CBSE Class 11 Mathematics Complex Numbers And Quadratic Equation Worksheet Set 03

CBSE Class 11 Mathematics Worksheet - Complex Numbers and Quadratic Equation (2). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

On Conjugates: \( \{ (a + ib)(a - ib) = a^2 + b^2 \} \)

Question. If \( x + iy = \sqrt{\frac{a+ib}{c+id}} \) show that \( (x^2 + y^2)^2 = \frac{a^2+b^2}{c^2+d^2} \).
Answer: We have \( x + iy = \sqrt{\frac{a+ib}{c+id}} \) ……… (i)
Taking conjugate on both sides
\( \Rightarrow \overline{x+iy} = \sqrt{\frac{\overline{a+ib}}{\overline{c+id}}} \)
\( \Rightarrow x - iy = \sqrt{\frac{a-ib}{c-id}} \) ……… (ii)
Equation (i) x (ii)
\( \Rightarrow (x + iy)(x - iy) = \sqrt{\frac{a+ib}{c+id}} \times \sqrt{\frac{a-ib}{c-id}} \)
\( \Rightarrow x^2 + y^2 = \sqrt{\frac{(a+ib)(a-ib)}{(c+id)(c-id)}} \)
\( \Rightarrow x^2 + y^2 = \sqrt{\frac{a^2+b^2}{c^2+d^2}} \) ……….. \( \{ (a + ib)(a - ib) = a^2 + b^2 \} \)
Squaring
\( (x^2 + y^2)^2 = \frac{a^2+b^2}{c^2+d^2} \) (proved)

Question. If \( \frac{(a+i)^2}{2a-i} = p + iq \) show that \( p^2 + q^2 = \frac{(a^2+1)^2}{4a^2+1} \).
Answer: We have \( \frac{(a+i)^2}{2a-i} = p + iq \) …………… (i)
Taking conjugate on both sides
\( \Rightarrow \overline{\left( \frac{(a+i)^2}{2a-i} \right)} = \overline{p+iq} \)
\( \Rightarrow \frac{(a-i)^2}{2a+i} = p - iq \) ………… (ii)
Equation (i) x (ii)
\( \Rightarrow \frac{(a+i)^2}{2a-i} \times \frac{(a-i)^2}{2a+i} = (p + iq) (p - iq) \)
\( \Rightarrow \frac{((a+i)(a-i))^2}{4a^2+1} = p^2 + q^2 \) …………….. \( \{ (a + ib)(a - ib) = a^2 + b^2 \} \)
Squaring
\( \frac{(a^2+1)^2}{4a^2+1} = p^2 + q^2 \) (proved)

Question. If \( (1 + i)(2 + 2i)(1 + 3i) \dots (1 + ni) = x + iy \) show that \( 2 \cdot 5 \cdot 10 \dots (1 + n^2) = x^2 + y^2 \).
Answer: We have \( (1 + i)(2 + 2i)(1 + 3i) \dots (1 + ni) = x + iy \) ………… (i)
Taking conjugate on both sides
\( \Rightarrow \overline{(1 + i)(2 + 2i)(1 + 3i) \dots (1 + ni)} = \overline{x + iy} \)
\( \Rightarrow (1 - i)(2 - 2i)(1 - 3i) \dots (1 - ni) = x - iy \) ………… (ii)
Equation (i) x (ii)
\( \Rightarrow (1 + i)(1 - i)(2 + 2i)(2 - 2i)(1 + 3i)(1 - 3i) \dots (1 + ni)(1 - ni) = (x + iy)(x - iy) \)
\( \Rightarrow (1 + 1)(1 + 4)(1 + 9)\dots(1 + n^2) = x^2 + y^2 \) ……... \( \{ (a + ib)(a - ib) = a^2 + b^2 \} \)
Squaring
\( 2 \cdot 5 \cdot 10 \dots (1 + n^2) = x^2 + y^2 \) (proved)

