Complex Number MCQ Questions & Answers in Algebra | Maths

$$\because \frac{{w - wz}}{{1 - z}}$$   is purely real
$$\eqalign{ & \therefore \,\,\overline {\left( {\frac{{w - \overline w z}}{{1 - z}}} \right)} = \left( {\frac{{w - \overline w z}}{{1 - z}}} \right) \cr & \Rightarrow \,\,\frac{{\overline w - w\overline z }}{{1 - \overline z }} = \frac{{w - \overline w z}}{{1 - z}} \cr & \Rightarrow \,\,\overline w - \overline w z - w\overline z + wz\overline z = w - w\overline z - \overline w z + \overline w z\overline z \cr & \Rightarrow \,\,w - \overline w = \left( {w - \overline w } \right){\left| z \right|^2} \cr & \Rightarrow \,\,{\left| z \right|^2} = 1\,\,\,\,\,\,\left( {\because \,\,w = \alpha + i\beta \,\,{\text{and }}\beta \ne {\text{0}}} \right) \cr & \Rightarrow \,\,\left| z \right| = 1\,\,{\text{also given }}z \ne 1 \cr} $$
∴ The required set is $$\left\{ {z:\left| z \right| = 1,z \ne 1} \right\} = 3\omega \left( {\omega - 1} \right)$$

$$\eqalign{ & {\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1 \cr & \Rightarrow \,\,{\left[ {\frac{{{{\left( {1 + i} \right)}^2}}}{{1 - {i^2}}}} \right]^x} = 1 \cr & \left( {\frac{{1 + {i^2} + 2i}}{{1 + 1}}} \right) = 1 \cr & \Rightarrow \,\,{\left( i \right)^x} = 1; \cr & \therefore \,\,x = 4n;\,\,\,\,n \in {I^ + } \cr} $$

As given $$\omega = \frac{z}{{z - \frac{1}{3}i}}$$
$$\eqalign{ & \Rightarrow \left| \omega \right| = \frac{{\left| z \right|}}{{\left| {z - \frac{1}{3}i} \right|}} = 1 \cr & \Rightarrow \left| z \right| = \left| {z - \frac{1}{3}i} \right| \cr} $$
⇒ distance of $$z$$ from origin and point
$$\left( {0,\frac{1}{3}} \right)$$  is same hence $$z$$ lies on bisector of the line joining point $$\left( {0,0} \right)$$  and $$\left( {0,\frac{1}{3}} \right).$$
Hence, $$z$$ lies on a straight line.

$$\left| {{z_1}} \right| = 12$$
$$ \Rightarrow \,{z_{1\,}}\,$$ lies on a circle with center $$\left( {0,0} \right)$$  and radius 12 unites, and $$\left| {{z_2} - 3 - 4i} \right| = 5$$
$$ \Rightarrow \,\,{z_2}$$  lies on a circle with center $$\left( {3,4} \right)$$  and radius 5 units.

From fig. it is clear that $$\left| {{z_1} - {z_2}} \right|$$   i.e., distance between $${{z_1}}$$ and $${{z_2}}$$ will be min when they lie at $$A$$ and $$B$$ resp. i.e., $$O, C, B, A$$   are collinear as shown. Then $${{z_1} - {z_2}}$$  $$= AB = OA$$  $$ - OB = 12 - 2(5) = 2.$$     As above is the min, value, we must have $$\left| {{z_1} - {z_2}} \right| \geqslant 2.$$

$$\eqalign{ & {\alpha ^4} = 1 \cr & \therefore \,\,{\alpha ^{4n - 1}} + {\alpha ^{4n - 2}} + {\alpha ^{4n - 3}} = {\alpha ^{ - 1}} + {\alpha ^{ - 2}} + {\alpha ^{ - 3}} \cr & {\alpha ^{4n - 1}} + {\alpha ^{4n - 2}} + {\alpha ^{4n - 3}} = {\alpha ^3} + {\alpha ^2} + \alpha \,\,\left( {\because \,\,{\alpha ^4} = 1} \right) \cr & {\alpha ^{4n - 1}} + {\alpha ^{4n - 2}} + {\alpha ^{4n - 3}} = \frac{{\alpha \left( {1 - {\alpha ^3}} \right)}}{{1 - \alpha }} = \frac{{\alpha - {\alpha ^4}}}{{1 - \alpha }} = \frac{{\alpha - 1}}{{1 - \alpha }} = - 1. \cr} $$

$$\eqalign{ & \left| {z - 4} \right|{\text{ < }}\left| {z - 2} \right| \cr & \Rightarrow \,\,\left| {\left( {x - 4} \right) + iy} \right| < \left| {\left( {x - 2} \right) + iy} \right| \cr & \Rightarrow \,\,{\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2} \cr & \Rightarrow \,\, - 8x + 16 < - 4x + 4 \cr & \Rightarrow \,\,4x - 12 > 0 \cr & \Rightarrow \,\,x > 3 \cr & \Rightarrow \,\,{\text{Re}}\left( z \right) > 3 \cr} $$

