GSEB Class 11 Maths Solutions Chapter 5 સંકર સંખ્યાઓ અને દ્વિઘાત સમીકરણો Exercise 5.1

નીચે પ્રશ્ન 1થી 10માં આપેલ દરેક સંકર સંખ્યાને a + ib સ્વરૂપમાં દર્શાવોઃ

Question 1. \( (5i)\left(-\frac{3}{5}i\right) \)
Answer: Let's simplify the given expression:
\( (5i)\left(-\frac{3}{5}i\right) \)
\( \implies 5 \cdot \left(-\frac{3}{5}\right) \cdot i^2 \)
\( \implies -3i^2 \)
\( \implies -3(-1) \) (since \( i^2 = -1 \))
\( \implies 3 \)
\( \implies 3 + 0i \)
This is in the form \( a + ib \); here, \( a = 3 \) and \( b = 0 \).
In simple words: We multiply the numbers and the 'i's. Because 'i' squared is -1, the expression simplifies to just 3. So, it's 3 plus zero 'i'.

Exam Tip: Focus on simplifying expressions involving the imaginary unit \( i \) by remembering that \( i^2 = -1 \), which is a key identity in complex number calculations.

Question 2. \( i^9 + i^{19} \)
Answer: Let's simplify the given expression:
\( i^9 + i^{19} \)
\( \implies i^{4 \cdot 2 + 1} + i^{4 \cdot 4 + 3} \)
\( \implies (i^4)^2 \cdot i^1 + (i^4)^4 \cdot i^3 \)
\( \implies (1)^2 \cdot i + (1)^4 \cdot (-i) \) (since \( i^4 = 1 \) and \( i^3 = -i \))
\( \implies i - i \)
\( \implies 0 \)
\( \implies 0 + 0i \)
This is in the form \( a + ib \); here, \( a = 0 \) and \( b = 0 \).
In simple words: We simplify the powers of 'i' using the pattern that 'i' to the power of 4 is 1. Then we add the simplified terms, which cancel each other out, leaving zero.

Exam Tip: Remember that powers of \( i \) cycle with a period of 4: \( i^1=i, i^2=-1, i^3=-i, i^4=1 \). Use this cycle to simplify higher powers by dividing the exponent by 4 and using the remainder.

Question 3. \( i^{-39} \)
Answer: Let's simplify the given expression:
\( i^{-39} \)
\( \implies \frac{1}{i^{39}} \)
\( \implies \frac{1}{i^{4 \cdot 9 + 3}} \)
\( \implies \frac{1}{(i^4)^9 \cdot i^3} \)
\( \implies \frac{1}{(1)^9 \cdot (-i)} \) (since \( i^4 = 1 \) and \( i^3 = -i \))
\( \implies \frac{1}{-i} \)
\( \implies \frac{1}{-i} \times \frac{i}{i} \)
\( \implies \frac{i}{-i^2} \)
\( \implies \frac{i}{-(-1)} \) (since \( i^2 = -1 \))
\( \implies \frac{i}{1} \)
\( \implies i \)
\( \implies 0 + 1i \)
This is in the form \( a + ib \); here, \( a = 0 \) and \( b = 1 \).
In simple words: To simplify a negative power of 'i', we turn it into a fraction with 'i' in the denominator. Then we simplify 'i' to the power of 39. After that, we multiply the top and bottom by 'i' to remove 'i' from the denominator, which simplifies to just 'i'.

Exam Tip: When simplifying negative powers of \( i \), convert them to positive powers by taking the reciprocal, and then use the cyclic property of \( i \) to reduce the exponent to its equivalent value within `\( i^1, i^2, i^3, i^4 \) `.

Question 4. \( 3(7+7i) + i(7+7i) \)
Answer: Let's simplify the given expression:
\( 3(7+7i) + i(7+7i) \)
\( \implies 21 + 21i + 7i + 7i^2 \)
\( \implies 21 + 28i + 7(-1) \) (since \( i^2 = -1 \))
\( \implies 21 + 28i - 7 \)
\( \implies (21-7) + 28i \)
\( \implies 14 + 28i \)
This is in the form \( a + ib \); here, \( a = 14 \) and \( b = 28 \).
In simple words: First, we multiply the numbers outside the brackets by everything inside. Then, we combine the parts with 'i' and replace 'i' squared with -1. Finally, we group the real numbers and the 'i' numbers to get the final answer.

