Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\text{Let }}z = x + iy \cr & \left| {z - 1} \right| = \left| {z + 1} \right|{\left( {x - 1} \right)^2} + {y^2} = {\left( {x + 1} \right)^2} + {y^2} \cr & \Rightarrow \,\,{\text{Re}}\,z = 0\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,x = 0 \cr & \left| {z - 1} \right| = \left| {z - i} \right|{\left( {x - 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2} \cr & \Rightarrow \,\,x = y \cr & \left| {z + 1} \right| = \left| {z - i} \right|{\left( {x + 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2} \cr} $$
Only $$(0, 0)$$  will satisfy all conditions.
⇒ Number of complex number $$z = 1$$

$$\eqalign{ & {\text{arg}}\left( z \right) < 0\,\left( {{\text{given}}} \right) \cr & \Rightarrow \,\,{\text{arg}}\left( z \right) = - \theta \cr & {\text{Now}} \cr} $$

$$\eqalign{ & z = r\cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right) \cr & \,\,\,\, = r\left[ {\cos \left( \theta \right) - i\sin \left( \theta \right)} \right] \cr & {\text{Again }} - z = - r\left[ {\cos \left( \theta \right) - i\sin \left( \theta \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = r\left[ {\cos \left( {\pi - \theta } \right) + i\sin \left( {\pi - \theta } \right)} \right] \cr & \therefore \,\,{\text{arg}}\left( { - z} \right) = \pi - \theta ; \cr & {\text{Thus arg}}\left( { - z} \right) - {\text{arg}}\left( z \right) = \pi - \theta - \left( { - \theta } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \pi - \theta + \theta = \pi \cr} $$

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

$${\text{Let }}z \in S{\text{ then }}z = \frac{{\alpha + i}}{{\alpha - i}}$$
Since, $$z$$ is a complex number and let $$z = x + iy$$
$$\eqalign{ & {\text{Then, }}x + iy = \frac{{{{\left( {\alpha + 1} \right)}^2}}}{{{\alpha ^2} + 1}}\left( {{\text{by rationalisation}}} \right) \cr & \Rightarrow \,\,x + iy = \frac{{\left( {{\alpha ^2} - 1} \right)}}{{{\alpha ^2} + 1}} + \frac{{i\left( {2\alpha } \right)}}{{{\alpha ^2} + 1}} \cr} $$
Then compare both sides
$$\eqalign{ & x = \frac{{{\alpha ^2} - 1}}{{{\alpha ^2} + 1}}\,\,\,\,\,\,\,.....\left( 1 \right) \cr & y = \frac{{2\alpha }}{{{\alpha ^2} + 1}}\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Now squaring and adding equations (1) and (2)
$$ \Rightarrow \,\,{x^2} + {y^2} = \frac{{{{\left( {{\alpha ^2} - 1} \right)}^2}}}{{{{\left( {{\alpha ^2} + 1} \right)}^2}}} + \frac{{4{\alpha ^2}}}{{{{\left( {{\alpha ^2} + 1} \right)}^2}}} = 1$$

$$\left| {iz + {z_0}} \right| = \left| {i\left( {z - i} \right) - 1 + 5 + 3i} \right| = \left| {i\left( {z - i} \right) + 4 + 3i} \right| \leqslant \left| i \right|\left| {z - i} \right| + \left| {4 + 3i} \right| \leqslant 1 \cdot 2 + 5 = 7.$$

Let the point $$A$$ represents the complex number $$z,$$ $$B$$ represents $$\omega z$$  and $$C$$ represents $$\bar \omega z.$$
$$\omega \,\,\& \,\,\bar \omega $$  are complex cube roots of unity clearly $$\omega z$$  means rotation of $$z$$ by $$\frac{{2\pi }}{3}$$ and $${\omega ^2}z\left( { = \bar \omega z} \right)$$   means rotation of $$\omega z$$  by $$\frac{{2\pi }}{3}.$$

$$\therefore \,\angle AOB = \angle BOC = \angle COA = \frac{{2\pi }}{3}$$        also $$OA = OB = OC = \left| z \right|.$$      That is the $$\Delta ABC$$  is equilateral.
Now, $$AC = 2AD + 2\left( {OA\,\cos \,{{30}^ \circ }} \right)$$
$$ = 2\left| z \right|\frac{{\sqrt 3 }}{2} = \sqrt 3 \left| z \right|$$
Area of $$\Delta ABC = \frac{{\sqrt 3 }}{2}{\left( {{\text{side}}} \right)^2} = \frac{{3\sqrt 3 }}{2}{\left| z \right|^2}$$