On Standard Form

Question. Find \( \theta \) such that \( \frac{3+2i \sin \theta}{1-2i \sin \theta} \) is purely real.
Answer: Let \( z = \frac{3+2i \sin \theta}{1-2i \sin \theta} \)
Rationalize
\( \Rightarrow z = \frac{3+2i \sin \theta}{1-2i \sin \theta} \times \frac{1+2i \sin \theta}{1+2i \sin \theta} \)
\( \Rightarrow z = \frac{3+6i \sin \theta+2i \sin \theta+4i^2 \sin^2 \theta}{1-4i^2 \sin^2 \theta} \)
\( \Rightarrow z = \frac{3+8i \sin \theta-4 \sin^2 \theta}{1+4 \sin^2 \theta} \)
\( \Rightarrow z = \frac{3-4 \sin^2 \theta}{1+4 \sin^2 \theta} + \frac{8i \sin \theta}{1+4 \sin^2 \theta} \)
It is given that \( z \) is purely real
\( \therefore \operatorname{Im}(z) = 0 \)
\( \Rightarrow \frac{8 \sin \theta}{1+4 \sin^2 \theta} = 0 \)
\( \Rightarrow 8 \sin \theta = 0 \)
\( \Rightarrow \sin \theta = 0 \)
\( \Rightarrow \theta = n\pi ; n \in \mathbb{Z} \) ans.

Question. If \( z \) is a complex number such that \( |z| = 1 \), prove that \( \frac{z-1}{z+1} \) is purely imaginary. What will be your conclusion of \( z = 1 \).
Answer: Let \( z = x + iy \)
\( |z| = 1 \) (given)
\( \Rightarrow \sqrt{x^2 + y^2} = 1 \)
\( \Rightarrow x^2 + y^2 = 1 \) …………. (i)
Now, \( \frac{z-1}{z+1} = \frac{x+iy-1}{x+iy+1} = \frac{(x-1)+iy}{(x+1)+iy} \)
\( \Rightarrow \frac{z-1}{z+1} = \frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy]} \) …………….(rationalize)
\( = \frac{(x^2-1)-iy(x-1)+iy(x+1)-i^2y^2}{(x+1)^2-i^2y^2} \)
\( = \frac{(x^2-1)+i(-yx+y+yx+y)+y^2}{(x+1)^2+y^2} \)
\( = \frac{(x^2+y^2-1)+2iy}{(x+1)^2+y^2} \)
\( = \frac{0+2iy}{(x+1)^2+y^2} \) …………. (\( x^2 + y^2 = 1 \) from eq. (i))
\( \Rightarrow \frac{z-1}{z+1} = \frac{2iy}{(x+1)^2+y^2} \) which is purely imaginary
Now when \( z = 1 \)
Then \( \frac{z-1}{z+1} = \frac{1-1}{1+1} = \frac{0}{2} = 0 \) which is purely real ans.

Natural Number

Question. Find the least +ve integral value of \( m \), if \( \left(\frac{1+i}{1-i}\right)^m = 1 \).
Answer: We have \( \left(\frac{1+i}{1-i}\right)^m = 1 \)
\( \Rightarrow \left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^m = 1 \)
\( \Rightarrow \left(\frac{1+i^2+2i}{1-i^2}\right)^m = 1 \)
\( \Rightarrow \left(\frac{1-1+2i}{1+1}\right)^m = 1 \)
\( \Rightarrow \left(\frac{2i}{2}\right)^m = 1 \)
\( \Rightarrow (i)^m = 1 \)
\( \dots \) the least +ve integral value of \( m = 4 \) ans. ……. (Since \( i^4 = 1 \))