Exam Tip: Always distribute carefully, combine like terms (real parts with real, imaginary parts with imaginary), and replace \( i^2 \) with \( -1 \) to simplify complex expressions correctly.

Question 5. \( (1-i) - (-1+6i) \)
Answer: Let's simplify the given expression:
\( (1-i) - (-1+6i) \)
\( \implies 1 - i + 1 - 6i \)
\( \implies (1+1) + (-i-6i) \)
\( \implies 2 - 7i \)
This is in the form \( a + ib \); here, \( a = 2 \) and \( b = -7 \).
In simple words: When we subtract complex numbers, we change the signs of the second number and then combine the real parts together and the 'i' parts together. This gives us 2 minus 7i.

Exam Tip: When subtracting complex numbers, remember to distribute the negative sign to all terms within the second parenthesis before combining the real and imaginary parts.

Question 6. \( \left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right) \)
Answer: Let's simplify the given expression:
\( \left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right) \)
\( \implies \frac{1}{5}+i \frac{2}{5}-4-i \frac{5}{2} \)
\( \implies \left(\frac{1}{5}-4\right) + i\left(\frac{2}{5}-\frac{5}{2}\right) \)
\( \implies \left(\frac{1-20}{5}\right) + i\left(\frac{4-25}{10}\right) \)
\( \implies \frac{-19}{5} + i\left(\frac{-21}{10}\right) \)
\( \implies \frac{-19}{5} - i\frac{21}{10} \)
This is in the form \( a + ib \); here, \( a = -\frac{19}{5} \) and \( b = -\frac{21}{10} \).
In simple words: To subtract these complex numbers, we group the real parts and the 'i' parts separately. Then we find a common bottom number for the fractions and subtract them. This gives us the answer in the form of a real number plus an 'i' number.

Exam Tip: To subtract complex numbers, group the real parts and imaginary parts separately, then find a common denominator for any fractions before performing the subtraction operations.

Question 7. \( \left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right) \)
Answer: Let's simplify the given expression:
\( \left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right) \)
First, simplify the terms inside the square brackets:
\( \left(\frac{1}{3}+4\right) + i\left(\frac{7}{3}+\frac{1}{3}\right) \)
\( \implies \left(\frac{1+12}{3}\right) + i\left(\frac{7+1}{3}\right) \)
\( \implies \frac{13}{3} + i\frac{8}{3} \)
Now, subtract the last complex number:
\( \left(\frac{13}{3} + i\frac{8}{3}\right) - \left(-\frac{4}{3}+i\right) \)
\( \implies \frac{13}{3} + i\frac{8}{3} + \frac{4}{3} - i \)
\( \implies \left(\frac{13}{3}+\frac{4}{3}\right) + i\left(\frac{8}{3}-1\right) \)
\( \implies \left(\frac{13+4}{3}\right) + i\left(\frac{8-3}{3}\right) \)
\( \implies \frac{17}{3} + i\frac{5}{3} \)
This is in the form \( a + ib \); here, \( a = \frac{17}{3} \) and \( b = \frac{5}{3} \).
In simple words: We first add the complex numbers inside the big bracket by adding their real parts and their 'i' parts. Then we subtract the last complex number from this result, making sure to change the signs when subtracting. Finally, we group the real parts and the 'i' parts to get the answer.

Exam Tip: When combining multiple complex numbers, first perform operations within parentheses, then combine the real and imaginary parts, paying close attention to signs, especially with subtraction, to ensure accuracy.

