Complex Number MCQ Questions & Answers in Algebra | Maths

Consider $$\left| {\overline z \omega } \right| = \left| {\overline z } \right|\left| \omega \right| = \left| z \right|\left| \omega \right| = \left| {z\omega } \right| = 1$$
Consider $${\text{Arg}}\left( {\overline z \omega } \right) = \arg \left( {\overline z } \right) + \arg \left( \omega \right) = - \arg \left( z \right) + \arg \omega = - \frac{\pi }{2}$$
$$\therefore \overline z \omega = - 1$$

$$\eqalign{ & {\text{Given, }}2x = 3 + 5i \cr & \Rightarrow x = \frac{{3 + 5i}}{2} \cr & {\text{Consider }}{x^3} = \frac{{27 + 125{i^3} + 225{i^2} + 135i}}{8} \cr & = \frac{{27 - 125i - 225 + 135i}}{8}\left( {\because {i^2} = - 1,{i^3} = - i} \right) \cr & = \frac{{ - 198 + 10i}}{8} \cr & = \frac{{ - 99 + 5i}}{4} \cr & {\text{and}}\,\,{x^2} = \frac{{9 + 25{i^2} + 30i}}{4} \cr & = \,\frac{{9 - 25 + 30i}}{4} \cr & = \frac{{ - 8 + 15i}}{2} \cr & {\text{Now, Consider }}2{x^3} + 2{x^2} - 7x + 72 \cr & = \left( {\frac{{ - 99 + 5i}}{2}} \right) + \left( { - 8 + 15i} \right) - \frac{{7\left( {3 + 5i} \right)}}{2} + 72 \cr & = - \frac{{99}}{2} + \frac{{5i}}{2} - 8 + 15i - \frac{{21}}{2} - \frac{{35}}{2}i + 72 \cr & = \left( { - \frac{{99}}{2} - 8 - \frac{{21}}{2} + 72} \right) + \left( {\frac{5}{2} + 15 - \frac{{35}}{2}} \right)i \cr & = \frac{{ - 99 - 16 - 21 + 144}}{2} \cr & = \frac{8}{2} \cr & = 4 \cr} $$

Given, $$\frac{{{{\left( {1 + i} \right)}^{4n + 5}}}}{{{{\left( {1 - i} \right)}^{4n + 3}}}}$$
$$\eqalign{ & = \frac{{{{\left( {1 + i} \right)}^{4n + 3}} \cdot {{\left( {1 + i} \right)}^2}}}{{{{\left( {1 - i} \right)}^{4n + 3}}}} = {\left( {\frac{{1 + i}}{{1 - i}}} \right)^{4n + 3}} \cdot {\left( {1 + i} \right)^2} \cr & = {\left[ {\frac{{\left( {1 + i} \right)\left( {1 + i} \right)}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}} \right]^{4n + 3}} \cdot \left( {1 + {i^2} + 2i} \right) \cr & = {\left[ {\frac{{1 + {i^2} + 2i}}{{1 + i}}} \right]^{4n + 3}} \cdot 2i = {\left( i \right)^{4n + 3}} \cdot 2i = 2{\left( i \right)^{4n + 4}} \cr & = 2 \cdot \left( {{i^{4\left( {n + 1} \right)}}} \right) = 2 \cr} $$

Roots of $${x^2} + x + 1 = 0$$    are $$x = \omega ,{\omega ^2}.$$   For both these values $${x^{3m}} + {x^{3n - 1}} + {x^{3r - 2}} = 0.$$

$$\eqalign{ & \left( {\frac{{\sqrt 3 }}{2} + \frac{i}{2}} \right) = - i\left( {\frac{{ - 1}}{2} + \frac{{i\sqrt 3 }}{2}} \right) = i\omega \cr & \frac{{\sqrt 3 }}{2} - \frac{i}{2} = i\left( {\frac{{ - 1}}{2} - \frac{{i\sqrt 3 }}{2}} \right) = i{\omega ^2} \cr & \therefore \,\,z = {\left( { - i\omega } \right)^5} + {\left( {i{\omega ^2}} \right)^5} = - i{\omega ^2} + i\omega \cr & = i\left( {\omega - {\omega ^2}} \right) = i\left( {i\sqrt 3 } \right) = - \sqrt 3 \cr & \Rightarrow \,\,{\text{Re}}\left( z \right) < 0\,\,{\text{and Im}}\left( z \right) = 0 \cr & \therefore \,\,\,\left( {\text{B}} \right)\,\,{\text{is the correct choice}}{\text{.}} \cr} $$

$$\left| z \right| = \frac{1}{2},$$   it represents circle with circle center $$\left( {0.0} \right)r = \frac{1}{2}$$
by drawing figure we get, since $${z_2}$$ is anticlockwise $${z_1} = - \frac{1}{2}$$
$${z_1} = \frac{1}{2} + \sqrt 3 \frac{i}{2}$$
and Argument $$\left( z \right) = \sqrt 3 $$
$$\eqalign{ & \tan \,\theta = \sqrt 3 \,\theta = {60^ \circ }\left| z \right| = \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} = 1 \cr & {\text{Similarly }}\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| \cr} $$
So, $$\left| z \right| = 1$$  and $$z$$ lies on the real axis of the positive direction.
∴ option D is correct.

$$\eqalign{ & \frac{{1 - i}}{{1 + i}} = \frac{{{{\left( {1 - i} \right)}^2}}}{{1 - {i^2}}} = \frac{{ - 2i}}{2} = - i \cr & \therefore \,\,{\left( {\frac{{1 - i}}{{1 + i}}} \right)^n} = {\left( { - i} \right)^n} = {\text{imaginary}} \cr & \Rightarrow \,\,n = 1,3,5,.....\,\,. \cr} $$

$$\eqalign{ & {\text{arg }}zw = \pi \cr & \Rightarrow \,\,{\text{arg }}z + {\text{arg }}w = \pi \,\,\,\,\,.....\left( 1 \right) \cr & \overline z + i\overline w = 0 \cr & \Rightarrow \,\,\overline z = - i\overline w \cr & \therefore \,\,\,z = iw \cr & \Rightarrow \,{\text{arg }}z = \frac{\pi }{2} + {\text{arg }}w \cr & \Rightarrow \,\,{\text{arg }}z = \frac{\pi }{2} + \pi - {\text{arg }}z\left( {{\text{from}}\left( 1 \right)} \right) \cr & \therefore \,\,{\text{arg }}z = \frac{{3\pi }}{4} \cr} $$