Question. Find the smallest +ve integer value of \( n \) for which \( \frac{(1+i)^n}{(1-i)^{n-2}} \) is a real number.
Answer: We have \( \frac{(1+i)^n}{(1-i)^{n-2}} \)
\( = \frac{(1+i)^n}{(1-i)^n(1-i)^{-2}} \)
\( = \left( \frac{1+i}{1-i} \right)^n (1-i)^2 \)
\( = \left( \frac{1+i}{1-i} \times \frac{1+i}{1+i} \right)^n (1-i)^2 \)
\( = \left( \frac{1+i^2+2i}{1-i^2} \right)^n (1-i^2-2i) \)
\( = \left( \frac{1-1+2i}{1+1} \right)^n (1-1-2i) \)
\( = (i)^n (-2i) \)
\( = -2(i)^{n+1} \)
For real number : \( n \) should be equal to 1
Since \( -2(i)^{1+1} = -2(i)^2 = -2(-1) = 2 \) which is a real number
\( \therefore n = 1 \) ans.

Question. Find the number of non-zero integral solutions if \( |1-i|^x = 2^x \).
Answer: We have \( |1-i|^x = 2^x \)
\( \Rightarrow (\sqrt{1+1})^x = 2^x \) ………. \( \{ |z| = \sqrt{a^2+b^2} \} \)
\( \Rightarrow (\sqrt{2})^x = 2^x \)
\( \Rightarrow (2)^{\frac{x}{2}} = 2^x \)
\( \Rightarrow \frac{x}{2} = x \)
\( \Rightarrow x = 2x \)
\( \Rightarrow 2x - x = 0 \)
\( \Rightarrow x = 0 \)
Hence, there is no non-zero integral solution.

Question. If \( \alpha \) & \( \beta \) are different complex numbers with \( |\beta| = 1 \). Find \( \left| \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right| \).
Answer: We have \( \left| \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right|^2 = \left( \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right) \overline{\left( \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right)} \) ………….. \( |z|^2 = z\bar{z} \)
\( = \left( \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right) \left( \frac{\bar{\beta}-\bar{\alpha}}{1-\alpha\bar{\beta}} \right) \) ………. \( \{ \overline{\bar{z}} = z \} \)
\( = \frac{\beta\bar{\beta}-\beta\bar{\alpha}-\alpha\bar{\beta}+\alpha\bar{\alpha}}{1-\alpha\bar{\beta}-\bar{\alpha}\beta+\alpha\bar{\alpha}\beta\bar{\beta}} \)
\( = \frac{|\beta|^2-\beta\bar{\alpha}-\alpha\bar{\beta}+|\alpha|^2}{1-\alpha\bar{\beta}-\bar{\alpha}\beta+|\alpha|^2|\beta|^2} \) …………… \( |z|^2 = z\bar{z} \)
\( = \frac{1-\beta\bar{\alpha}-\alpha\bar{\beta}+|\alpha|^2}{1-\alpha\bar{\beta}-\bar{\alpha}\beta+|\alpha|^2} \) …………… given \( |\beta| = 1 \)
\( \Rightarrow \left| \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right|^2 = 1 \)
\( \Rightarrow \left| \frac{\beta-\alpha}{1-\bar{\alpha}\beta} \right| = 1 \) ans.

Question. If \( z_1 \) and \( z_2 \) are any two complex numbers then show that \( \operatorname{Re}(z_1z_2) = \operatorname{Re}(z_1)\operatorname{Re}(z_2) - \operatorname{Im}(z_1)\operatorname{Im}(z_2) \).
Answer: Let \( z_1 = a + ib \) and \( z_2 = c + id \)
\( \therefore \operatorname{Re}(z_1) = a \) and \( \operatorname{Re}(z_2) = c \)
\( \operatorname{Im}(z_1) = b \) and \( \operatorname{Im}(z_2) = d \)
Now, \( z_1z_2 = (a+ib)(c+id) \)
\( = ac + iad + ibc + i^2bd \)
\( = ac + iad + ibc - bd \)
\( z_1z_2 = (ac-bd) + i(ad+bc) \)
Here, \( \operatorname{Re}(z_1z_2) = ac - bd \)
\( = \operatorname{Re}(z_1)\operatorname{Re}(z_2) - \operatorname{Im}(z_1)\operatorname{Im}(z_2) \) RHS (proved)