Question 8. \( (1-i)^4 \)
Answer: Let's simplify the given expression:
\( (1-i)^4 \)
We can write this as \( ((1-i)^2)^2 \).
First, calculate \( (1-i)^2 \):
\( (1-i)^2 = 1^2 - 2(1)(i) + i^2 \)
\( \implies 1 - 2i + (-1) \) (since \( i^2 = -1 \))
\( \implies 1 - 2i - 1 \)
\( \implies -2i \)
Now, square the result:
\( (-2i)^2 \)
\( \implies (-2)^2 \cdot i^2 \)
\( \implies 4i^2 \)
\( \implies 4(-1) \) (since \( i^2 = -1 \))
\( \implies -4 \)
\( \implies -4 + 0i \)
This is in the form \( a + ib \); here, \( a = -4 \) and \( b = 0 \).
In simple words: To solve (1-i) to the power of 4, we first square (1-i), which gives us -2i. Then we square -2i, which results in -4 because i-squared is -1. So, the answer is -4.

Exam Tip: When raising complex numbers to higher powers, it is often simpler to square the complex number first, then square the result, or use binomial expansion and simplify powers of \( i \) carefully.

Question 9. \( \left(\frac{1}{3}+3i\right)^3 \)
Answer: Let's simplify the given expression using the binomial expansion `\( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)`:
\( \left(\frac{1}{3}+3i\right)^3 \)
\( \implies \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3\left(\frac{1}{3}\right)(3i)^2 + (3i)^3 \)
\( \implies \frac{1}{27} + 3\left(\frac{1}{9}\right)(3i) + 3\left(\frac{1}{3}\right)(9i^2) + 27i^3 \)
\( \implies \frac{1}{27} + \left(\frac{1}{3}\right)(3i) + (1)(9i^2) + 27i^3 \)
\( \implies \frac{1}{27} + i + 9i^2 + 27i^3 \)
Now substitute \( i^2 = -1 \) and \( i^3 = -i \):
\( \implies \frac{1}{27} + i + 9(-1) + 27(-i) \)
\( \implies \frac{1}{27} + i - 9 - 27i \)
Group the real and imaginary parts:
\( \implies \left(\frac{1}{27}-9\right) + (1-27)i \)
\( \implies \left(\frac{1-243}{27}\right) - 26i \)
\( \implies \frac{-242}{27} - 26i \)
This is in the form \( a + ib \); here, \( a = -\frac{242}{27} \) and \( b = -26 \).
In simple words: We expand the expression using the cube formula for two terms. Then, we replace 'i' squared with -1 and 'i' cubed with -i. Finally, we combine all the real numbers and all the 'i' numbers to get the answer in the correct form.

Exam Tip: When expanding `\( (a+b)^3 \)`, use the binomial theorem: `\( a^3 + 3a^2b + 3ab^2 + b^3 \)`, and simplify powers of \( i \) carefully at each step to avoid calculation errors.

Question 10. \( \left(-2-\frac{1}{3}i\right)^3 \)
Answer: Let's simplify the given expression:
\( \left(-2-\frac{1}{3}i\right)^3 \)
We can factor out -1 from the base:
\( \implies \left(-\left(2+\frac{1}{3}i\right)\right)^3 \)
\( \implies (-1)^3 \left(2+\frac{1}{3}i\right)^3 \)
\( \implies -1 \cdot \left(2+\frac{1}{3}i\right)^3 \)
Now, expand `\( \left(2+\frac{1}{3}i\right)^3 \)` using `\( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)`:
\( \implies -\left[2^3 + 3(2^2)\left(\frac{1}{3}i\right) + 3(2)\left(\frac{1}{3}i\right)^2 + \left(\frac{1}{3}i\right)^3\right] \)
\( \implies -\left[8 + 3(4)\left(\frac{1}{3}i\right) + 6\left(\frac{1}{9}i^2\right) + \frac{1}{27}i^3\right] \)
\( \implies -\left[8 + 4i + \frac{2}{3}i^2 + \frac{1}{27}i^3\right] \)
Substitute \( i^2 = -1 \) and \( i^3 = -i \):
\( \implies -\left[8 + 4i + \frac{2}{3}(-1) + \frac{1}{27}(-i)\right] \)
\( \implies -\left[8 + 4i - \frac{2}{3} - \frac{1}{27}i\right] \)
Group the real and imaginary parts inside the bracket:
\( \implies -\left[\left(8 - \frac{2}{3}\right) + \left(4 - \frac{1}{27}\right)i\right] \)
\( \implies -\left[\left(\frac{24-2}{3}\right) + \left(\frac{108-1}{27}\right)i\right] \)
\( \implies -\left[\frac{22}{3} + \frac{107}{27}i\right] \)
Finally, distribute the negative sign:
\( \implies -\frac{22}{3} - i\frac{107}{27} \)
This is in the form \( a + ib \); here, \( a = -\frac{22}{3} \) and \( b = -\frac{107}{27} \).
In simple words: First, we take out the minus sign from the bracket because it's a cube. Then we expand the remaining term using the cube formula. We substitute 'i' squared with -1 and 'i' cubed with -i. After simplifying and grouping, we distribute the minus sign back to get the final 'a + ib' form.

