}{2} = x \)
To solve for \( x \), subtract \( \frac{x}{2} \) from both sides:
\( x - \frac{x}{2} = 0 \)
\( \frac{x}{2} = 0 \)
\( x = 0 \)
The only solution to the equation is \( x = 0 \).
The question asks for the number of *non-zero* integral solutions. Since the only integral solution found is \( 0 \), there are no non-zero integral solutions.
In simple words: We first found the length (modulus) of the complex number \( 1-i \), which was \( \sqrt{2} \). We rewrote \( \sqrt{2} \) as \( 2^{1/2} \). Plugging this into the equation, we got \( 2^{x/2} = 2^x \). For this to be true, the powers must be equal, so \( x/2 = x \), which means \( x \) must be 0. Since the problem asks for solutions that are not zero, there are no such solutions.

Exam Tip: When solving exponential equations, try to express both sides with the same base. Remember that \( \sqrt{A} = A^{1/2} \). Carefully read the question to distinguish between all solutions and solutions that meet specific criteria (e.g., non-zero, integral, real).

Question 19. If \( (a + ib)(c + id)(e + if)(g + ih) = A + iB \), then show that \( (a^2 + b^2)(c^2 + d^2)(e^2 + f^2)(g^2 + h^2) = A^2 + B^2 \).
Answer: We are given the equation:
\( (a + ib)(c + id)(e + if)(g + ih) = A + iB \) ... (1)
To prove the desired identity, we can take the modulus of both sides of equation (1).
Recall that for complex numbers \( Z_1, Z_2, \dots, Z_n \), the modulus of their product is the product of their moduli: \( |Z_1 Z_2 \dots Z_n| = |Z_1| |Z_2| \dots |Z_n| \).
Taking the modulus of the left side of equation (1):
\( |(a + ib)(c + id)(e + if)(g + ih)| = |a + ib| |c + id| |e + if| |g + ih| \)
\( = \sqrt{a^2+b^2} \sqrt{c^2+d^2} \sqrt{e^2+f^2} \sqrt{g^2+h^2} \)
Taking the modulus of the right side of equation (1):
\( |A + iB| = \sqrt{A^2+B^2} \)
Equating the moduli of both sides:
\( \sqrt{a^2+b^2} \sqrt{c^2+d^2} \sqrt{e^2+f^2} \sqrt{g^2+h^2} = \sqrt{A^2+B^2} \)
Now, square both sides of this equation to remove the square roots:
\( (\sqrt{a^2+b^2} \sqrt{c^2+d^2} \sqrt{e^2+f^2} \sqrt{g^2+h^2})^2 = (\sqrt{A^2+B^2})^2 \)
\( (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2 \)
This proves the desired result.

Alternatively, using conjugates:
Given \( (a + ib)(c + id)(e + if)(g + ih) = A + iB \) ... (1)
Take the conjugate of both sides:
\( \overline{(a + ib)(c + id)(e + if)(g + ih)} = \overline{A + iB} \)
\( (a - ib)(c - id)(e - if)(g - ih) = A - iB \) ... (2)
Multiply equation (1) by equation (2):
\( [(a+ib)(c+id)(e+if)(g+ih)][(a-ib)(c-id)(e-if)(g-ih)] = (A+iB)(A-iB) \)
Group the conjugate pairs on the left side:
\( [(a+ib)(a-ib)][(c+id)(c-id)][(e+if)(e-if)][(g+ih)(g-ih)] = A^2 - (iB)^2 \)
Using \( (X+iY)(X-iY) = X^2+Y^2 \):
\( (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 - i^2B^2 \)
Since \( i^2 = -1 \):
\( (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2 + B^2 \)
Both methods achieve the same proof.
In simple words: We started with the given equation where a product of four complex numbers equals another complex number. One way to solve this was to take the length (modulus) of both sides. The length of a product is the product of lengths. Squaring both sides then gave us the desired result. Another method involved taking the conjugate of both sides and then multiplying the original equation by its conjugate. This also simplified to the same proven equation.

Exam Tip: This is a standard identity. You can prove it either by taking the modulus of both sides (and squaring) or by taking the conjugate of the expression and multiplying it by the original expression. Both methods effectively use the property \( |Z|^2 = Z\overline{Z} \).

Question 20. If \( (\frac{1+i}{1-i})^m = 1 \), then find the least integral value of \( m \)?
Answer: First, simplify the complex fraction \( \frac{1+i}{1-i} \):
\( \frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} \)
\( = \frac{(1+i)^2}{1^2 - i^2} \)
\( = \frac{1 + 2i + i^2}{1 - (-1)} \)
\( = \frac{1 + 2i - 1}{1 + 1} \)
\( = \frac{2i}{2} \)
\( = i \)
Now, substitute this simplified form back into the given equation:
\( (i)^m = 1 \)
We need to find the least positive integral value of \( m \) for which \( i^m = 1 \).
Let's look at the powers of \( i \):
\( i^1 = i \)
\( i^2 = -1 \)
\( i^3 = -i \)
\( i^4 = 1 \)
The powers of \( i \) repeat in a cycle of 4. Therefore, for \( i^m = 1 \), \( m \) must be a multiple of 4.
The least positive integral value for \( m \) that is a multiple of 4 is \( 4 \).
In simple words: We first simplified the fraction inside the bracket, which turned out to be just \( i \). So the problem became \( i^m = 1 \). We know that powers of \( i \) repeat every four times (\( i^1=i, i^2=-1, i^3=-i, i^4=1 \)). The smallest whole number \( m \) that makes \( i^m \) equal to 1 is 4.

Exam Tip: Always simplify complex fractions before raising them to a power. Remember the cycle of powers of \( i \) ( \( i, -1, -i, 1 \)) which repeats every four exponents. This is crucial for solving equations like \( i^m = 1 \).