Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \left| z \right| = \left| {z - 1} \right| \cr & \Rightarrow \,\,{\left| z \right|^2} = {\left| {z - 1} \right|^2} \cr & \Rightarrow \,\,z\overline z = \left( {z - 1} \right)\left( {\overline z - 1} \right) \cr & \Rightarrow \,\,z\overline z = z\overline z - z - \overline z + 1 \cr & \therefore \,\,z + \overline z = 1 \cr & \left| z \right| = \left| {z + 1} \right| \cr & \Rightarrow \,\,z + \overline z = - 1\left( {{\text{similarly}}} \right) \cr & \therefore \,\,\left| {z + \overline z } \right| = 1. \cr} $$

$$\eqalign{ & {z^2} + z + 1 = 0 \cr & \Rightarrow \,\,z = \omega \,\,{\text{or }}{\omega ^2} \cr & {\text{So, }}z + \frac{1}{z} = \omega + {\omega^2} = - 1 \cr & {z^2} + \frac{1}{{{z^2}}} = {\omega ^2} + \omega = - 1,{z^3} + \frac{1}{{{z^3}}} = {\omega ^3} + {\omega ^3} = 2 \cr & {z^4} + \frac{1}{{{z^4}}} = - 1,{z^5} + \frac{1}{{{z^5}}} = - 1\,\,{\text{and }}{z^6} + \frac{1}{{{z^6}}} = 2 \cr} $$
∴ The given sum = 1 + 1 + 4 + 1 + 1 + 4 = 12

$$\eqalign{ & {\text{Let, }}{z_1} = 1 + 4i,{z_2} = 3 + i,{z_3} = 1 - i{\text{ and }}{z_4} = 2 - 3i \cr & \therefore {m_1} = \left| {{z_1}} \right|,{m_2} = \left| {{z_2}} \right|,{m_3} = \left| {{z_3}} \right|{\text{ and }}{m_4} = \left| {{z_4}} \right| \cr & \Rightarrow {m_1} = \sqrt {17} ,{m_2} = \sqrt {10} ,{m_3} = \sqrt 2 {\text{ and }}{m_4} = \sqrt {13} \cr & \Rightarrow {m_3} < {m_2} < {m_4} < {m_1}. \cr} $$

$$\eqalign{ & z = \left| z \right|\left( {\cos \,\theta + i\,\sin \,\theta } \right) \cr & {\text{where }}\theta = {\text{Arg }}z \cr & {\text{if }}{\theta _1} = {\text{Arg }}w{\text{ then }}\theta = \pi - {\theta _1} \cr & \therefore \,z = \left| w \right|\left\{ {\cos \left( {\pi - {\theta _1}} \right) + i\,\sin \left( {\pi - {\theta _1}} \right)} \right\} \cr & \Rightarrow z = \left| w \right|\left( { - \cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right) \cr & \Rightarrow z = - \left| w \right|\left( { - \cos \,{\theta _1} + i\,\sin \,{\theta _1}} \right) \cr & \Rightarrow z = - \left| w \right|\left( {\cos \,{\theta _1} - i\,\sin \,{\theta _1}} \right) \cr & \Rightarrow z = - \left| {\overline w } \right| \cr} $$

$$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$
$$ \Rightarrow \,\,{z_1}\,{\text{and }}{z_2}$$   are collinear and are to the same side of origin; hence $${\text{arg }}{z_1} - {\text{arg }}{z_2} = 0.$$

$$\eqalign{ & \frac{{z - 1}}{{z + 1}} = \frac{{x + iy - 1}}{{x + iy + 1}} \cr & \frac{{z - 1}}{{z + 1}} = \frac{{{x^2} + {y^2} - 1 + 2iy}}{{{x^2} + {y^2} + 2x + 1}} \cr & \Rightarrow \,\operatorname{Re} \left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{{{x^2} + {y^2} - 1}}{{{x^2} + {y^2} + 2x + 1}} = 0 \cr & \Rightarrow \,{x^2} + {y^2} - 1 = 0 \cr & \Rightarrow \,{x^2} + {y^2} = 1 \cr & {\text{Also}},\,\,z\bar z = {x^2} + {y^2} = 1 \cr & {\text{and}}\,\,z\bar z = {\left| z \right|^2} \cr & \Rightarrow \,{\left| z \right|^2} = 1 \cr & \Rightarrow \,\left| z \right| = 1 \cr} $$

$$\eqalign{ & \left| \omega \right| = 1 \cr & \Rightarrow \,\,\left| {\frac{{1 - iz}}{{z - i}}} \right| = 1 \cr & \Rightarrow \,\,\left| {1 - iz} \right| = \left| {z - i} \right| \cr & \Rightarrow \,\,\left| {1 - i\left( {x + iy} \right)} \right| = \left| {x + iy - i} \right| \cr & \Rightarrow \,\,\left| {\left( {y + 1} \right) - ix} \right|\, = \left| {x + i\left( {y - 1} \right)} \right| \cr & \Rightarrow \,\,{x^2} + {\left( {y + 1} \right)^2} = {x^2} + {\left( {y - 1} \right)^2} \cr & \Rightarrow \,\,4y = 0 \cr & \Rightarrow \,\,y = 0 \cr & \Rightarrow \,\,z\,\,{\text{lies on real axis}}\, \cr} $$

Clearly, $$\left| z \right| = \left| {{z_1}} \right|$$
$$\eqalign{ & \therefore \,\,z\overline z = {z_1}{\overline z _1} \cr & {\text{or, }}\frac{z}{{{z_1}}} = \frac{{{{\overline z }_1}}}{{\overline z }} = \overline {\left( {\frac{{{z_1}}}{z}} \right)} . \cr} $$

$$\eqalign{ & z = 1 + 2i \cr & \Rightarrow \left| z \right| = \sqrt {1 + 4} = \sqrt 5 \cr & \therefore \,f\left( z \right) = \frac{{7 - z}}{{1 - {z^2}}} = \frac{{7 - 1 - 2i}}{{1 - {{\left( {1 + 2i} \right)}^2}}} \cr & = \,\frac{{6 - 2i}}{{1 - \left( {1 - 4 + 4i} \right)}} = \frac{{6 - 2i}}{{4 - 4i}} = \frac{{3 - i}}{{2 - 2i}} \cr & \Rightarrow \,\left| {f\left( z \right)} \right| = \left| {\frac{{3 - i}}{{2 - 2i}}} \right| = \frac{{\left| {3 - i} \right|}}{{\left| {2 - 2i} \right|}} \cr & = \,\frac{{\sqrt {9 + 1} }}{{\sqrt {4 + 4} }} = \frac{{\sqrt 5 }}{2} = \frac{{\left| z \right|}}{2} \cr} $$

$$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$     can hold when $$0,{z_1},{z_2}$$   are collinear with $$0$$ at one end.