Exam Tip: When cubing a complex number with a negative real part, factoring out the negative sign first can simplify the expansion process. Remember to distribute the negative sign to both the real and imaginary components at the end.

નીચે પ્રશ્ન 11થી 13માં આપેલ દરેક સંકર સંખ્યાનો ગુણાકાર માટેનો વ્યસ્ત શોધો :

Question 11. Find the multiplicative inverse of \( 4-3i \).
Answer: Let the complex number be \( z = 4-3i \).
The multiplicative inverse of a complex number \( z \) is given by `\( z^{-1} = \frac{\bar{z}}{|z|^2} \)`.
First, find the conjugate \( \bar{z} \):
\( \bar{z} = 4+3i \)
Next, find the modulus squared \( |z|^2 \):
\( |z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25 \)
Now, substitute these values into the formula:
\( z^{-1} = \frac{4+3i}{25} \)
\( \implies z^{-1} = \frac{4}{25} + i\frac{3}{25} \)
Alternatively, we can find the inverse by rationalizing the denominator:
\( z^{-1} = \frac{1}{4-3i} \)
Multiply the numerator and denominator by the conjugate `\( (4+3i) \)`:
\( z^{-1} = \frac{1}{4-3i} \times \frac{4+3i}{4+3i} \)
\( \implies \frac{4+3i}{4^2 - (3i)^2} \)
\( \implies \frac{4+3i}{16 - 9i^2} \)
\( \implies \frac{4+3i}{16 - 9(-1)} \) (since \( i^2 = -1 \))
\( \implies \frac{4+3i}{16+9} \)
\( \implies \frac{4+3i}{25} \)
\( \implies \frac{4}{25} + i\frac{3}{25} \)
In simple words: To find the multiplicative inverse, we use the formula of conjugate divided by modulus squared. We find the conjugate (change the sign of the 'i' part) and the modulus squared (square the real and 'i' parts and add them). Then we divide the conjugate by the modulus squared to get the answer.

Exam Tip: The multiplicative inverse of a complex number \( z \) can be found using the formula `\( \frac{\bar{z}}{|z|^2} \)` or by rationalizing the denominator of `\( \frac{1}{z} \)` by multiplying the numerator and denominator by the conjugate.

Question 12. Find the multiplicative inverse of \( \sqrt{5}+3i \).
Answer: Let the complex number be \( z = \sqrt{5}+3i \).
The multiplicative inverse is `\( z^{-1} = \frac{1}{\sqrt{5}+3i} \)`.
To express this in the form \( a+ib \), we multiply the numerator and denominator by the conjugate of the denominator, `\( (\sqrt{5}-3i) \)`:
\( z^{-1} = \frac{1}{\sqrt{5}+3i} \times \frac{\sqrt{5}-3i}{\sqrt{5}-3i} \)
\( \implies \frac{\sqrt{5}-3i}{(\sqrt{5})^2 - (3i)^2} \)
\( \implies \frac{\sqrt{5}-3i}{5 - 9i^2} \)
\( \implies \frac{\sqrt{5}-3i}{5 - 9(-1)} \) (since \( i^2 = -1 \))
\( \implies \frac{\sqrt{5}-3i}{5+9} \)
\( \implies \frac{\sqrt{5}-3i}{14} \)
\( \implies \frac{\sqrt{5}}{14} - i\frac{3}{14} \)
In simple words: To find the inverse, we write 1 over the given complex number. Then, we multiply the top and bottom by the complex conjugate (changing the sign of the 'i' part) to clear the 'i' from the bottom. This gives us the answer with the square root on top.

