Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \sum\limits_{k = 33}^{65} {\left( {\sin \frac{{2k\pi }}{8} - i\cos \frac{{2k\pi }}{8}} \right)} \cr & = \left[ {\sin \frac{{33\pi }}{4} + \sin \frac{{34\pi }}{4} + ..... + \sin \frac{{65\pi }}{4}} \right] - i\left[ {\cos \frac{{33\pi }}{4} + \cos \frac{{34\pi }}{4} + .... + \cos \frac{{65\pi }}{4}} \right] \cr & = \sin \frac{\pi }{4} - i\cos \frac{\pi }{4}\sin \alpha + \sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha + 2\beta } \right) + ..... + \sin \left[ {\alpha + \left( {n - 1} \right)\beta } \right] \cr & = \frac{{\sin \left\{ {\alpha + \left( {n - 1} \right)\frac{\beta }{2}} \right\} \cdot \sin \frac{{n\beta }}{2}}}{{\sin \frac{\beta }{2}}} \cr & {\text{and }}\cos \left( \alpha \right) + \cos \left( {\alpha + \beta } \right) + ..... + \cos \left( {\alpha + \left( {n - 1} \right)\beta } \right) \cr & = \frac{{\cos \left\{ {\alpha + \left( {n - 1} \right)\frac{\beta }{2}} \right\} \cdot \sin \left( {\frac{{n\beta }}{2}} \right)}}{{\sin \frac{\beta }{2}}} \cr & = - \left( {\frac{{1 + i}}{{\sqrt 2 }}} \right) = \frac{{1 - i}}{{\sqrt 2 }} \cr} $$

$$\eqalign{ & \left( {1 + i} \right)\left( {x + iy} \right) = \left( {1 - i} \right)\left( {x - iy} \right) \cr & \Rightarrow \,\,\left( {x - y} \right) + i\left( {x + y} \right) = \left( {x - y} \right) - i\left( {x + y} \right) \cr & \therefore \,\,x + y = 0 \cr & \therefore \,\,z = x - ix = x\left( {1 - i} \right). \cr} $$

Given that $$a, b, c$$  are integers not all equal, $$\omega $$ is cube root of unity $${ \ne 1},$$ then $$\left| {a + b\omega + c{\omega ^2}} \right|\,$$
$$\eqalign{ & = \left| {a + b\left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right) + c\left( {\frac{{ - 1 - i\sqrt 3 }}{2}} \right)} \right| \cr & = \left| {\left( {\frac{{2a - b - c}}{2}} \right) + i\left( {\frac{{b\sqrt 3 - c\sqrt 3 }}{2}} \right)} \right| \cr & = \frac{1}{2}\sqrt {{{\left( {2a - b - c} \right)}^2} + 3{{\left( {b - c} \right)}^2}} \cr & = \sqrt {\frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]} \cr} $$
R.H.S. will be min. when $$a = b = c,$$  but we cannot take $$a = b = c$$   as per question.
∴ The min value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1. Thus $$b = c = 0,$$  $$a = 1$$  gives us the minimum value 1.

Operating $${R_1} + {R_2} + {R_3},$$   we get
\[\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\, - 1 - {\omega ^2}\,\,\,\,\,\,\,{\omega ^2}\,\,\\ 1\,\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^4} \end{array} \right|\]
$$\eqalign{ & = 3\left[ { - \omega - 1 - \omega } \right] \cr & = 3\left( {{\omega ^2} - \omega } \right) \cr} $$

$$\eqalign{ & \frac{{z - 2}}{{2z + 3i}} = \frac{{\left( {x - 2} \right) + iy}}{{2x + i\left( {2y + 3} \right)}} \cr & \therefore \,\,{\text{amp}}\frac{{z - 2}}{{2z + 3i}} = {\text{amp}}\left\{ {\left( {x - 2} \right) + iy} \right\} - {\text{amp}}\left\{ {2x + i\left( {2y + 3} \right)} \right\} \cr & {\text{or, }}0 = {\tan ^{ - 1}}\frac{y}{{x - 2}} - {\tan ^{ - 1}}\frac{{2y + 3}}{{2x}} \cr & \Rightarrow \,\,\frac{y}{{x - 2}} = \frac{{2y + 3}}{{2x}} \cr & \Rightarrow \,\,3x - 4y = 6. \cr & {\text{Now, }}{z_0}z = \left( {3 + 4i} \right)\left( {x + iy} \right) = \left( {3x - 4y} \right) + i\left( {4x + 3y} \right) \cr & {z_0}\overline z = \left( {3 + 4i} \right)\left( {x - iy} \right) = \left( {3x + 4y} \right) + i\left( {4x - 3y} \right) \cr & {\overline z _0}z = \left( {3 - 4i} \right)\left( {x + iy} \right) = \left( {3x + 4y} \right) + i\left( { - 4x + 3y} \right) \cr & {\overline z _0}\overline z = \left( {3 - 4i} \right)\left( {x - iy} \right) = \left( {3x - 4y} \right) + i\left( { - 4x - 3y} \right). \cr} $$
Check the correct answer.