The initial position of point is $${Z_0} = 1 + 2i$$
$$\therefore \,\,{Z_1} = \left( {1 + 5} \right) + \left( {2 + 3} \right)i = 6 + 5i$$
Now $${Z_1}$$ is moved through a distance of $$\sqrt 2 $$ units in the direction $$\hat i + \hat j.$$   (i.e. by $$1 + i\,\,$$)
∴ It becomes $${Z_1}' = {Z_1} + \left( {1 + i} \right) = 7 + 6i$$
Now $$O{Z_1}'$$ is rotated through an angle $$\frac{\pi }{2}$$ in anticlockwise direction, therefore $${Z_2} = i{Z_1}' = - 6 + 7i$$

$$\eqalign{ & \operatorname{Im} \left( {{z_1}{{\overline z }_2}} \right) = 1 \cr & \Rightarrow \,\,bp - aq = 1 \cr & {\omega _1}{\overline \omega _2} = \left( {a + ip} \right)\left( {b - iq} \right) = \left( {ab + pq} \right) + i\left( {bp - aq} \right) \cr & \therefore \,\,\operatorname{Im} \left( {{\omega _1}{{\overline \omega }_2}} \right) = 1. \cr} $$

$$\eqalign{ & {\left( {1 + i} \right)^{{n_1}}} + {\left( {1 + {i^3}} \right)^{{n_1}}} + {\left( {1 + {i^5}} \right)^{{n_2}}} + {\left( {1 + {i^7}} \right)^{{n_2}}} \cr & = {\left( {1 + i} \right)^{{n_1}}} + {\left( {1 - i} \right)^{{n_1}}} + {\left( {1 + i} \right)^{{n_2}}} + {\left( {1 - i} \right)^{{n_2}}} \cr & {\text{Using }}1 + i = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right) \cr & {\text{and 1}} - i = \sqrt 2 \left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right) \cr} $$
We get the given expression as
$$\eqalign{ & = {\left( {\sqrt 2 } \right)^{{n_1}}}\left[ {\cos \frac{{{n_1}\pi }}{4} + i\sin \frac{{{n_1}\pi }}{4}} \right] + {\left( {\sqrt 2 } \right)^{{n_2}}}\left[ {\cos \frac{{{n_2}\pi }}{4} + i\sin \frac{{{n_2}\pi }}{4}} \right] + {\left( {\sqrt 2 } \right)^{{n_2}}}\left[ {\cos \frac{{{n_2}\pi }}{4} - i\sin \frac{{{n_2}\pi }}{4}} \right] \cr & = {\left( {\sqrt 2 } \right)^{{n_1}}}\left[ {2\cos \frac{{{n_1}\pi }}{4}} \right] + {\left( {\sqrt 2 } \right)^{{n_2}}}\left[ {2\cos \frac{{{n_2}\pi }}{4}} \right] \cr} $$
= real number irrespective the values of $${{n_1}}$$ and $${{n_2}}$$
∴ $$(D)$$ is the most appropriate answer.

Let the vertices be $${z_0},{z_1},.....,{z_5}$$   w.r.t. center $$O$$ at origin and $$\left| {{z_0}} \right| = \sqrt 5 .$$

$$\eqalign{ & \Rightarrow \,{A_0}{A_1} = \left| {{z_1} - {z_0}} \right| = \left| {{z_0}{e^{i\theta }} - {z_0}} \right| \cr & = \,\left| {{z_0}} \right|\left| {\cos \,\theta + i\,\sin \,\theta - 1} \right| \cr & = \,\sqrt 5 \sqrt {{{\left( {\cos \,\theta - 1} \right)}^2} + {{\sin }^2}\,\theta } \cr & = \,\sqrt 5 \sqrt {2\left( {1 - \cos \,\theta } \right)} = \sqrt 5 \,2\sin \left( {\frac{\theta }{2}} \right) \cr & \Rightarrow \,{A_0}{A_1} = \sqrt 5 .\,2\sin \left( {\frac{\pi }{6}} \right) = \sqrt 5 \cr & \left( {\because \,\theta = \frac{{2\pi }}{6} = \frac{\pi }{3}} \right) \cr} $$
Similarly, $${A_1}{A_2} = {A_2}{A_3} = {A_3}{A_4} = {A_4}{A_5} + {A_5}{A_0} = 6\sqrt 5 .$$
Hence the perimeter of regular polygon is
$$ = \,{A_0}{A_1} + {A_1}{A_2} + {A_2}{A_3} + {A_3}{A_4} + {A_4}{A_5} + {A_5}{A_0} = 6\sqrt 5 .$$