Complex Number MCQ Questions & Answers in Algebra | Maths

Using the result $$\left| {{z_1} + {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|,$$
we get, $$\left| z \right| + \left| {z - i} \right| = \left| z \right| + \left| {i - z} \right|\,\,\,\,\left[ {{\text{since}}\,\left| z \right| = \left| { - z} \right|} \right]$$
$$ \leqslant \left| {z + i - z} \right| = \left| i \right| = 1$$
∴ minimum value of $$\left| z \right| + \left| {z - i} \right|\,\,{\text{is}}\,\,1$$

$$\frac{{z - 1}}{i}$$  must be real. Now, $$\frac{{z - 1}}{i} = \frac{{x - 1 + iy}}{i} = y - \left( {x - 1} \right)i$$
$$\therefore \,\,x - 1 = 0.$$   Then $${\sin ^{ - 1}}\left\{ {\frac{1}{i}\left( {z - 1} \right)} \right\} = {\sin ^{ - 1}}y.\,{\text{So, }} - 1 \leqslant y \leqslant 1.$$

$$\eqalign{ & {\text{Let }}z = x + iy \cr & \therefore \,\,{z^2} = {x^2} - {y^2} + 2ixy \cr & {\text{Now }}\frac{{{z^2}}}{{z - 1}}\,{\text{is real}} \cr & \Rightarrow \,\,{\text{Im}}\left( {\frac{{{z^2}}}{{z - 1}}} \right) = 0 \cr & \Rightarrow \,\,{\text{Im}}\left( {\frac{{{x^2} - {y^2} + 2ixy}}{{\left( {x - 1} \right) + iy}}} \right) = 0 \cr & \Rightarrow \,\,{\text{Im}}\left[ {\left( {{x^2} - {y^2} + 2ixy} \right)\left( {x - 1} \right) - iy} \right] = 0 \cr & \Rightarrow \,\,2xy\left( {x - 1} \right) - y\left( {{x^2} - {y^2}} \right) = 0 \cr & \Rightarrow \,\,y\left( {{x^2} + {y^2} - 2x} \right) = 0 \cr & \Rightarrow \,\,y = 0;\,{x^2} + {y^2} - 2x = 0 \cr} $$
∴ $$z$$ lies either on real axis or on a circle through origin.

$$\eqalign{ & {\text{Here, }}\left| {0 - {z_1}} \right|\left| {0 - z} \right| = 1. \cr & \therefore \,\,\left| {{z_1}} \right|\left| z \right| = 1 \cr & \therefore \,\,\left| {{z_1}} \right| = \frac{1}{{\left| z \right|}}. \cr & {\text{Also, amp}}\frac{{{z_1} - 0}}{{z - 0}} = 0\,\,\,\,\,\therefore \,\,{\text{amp}}\frac{{{z_1}}}{z} = 0\,\,\,\,\therefore \,\,{\text{amp}}\,{z_1} = {\text{amp }}z. \cr & \therefore \,\,{z_1} = \frac{1}{{\left| z \right|}}\left\{ {\cos \left( {{\text{amp }}{z_1}} \right) + i\sin \left( {{\text{amp }}{z_1}} \right)} \right\} \cr & {z_1} = \frac{1}{{{{\left| z \right|}^2}}} \cdot \left| z \right|\left\{ {\cos \left( {{\text{amp }}z} \right) + i\sin \left( {{\text{amp }}z} \right)} \right\} \cr & {z_1} = \frac{1}{{{{\left| z \right|}^2}}} \cdot z = \frac{z}{{z\overline z }} = \frac{1}{{\overline z }}. \cr} $$

$$\eqalign{ & z = \frac{\pi }{4}{\left( {1 + i} \right)^4}\left( {\frac{{1 - \sqrt \pi i}}{{\sqrt \pi + i}} + \frac{{\sqrt \pi - i}}{{1 + \sqrt \pi i}}} \right) \cr & = \frac{\pi }{4}{\left( {1 + i} \right)^4}\left[ {\frac{{1 + \pi + \pi + 1}}{{\left( {\sqrt \pi + i} \right)\left( {1 + \sqrt \pi i} \right)}}} \right] = \frac{\pi }{4}{\left( {1 + i} \right)^4}\frac{2}{i} \cr & = \frac{\pi }{4}{\left( {2i} \right)^2}\frac{2}{i} = 2\pi i \cr & \therefore \left( {\frac{{\left| z \right|}}{{amp\left( z \right)}}} \right) = \frac{{2\pi }}{{\frac{\pi }{2}}} = 4 \cr} $$

