Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1 \cr & \Rightarrow \,\,{\left| {{z^2} - 1} \right|^2} = {\left( {z\overline z + 1} \right)^2} \cr & \Rightarrow \,\,\left( {{z^2} - 1} \right)\left( {{{\overline z }^2} - 1} \right) = {\left( {z\overline z + 1} \right)^2} \cr & \Rightarrow \,\,{z^2}{\overline z ^2} - {z^2} - {\overline z ^2} + 1 = \,{z^2}{\overline z ^2} + 2z\overline z + 1 \cr & \Rightarrow \,\,{z^2} + 2z\overline z + {\overline z ^2} = 0 \cr & \Rightarrow \,\,{\left( {z + \overline z } \right)^2} = 0 \cr & \Rightarrow \,\,z = - \overline z \cr & \Rightarrow \,\,z\,\,{\text{is purely imaginary}} \cr} $$

$$\eqalign{ & {z^2} + az + b = 0\,;\,\,\,{z_1} + {z_2} = - a \, \& \, {z_1}{z_2} = b \cr & 0,{z_1},{z_2}\,\,{\text{from}}\,{\text{an}}\,{\text{equilateral}}\,\,\Delta \cr & \therefore {0^2} + {z_1}^2 + {z_2}^2 = 0.{z_1} + {z_1}.{z_2} + {z_2}.0 \cr} $$
$$\left( {{\text{For}}\,{\text{an}}\,{\text{equilateral}}\,{\text{triangle}},\,\,{z_1}^2 + {z_2}^2 + {z_3}^2 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}} \right)$$
$$\eqalign{ & \Rightarrow \,\,{z_1}^2 + {z_2}^2 = {z_1}{z_2} \cr & \Rightarrow \,\,{\left( {{z_1} + {z_2}} \right)^2} = 3{z_1}{z_2} \cr & \therefore \,\,{a^2} = 3b \cr} $$

$$\eqalign{ & {e^{iA}} \cdot {e^{iB}} \cdot {e^{iC}} = \left( {\cos A + i\sin A} \right)\left( {\cos B + i\sin B} \right)\left( {\cos C + i\sin C} \right) \cr & {e^{iA}} \cdot {e^{iB}} \cdot {e^{iC}} = \cos \left( {A + B + C} \right) + i\sin \left( {A + B + C} \right) = \cos \pi + i\sin \pi . \cr} $$

$$\eqalign{ & z = {\log _2}\left( {1 + i} \right) \cr & = {\log _2}\left( {\sqrt 2 {e^{\frac{{i\pi }}{4}}}} \right) \cr & = \frac{1}{2} + i\frac{\pi }{4}{\log _2}e \cr & \therefore z + \bar z = 1{\text{ and }}z - \bar z = i\frac{\pi }{2}{\log _2}e \cr & {\text{Hence, }}\left( {z + \bar z} \right) + i\left( {z - \bar z} \right) \cr & = 1 - \frac{\pi }{2}{\log _2}e \cr & = 1 - \frac{\pi }{{2\ln 2}} \cr & = \frac{{\ln 4 - \pi }}{{\ln 4}} \cr} $$

$$z = 1 + \cos \frac{{11\pi }}{9} + i\sin \frac{{11\pi }}{9}$$
$$\operatorname{Re} \left( z \right) > 0$$   and $$\operatorname{Im} \left( z \right) < 0,$$   so the number lies in the fourth quadrant. Also
$$\eqalign{ & z = 2\cos \frac{{11\pi }}{{18}}\left\{ {\cos \frac{{11\pi }}{{18}} + i\sin \frac{{11\pi }}{{18}}} \right\} \cr & = 2\cos \frac{{11\pi }}{{18}}\left\{ {\cos \left( { - \frac{{7\pi }}{{18}}} \right) + i\sin \left( { - \frac{{7\pi }}{{18}}} \right)} \right\} \cr & \therefore \arg \left( z \right) = - \frac{{7\pi }}{{18}} \cr & \left| z \right| = \left| {2\cos \frac{{11\pi }}{{18}}} \right| = - 2\cos \frac{{11\pi }}{{18}} \cr} $$