Exam Tip: To find the multiplicative inverse of complex numbers involving square roots, remember to rationalize the denominator using the conjugate, applying the identity `\( (a+bi)(a-bi) = a^2+b^2 \)`.

Question 13. Find the multiplicative inverse of \( -i \).
Answer: Let the complex number be \( z = -i \).
The multiplicative inverse is `\( z^{-1} = \frac{1}{-i} \)`.
To express this in the form \( a+ib \), we multiply the numerator and denominator by \( i \):
\( z^{-1} = \frac{1}{-i} \times \frac{i}{i} \)
\( \implies \frac{i}{-i^2} \)
\( \implies \frac{i}{-(-1)} \) (since \( i^2 = -1 \))
\( \implies \frac{i}{1} \)
\( \implies i \)
\( \implies 0 + 1i \)
In simple words: To find the inverse of -i, we write 1 divided by -i. Then we multiply both the top and bottom by 'i' to remove 'i' from the bottom. Since i-squared is -1, this simplifies to just 'i'.

Exam Tip: Finding the inverse of pure imaginary numbers is straightforward: for `\( -i \)`, multiply numerator and denominator by `\( i \)` to eliminate `\( i^2 \)` in the denominator, which simplifies to `\( i \)`. For `\( i \)`, the inverse is `\( -i \)`.

Question 14. Express the following expression in the form a + ib: \( \frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+i \sqrt{2})-(\sqrt{3}-i \sqrt{2})} \)
Answer: Let's simplify the given expression:
\( \frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+i \sqrt{2})-(\sqrt{3}-i \sqrt{2})} \)
First, simplify the numerator using `\( (a+b)(a-b) = a^2-b^2 \)`:
Numerator \( = (3)^2 - (i\sqrt{5})^2 \)
\( = 9 - i^2(\sqrt{5})^2 \)
\( = 9 - (-1)(5) \) (since \( i^2 = -1 \))
\( = 9 + 5 \)
\( = 14 \)
Next, simplify the denominator:
Denominator \( = (\sqrt{3}+i \sqrt{2}) - (\sqrt{3}-i \sqrt{2}) \)
\( = \sqrt{3}+i \sqrt{2} - \sqrt{3} + i \sqrt{2} \)
\( = (\sqrt{3}-\sqrt{3}) + (i\sqrt{2}+i\sqrt{2}) \)
\( = 0 + 2i\sqrt{2} \)
\( = 2i\sqrt{2} \)
Now, combine the simplified numerator and denominator:
\( \frac{14}{2i\sqrt{2}} \)
\( \implies \frac{7}{i\sqrt{2}} \)
To express this in \( a+ib \) form, multiply the numerator and denominator by \( i \):
\( \implies \frac{7}{i\sqrt{2}} \times \frac{i}{i} \)
\( \implies \frac{7i}{i^2\sqrt{2}} \)
\( \implies \frac{7i}{(-1)\sqrt{2}} \) (since \( i^2 = -1 \))
\( \implies \frac{-7i}{\sqrt{2}} \)
Rationalize the denominator by multiplying by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( \implies \frac{-7i}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( \implies \frac{-7\sqrt{2}i}{2} \)
\( \implies 0 - i\frac{7\sqrt{2}}{2} \)
This is in the form \( a + ib \); here, \( a = 0 \) and \( b = -\frac{7\sqrt{2}}{2} \).
In simple words: We first simplify the top part using the difference of squares formula, remembering that 'i' squared is -1. Then we simplify the bottom part by distributing the minus sign and combining like terms. After dividing the simplified top by the simplified bottom, we multiply by 'i' over 'i' to remove 'i' from the bottom, and finally rationalize the square root in the denominator.

Exam Tip: When simplifying complex fractions, always use the conjugate to rationalize denominators and simplify terms like `\( (a+bi)(a-bi) = a^2+b^2 \)` in the numerator, then combine real and imaginary parts carefully.