The required angle $$ = {\text{amp}}\frac{1}{{{{\left( {\sqrt 3 - i} \right)}^{25}}}}.$$
$$\eqalign{ & {\text{Now}},{\left( {\sqrt 3 - i} \right)^{25}} = {\left( {\frac{{ - 1 - \sqrt {3}i }}{2} \cdot \frac{2}{{ - i}}} \right)^{25}} = {\left( {\frac{{2{\omega ^2}}}{{ - i}}} \right)^{25}} = {2^{25}} \cdot \frac{{{\omega ^2}}}{{ - i}} \cr & {\left( {\sqrt 3 - i} \right)^{25}} = {2^{25}} \cdot \left( {{\omega ^2} \cdot i} \right) \cr} $$
∴ the required angle
$$\eqalign{ & = {\text{amp}}\frac{1}{{{2^{25}} \cdot {\omega ^2} \cdot i}} = {\text{amp}}\left( {\frac{{ - \omega i}}{{{2^{25}}}}} \right) \cr & = {\text{amp}}\frac{{ - i \cdot \frac{{ - 1 + \sqrt {3}i }}{2}}}{{{2^{25}}}} \cr & = {\text{amp}}\frac{{\sqrt 3 + i}}{{{2^{26}}}} = {\tan ^{ - 1}}\frac{1}{{\sqrt 3 }} = \frac{\pi }{6}. \cr} $$

$$\eqalign{ & \left| {{z_1}} \right| = \left| {{z_2}} \right|\, = \left| {{z_3}} \right| = 1\left( {{\text{given}}} \right) \cr & {\text{Now, }}\left| {{z_1}} \right| = 1 \cr & \Rightarrow \,\,{\left| {{z_1}} \right|^2} = 1 \cr & \Rightarrow \,\,{z_1}\overline {{z_1}} = 1 \cr & {\text{Similarly }}{z_2}\overline {{z_2}} = 1,\,\,{z_3}\overline {{z_3}} = 1 \cr & {\text{Now,}}\,\,\,\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right| = 1 \cr & \Rightarrow \,\,\left| {\overline {{z_1}} + \overline {{z_2}} + \overline {{z_3}} } \right| = 1 \cr} $$
$$ \Rightarrow \,\,\left| {\overline {{z_1} + {z_2} + {z_3}} } \right| = 1$$       NOTE THIS STEP
$$ \Rightarrow \,\,\left| {{z_1} + {z_2} + {z_3}} \right| = 1$$

A quadratic equation $$p{x^2} + qx + r = 0$$    can have a pair of conjugate complex roots if all coefficients are real and $$D < 0.$$
Here, $$p = 1,r = 5.\,{\text{So, }}i\left( {a - 1} \right)$$     must be real. Hence, $$a = 1$$  and then $$D < 0$$  also.

$$\eqalign{ & {x^4} = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + i\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \cos \frac{{5\pi }}{{12}} + i\sin \frac{{5\pi }}{{12}} \cr & {\text{So, }}x = \cos \left\{ {\frac{{k\pi }}{2} + \frac{{5\pi }}{{48}}} \right\} + i\sin \left\{ {\frac{{k\pi }}{2} + \frac{{5\pi }}{{48}}} \right\} \cr & k = 0,1,2,3 \cr & \therefore {\text{Roots are}} \cr & \cos \frac{{5\pi }}{{48}} + i\sin \frac{{5\pi }}{{48}}{\text{ for }}k = 0 \cr & \cos \frac{{29\pi }}{{48}} + i\sin \frac{{29\pi }}{{48}}{\text{ for }}k = 1 \cr & \cos \frac{{53\pi }}{{48}} + i\sin \frac{{53\pi }}{{48}} = - \left( {\cos \frac{{5\pi }}{{48}} + i\sin \frac{{5\pi }}{{48}}} \right){\text{ for }}k = 2 \cr & \cos \frac{{77\pi }}{{48}} + i\sin \frac{{77\pi }}{{48}} = - \left( {\cos \frac{{29\pi }}{{48}} + i\sin \frac{{29\pi }}{{48}}} \right){\text{ for }}k = 3 \cr} $$