Quadratic Equation

Question. Solve \( 2x^2 + 3ix + 2 = 0 \).
Answer: Here \( a = 2, b = 3i, c = 2 \)
By quadratic formula,
\( \Rightarrow x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( \Rightarrow x = \frac{-3i \pm \sqrt{9i^2-4(2)(2)}}{4} \)
\( \Rightarrow x = \frac{-3i \pm \sqrt{-9-16}}{4} \)
\( \Rightarrow x = \frac{-3i \pm \sqrt{-25}}{4} \)
\( \Rightarrow x = \frac{-3i \pm 5i}{4} \)
\( \Rightarrow x = \frac{-3i+5i}{4} \) and \( x = \frac{-3i-5i}{4} \)
\( \Rightarrow x = \frac{i}{2} \) and \( x = -2i \) ans.

Question. Solve \( 2x^2 - (3 + 7i)x - (3 - 9i) = 0 \).
Answer: Here \( a = 2, b = -(3 + 7i), c = -(3 - 9i) \)
By quadratic formula,
\( \Rightarrow x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)
\( \Rightarrow x = \frac{(3+7i) \pm \sqrt{(3+7i)^2-4(2)(-(3-9i))}}{4} \)
\( \Rightarrow x = \frac{(3+7i) \pm \sqrt{9+49i^2+42i+24-72i}}{4} \)
\( \Rightarrow x = \frac{3+7i \pm \sqrt{9-49-30i+24}}{4} \)
\( \Rightarrow x = \frac{3+7i \pm \sqrt{-16-30i}}{4} \)
Let \( a + ib = \sqrt{-16 - 30i} \)
Squaring,
\( \Rightarrow a^2 + i^2b^2 + 2iab = -16 - 30i \)
\( \Rightarrow (a^2 - b^2) + 2iab = -16 - 30i \)
\( \Rightarrow a^2 - b^2 = -16 \) ………….. (i)
And \( 2ab = -30 \)
Now \( (a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2 \)
\( = 256 + 900 \)
\( = 1156 \)
\( \Rightarrow a^2 + b^2 = 34 \) ………… (ii)
Adding (i) & (ii)
\( \Rightarrow a^2 = 18 \)
\( \Rightarrow a^2 = 9 \)
\( \Rightarrow a = \pm 3 \)
Put \( a^2 = 9 \) in eq.(ii)
\( \Rightarrow 9 + b^2 = 34 \)
\( \Rightarrow b^2 = 25 \)
\( \Rightarrow b = \pm 5 \)
\( \dots \sqrt{-16 - 30i} = \pm (3 - 5i) \)
\( \therefore \) value of \( x \) becomes
\( \Rightarrow x = \frac{3+7i \pm (3-5i)}{4} \)
\( \Rightarrow x = \frac{3+7i + (3-5i)}{4} \) and \( x = \frac{3+7i - (3-5i)}{4} \)
\( \Rightarrow x = \frac{6+2i}{4} \) and \( x = \frac{12i}{4} \)
\( \Rightarrow x = \frac{3}{2} + \frac{1}{2}i \) and \( x = 3i \) ans.