$$\eqalign{ & {\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}} = {\left\{ {2\left( {\frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right)} \right\}^{\frac{n}{2}}} \cr & {\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}} = {2^{\frac{n}{2}}}{\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right)^{\frac{n}{2}}} \cr & {\left( {1 + \sqrt {3}i } \right)^{\frac{n}{2}}} = {2^{\frac{n}{2}}}{\left( {\cos \frac{{n\pi }}{3} + i\sin \frac{{n\pi }}{3}} \right)^{\frac{1}{2}}}. \cr} $$
Clearly, the least positive integral value of $$n$$ for which $$\cos\frac{{n\pi }}{3} + i\sin \frac{{n\pi }}{3}$$    is positive real is $$6.$$

$$\eqalign{ & {\text{Given that }}\left| {z + i\omega } \right| = \left| {z - i\overline \omega } \right| \cr & \Rightarrow \,\,\left| {z - \left( { - i\omega } \right)} \right| = \left| {z - \left( { - \overline {i\omega } } \right)} \right| \cr} $$
⇒ $$z$$ lies on perpendicular bisector of the line segment joining $${\left( { - i\omega } \right)}$$  and $${\left( { - \overline {i\omega } } \right)}$$ , which is real axis, $${\left( { - i\omega } \right)}\,$$ and $${\left( { - \overline {i\omega } } \right)}\,$$ being mirror images of each other.
$$\eqalign{ & \therefore \,\,{\text{Im}}\left( z \right) = 0. \cr & {\text{If }}z = x\,\,{\text{then }}\left| z \right| \leqslant 1 \cr & \Rightarrow \,\,{x^2} \leqslant 1 \cr & \Rightarrow \,\, - 1 \leqslant x \leqslant 1 \cr} $$
∴ $$(C)$$ is the correct option.

$$\eqalign{ & {z^{\frac{1}{3}}} = p + iq \cr & \Rightarrow \,\,z = {p^3} + {\left( {iq} \right)^3} + 3p\left( {iq} \right)\left( {p + iq} \right) \cr & \Rightarrow \,\,x - iy = {p^3} - 3p{q^2} + i\left( {3{p^2}q - {q^3}} \right) \cr & \therefore \,\,x = {p^3} - 3p{q^2} \cr & \Rightarrow \,\,\frac{x}{p} = {p^2} - 3{q^2} \cr & y = {q^3} - 3{p^2}q \cr & \Rightarrow \,\,\frac{y}{q} = {q^2} - 3{p^2} \cr & \therefore \,\,\frac{x}{p} + \frac{y}{q} = - 2{p^2}\, - \,2{q^2} \cr & \therefore \,\,\frac{{\left( {\frac{x}{p} + \frac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = - 2 \cr} $$

$$\eqalign{ & {\left( {1 + \omega } \right)^7} = A + B\omega \cr & \Rightarrow \,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because 1 + \omega + {\omega ^2} = 0} \right) \cr & \Rightarrow \,\, - {\omega ^{14}} = A + B\omega \cr & \Rightarrow \,\, - {\omega ^2} = A + B\omega \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because {\omega ^3} = 1} \right) \cr & \Rightarrow \,\,1 + \omega = A + B\omega \cr & \Rightarrow \,\,A = 1,B = 1 \cr} $$

$$\eqalign{ & \left| {z - 2} \right| = \min \left\{ {\left| {z - 1} \right|,\left| {z - 5} \right|} \right\} \cr & {\text{i}}{\text{.e}}{\text{., }}\left| {z - 2} \right| = \left| {z - 1} \right|,{\text{ where }}\left| {z - 1} \right| < \left| {z - 5} \right| \cr & \Rightarrow \operatorname{Re} \left( z \right) = \frac{3}{2}{\text{ which satisfy }}\left| {z - 1} \right| < \left| {z - 5} \right| \cr & {\text{Also, }}\left| {z - 2} \right| = \left| {z - 5} \right|,{\text{ where }}\left| {z - 5} \right| < \left| {z - 1} \right| \cr & \Rightarrow \operatorname{Re} \left( z \right) = \frac{7}{2}{\text{ which satisfy }}\left| {z - 5} \right| < \left| {z - 1} \right| \cr} $$