$$\because \,\,{\text{Im}}\left( z \right) \ne 0$$
⇒ $$z$$ is non real and equation $${z^2} + z + \left( {1 - a} \right) = 0$$     will have non real roots, if $$D < 0$$
$$\eqalign{ & \Rightarrow \,\,1 - 4\left( {1 - a} \right) < 0 \cr & \Rightarrow \,\,4a < 3 \cr & \Rightarrow \,\,a < \frac{3}{4} \cr} $$
∴ $$a$$ can not take the value $$\frac{3}{4}$$

$$\eqalign{ & {x^2} + {y^2} = 2{k^2},\operatorname{Re} \left\{ {{{\left( {x + iy} \right)}^2}} \right\} = 0 \cr & {\text{i}}{\text{.e}}{\text{., }}{x^2} - {y^2} = 0 \cr} $$
Solving these, $$x = \pm k,y = \pm k.$$

Consider,
$$\eqalign{ & {\left( { - \sqrt { - 1} } \right)^{4n + 3}} + {\left( {{i^{41}} + {i^{ - 257}}} \right)^9} \cr & = \,{\left( { - i} \right)^{4n + 3}} + {\left[ {{{\left( {{i^4}} \right)}^{10}}.{i^1} + {{\left( {{i^3}} \right)}^{ - 85}}.{i^{ - 2}}} \right]^9} \cr & = \,{\left( { - i} \right)^{4n + 3}} + {\left[ {i + \frac{1}{{{{\left( {{i^3}} \right)}^{85}}}}.\frac{1}{{{i^2}}}} \right]^9} \cr & = \,{\left( { - i} \right)^{4n + 3}} + {\left( {i + \frac{1}{i}} \right)^9} \cr & = \, - {\left( { - 1} \right)^{4n + 3}}{\left( i \right)^{4n}}{\left( i \right)^3} + {\left( {i - i} \right)^9} \cr & = - \left( 1 \right)\left( { - i} \right) + 0 \cr & = i \cr} $$

$$\eqalign{ & {\left| {z - \omega } \right|^2} + {\left| {z - {\omega ^2}} \right|^2} = \lambda \cr & \Rightarrow \,\,\lambda = {\left| {\omega - {\omega ^2}} \right|^2} \cr} $$

$$\eqalign{ & {\text{If}}\,\,\left| {z - 1} \right| > \left| {z + 1} \right|,{\text{then}}\,{\text{max}}\,\left\{ {\left| {z - 1} \right|,\left| {z + 1} \right|} \right\} = \left| {z - 1} \right| \cr & \Rightarrow \,{\text{If}}\,\,{\left| z \right|^2} + 1 - z - \bar z > {\left| z \right|^2} + 1 + z + \bar z\,\,{\text{then}}\,\,\left| z \right| = \left| {z - 1} \right| \cr & \Rightarrow \,{\text{If}}\,\,z + \bar z\,{\text{ < }}\,{\text{0}}\,\,{\text{then}}\,\,{\left| z \right|^2} = {\left| z \right|^2} + 1 - z - \bar z \cr & \Rightarrow \,{\text{If}}\,\,z + \bar z\,{\text{ < }}\,{\text{0}}\,\,{\text{then}}\,\,z + \bar z = 1, \cr} $$
which is not possible.
$$\eqalign{ & {\text{Again,}}\,\,{\text{If}}\,\,\left| {z + 1} \right| > \left| {z - 1} \right|\,\,{\text{then}}\,\,{\text{max}}\,\left\{ {\left| {z - 1} \right|,\left| {z + 1} \right|} \right\} = \left| {z + 1} \right| \cr & \Rightarrow \,{\text{If}}\,{\left| z \right|^2} + 1 + z + \bar z > {\left| z \right|^2} + 1 - z - \bar z\,\,{\text{then}}\,\,\left| z \right| = \left| {z + 1} \right| \cr & \Rightarrow \,{\text{If}}\,\,z + \bar z > 0\,\,{\text{then}}\,\,{\left| z \right|^2} = {\left| z \right|^2} + 1 + z + \bar z \cr & \Rightarrow \,{\text{If}}\,\,z + \bar z > 0\,\,{\text{then}}\,\,z + \bar z = - 1\,\,{\text{Not}}\,{\text{possible}}\,{\text{again}}. \cr} $$
Therefore the given result cannot hold.