Question. If \( z = x + iy \) and \( w = \frac{1-iz}{z-i} \), If \( |w| = 1 \) show that \( z \) is purely real.
Answer: We have \( z = x + iy \), \( w = \frac{1-iz}{z-i} \) and \( |w| = 1 \)
\( \Rightarrow \left| \frac{1-iz}{z-i} \right| = 1 \)
\( \Rightarrow \frac{|1-iz|}{|z-i|} = 1 \) ………. \( \left\{ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} \right\} \)
\( \Rightarrow |1 - iz| = |z - i| \)
\( \Rightarrow |1 - i(x + iy)| = |x + iy - i| \)
\( \Rightarrow |1 - ix - i^2y| = |x + i(y - 1)| \)
\( \Rightarrow |(1 + y) - ix| = |x + i(y - 1)| \)
\( \Rightarrow \sqrt{(1 + y)^2 + x^2} = \sqrt{x^2 + (y - 1)^2} \)
Squaring
\( \Rightarrow 1 + y^2 + 2y + x^2 = x^2 + y^2 + 1 - 2y \)
\( \Rightarrow 4y = 0 \)
\( \Rightarrow y = 0 \)
Since \( z = x + iy \)
\( \Rightarrow z = x + i(0) \)
\( \Rightarrow z = x \) which is purely real (proved)

Question. Convert into polar form \( z = \sin \left(\frac{\pi}{5}\right) + i \left(1 - \cos \frac{\pi}{5}\right) \).
Answer: We have \( z = \sin \left(\frac{\pi}{5}\right) + i \left(1 - \cos \frac{\pi}{5}\right) \)
Here \( a = \sin \left(\frac{\pi}{5}\right) \) and \( b = 1 - \cos \frac{\pi}{5} \)
\( \Rightarrow r = \sqrt{a^2 + b^2} = \sqrt{\sin^2 \frac{\pi}{5} + \left(1 - \cos \frac{\pi}{5}\right)^2} \)
\( = \sqrt{\sin^2 \frac{\pi}{5} + 1 + \cos^2 \frac{\pi}{5} - 2 \cos \frac{\pi}{5}} \)
\( = \sqrt{1 + 1 - 2 \cos \frac{\pi}{5}} \)
\( = \sqrt{2 - 2 \cos \frac{\pi}{5}} \)
\( = \sqrt{2}\sqrt{1 - \cos \frac{\pi}{5}} \)
\( = \sqrt{2}\sqrt{2 \sin^2 \left(\frac{\pi}{10}\right)} \)
\( \Rightarrow r = 2 \sin \left(\frac{\pi}{10}\right) \)
Now \( \tan \alpha = \left| \frac{b}{a} \right| = \left| \frac{1-\cos\frac{\pi}{5}}{\sin\frac{\pi}{5}} \right| \)
\( \Rightarrow \tan \alpha = \left| \frac{2 \sin^2 \left(\frac{\pi}{10}\right)}{2 \sin \left(\frac{\pi}{10}\right) \cos \left(\frac{\pi}{10}\right)} \right| \)
\( \Rightarrow \tan \alpha = \tan \left(\frac{\pi}{10}\right) \)
\( \Rightarrow \alpha = \frac{\pi}{10} \)
Since \( z \) is in 1st quadrant (\( a \) +ve, \( b \) +ve)
\( \therefore \theta = \alpha \)
\( \Rightarrow \theta = \frac{\pi}{10} \) (argument is also called AMPLITUDE)
Polar form is given by
\( \Rightarrow z = r[\cos(\theta) + i \sin(\theta)] \)
\( \Rightarrow z = 2 \sin \left(\frac{\pi}{10}\right) \left[ \cos \left(\frac{\pi}{10}\right) + i \sin \left(\frac{\pi}{10}\right) \right] \) ans.

Question. Evaluate: \( (1 + i)^6 + (1 - i)^3 \).
Answer: \( \Rightarrow \left((1 + i)^2\right)^3 + (1 - i)^3 \)
\( \Rightarrow (1 + i^2 + 2i)^3 + (1 - i)^3 \)
\( \Rightarrow (2i)^3 + 1^3 - i^3 - 3i + 3i^2 \)
\( \Rightarrow 8i^3 + 1 + i - 3i - 3 \)
\( \Rightarrow -8i + 1 + i - 3i - 3 \)
\( \Rightarrow -2 - 10i \) ans.