$$\eqalign{ & \frac{\alpha }{a} + \frac{\beta }{b} + \frac{\gamma }{c} = 1 + i{\text{ squaring}} \cr & \frac{{{\alpha ^2}}}{{{a^2}}} + \frac{{{\beta ^2}}}{{{b^2}}} + \frac{{{\gamma ^2}}}{{{c^2}}} + 2\left( {\frac{{\alpha \beta }}{{ab}} + \frac{{\beta \gamma }}{{bc}} + \frac{{\gamma \alpha }}{{ac}}} \right) = 2i \cr & {\text{or }}\frac{{{\alpha ^2}}}{{{a^2}}} + \frac{{{\beta ^2}}}{{{b^2}}} + \frac{{{\gamma ^2}}}{{{c^2}}} + \frac{{2\alpha \beta \gamma }}{{abc}}\left( {\frac{c}{\gamma } + \frac{a}{\alpha } + \frac{b}{\beta }} \right) = 2i \cr & \therefore \frac{{{\alpha ^2}}}{{{a^2}}} + \frac{{{\beta ^2}}}{{{b^2}}} + \frac{{{\gamma ^2}}}{{{c^2}}} = 2i \cr} $$

$$\eqalign{ & {\text{Given,}}\,x\, = {\omega ^2} - \omega - 2 \cr & \Rightarrow \,x + 2 = {\omega ^2} - \omega \cr} $$
On squaring both sides, we get
$$\eqalign{ & {\left( {x + 2} \right)^2} = {\left( {{\omega ^2} - \omega } \right)^2} \cr & \Rightarrow \,{x^2} + 4x + 4 = {\omega ^4} + {\omega ^2} - 2{\omega ^3} \cr} $$
Add 3 on both side
$$\eqalign{ & \Rightarrow \,{x^2} + 4x + 4 + 3 = \omega + {\omega ^2} - 2 + 3\,\,\left( {\because \,{\omega ^3} = 1} \right) \cr & \Rightarrow \,{x^2} + 4x + 7 = 1 + \omega + {\omega ^2} \cr & \Rightarrow \,{x^2} + 4x + 7 = 0\,\,\left( {\because \,1 + \omega + {\omega ^2} = 0} \right) \cr} $$

$$\eqalign{ & \because \,\,\left| z \right| = \left| \omega \right|\,\,{\text{and arg }}z = \pi - \,{\text{arg }}\omega \cr & {\text{Let }}\omega = r{e^{i\theta }}\,\,{\text{then }}z = r{e^{i\left( {\pi - \theta } \right)}} \cr & \Rightarrow \,\,z = r{e^{i\pi }}.{e^{ - i\theta }} \cr & = \left( {r{e^{ - i\theta }}} \right)\left( {\cos \pi + i\sin \pi } \right) = \overline \omega \left( { - 1} \right) = - \overline \omega \cr} $$

$$\eqalign{ & {\text{amp}}\,{z_1} = {\text{amp}}\left( {1 + \sqrt {3}i } \right) = {\tan ^{ - 1}}\sqrt 3 = \frac{\pi }{3}. \cr & {\text{amp}}\,{z_2} = \frac{\pi }{3} - \frac{{2\pi }}{3} = - \frac{\pi }{3}\,\,{\text{and }}\left| {{z_2}} \right| = 2 \cr & \therefore \,\,{z_2} = 2\left\{ {\cos \left( { - \frac{\pi }{3}} \right) + i\sin \left( { - \frac{\pi }{3}} \right)} \right\}. \cr & {\text{amp }}{z_3} = \frac{\pi }{3} + \frac{{2\pi }}{3} = \pi \,\,{\text{and }}\left| {{z_3}} \right| = 2 \cr & \therefore \,\,{z_3} = 2\left\{ {\cos \pi + i\sin \pi } \right\}. \cr} $$

$$\eqalign{ & {\left( {x - 1} \right)^3} + 8 = 0 \cr & \Rightarrow \,\,{\left( {x - 1} \right)^3} = - 8 = {\left( { - 2} \right)^3} \cr & \Rightarrow \,\,x - 1 = - 2 \cr & {\text{or}}\,\,\, - 2\omega \,\,\,{\text{or }} - 2{\omega ^2} \cr & \Rightarrow \,\,x = - 1,1 - 2\omega ,1 - 2{\omega ^2} \cr} $$