Complex Number MCQ Questions & Answers in Algebra | Maths

For the non-real roots of the equation $$2{z^2} + 2z + \lambda = 0\,\,\,\,\,.....\left( {\text{i}} \right)$$
$$\eqalign{ & {\text{discriminant}} < 0. \cr & {\text{That}}\,{\text{is}}\,\,4 - 8\lambda < 0 \cr & \Rightarrow \,\lambda > \frac{1}{2}\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Let the roots of (i) be $${z_1}\,\,\& \,\,{z_2}$$
$${\text{Then}}\,\,{z_1} + {z_2} = - \frac{2}{2} = - 1,\,{z_1}{z_2} = \frac{\lambda }{2}$$
$$\eqalign{ & {z^2} + z_2^2 - {z_1}{z_2} = 0 \cr & \Rightarrow \,{\left( {{z_1} + {z_2}} \right)^2} = 3{z_1}{z_2} \cr & \Rightarrow \,{\left( { - 1} \right)^2} = 3\frac{\lambda }{2} \cr & \Rightarrow \,\lambda = \frac{2}{3} \cr} $$
$$\lambda = \frac{2}{3}\left( { > \frac{1}{2}} \right)$$   satisfies the condition (ii).
Hence, it is the required result.

$$\eqalign{ & \left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = 1\,\,\left( {{\text{given}}} \right) \cr & {\text{Now, }}\left| {{z_1}} \right| = 1 \cr & \Rightarrow {\left| {{z_1}} \right|^2} = 1 \cr & \Rightarrow {z_1}{{\bar z}_1} = 1 \cr & {\text{Similarly, }}{z_2}{{\bar z}_2} = 1,{z_3}{{\bar z}_3} = 1 \cr & {\text{Now, }}\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right| = 1 \cr & \Rightarrow \left| {{{\bar z}_1} + {{\bar z}_2} + {{\bar z}_3}} \right| = 1 \cr & \Rightarrow \left| {\overline {{z_1} + {z_2} + {z_3}} } \right| = 1 \cr & \Rightarrow \left| {{z_1} + {z_2} + {z_3}} \right| = 1 \cr} $$

$$\eqalign{ & {z_1} + {z_2} = - 1,{z_1}{z_2} = \frac{b}{3} \cr & {0^2} + {z_1}^2 + {z_2}^2 = 0 \times {z_1} + 0 \times {z_2} + {z_1}{z_2} \cr & \Rightarrow {\left( {{z_1} + {z_2}} \right)^2} - 2{z_1}{z_2} = {z_1}{z_2} \cr & \Rightarrow 1 = 3{z_1}{z_2} = 3\frac{b}{3} \cr & \Rightarrow b = 1. \cr} $$

Let $$z_1 , z_2 , z_3$$  and $$z_4$$ the points in complex plane be the vertices of a parallelogram taken in order.

Since, the diagonals of a parallelogram bisect,
hence, the mid points of $$AC$$  and $$BD$$  must coincide
i.e.,
$$\eqalign{ & \frac{{{z_1} + {z_3}}}{2} = \frac{{{z_2} + {z_4}}}{2} \cr & \Rightarrow {z_1} + {z_3} = {z_2} + {z_4} \cr} $$

We know minimum value of $$\left| {{Z_1} + {Z_2}} \right|{\text{ is }}\left| {\left| {{Z_1}} \right| - \left| {{Z_2}} \right|} \right|$$
Thus minimum value of $$\left| {Z + \frac{1}{2}} \right|{\text{ is }}\left| {\left| Z \right| - \frac{1}{2}} \right| \leqslant \left| {Z + \frac{1}{2}} \right| \leqslant \left| Z \right| + \frac{1}{2}$$
$$\eqalign{ & {\text{Since, }}\left| Z \right| \geqslant 2{\text{ therefore 2}} - \frac{1}{2} < \left| {Z + \frac{1}{2}} \right| < 2 + \frac{1}{2} \cr & \Rightarrow \,\,\frac{3}{2} < \left| {Z + \frac{1}{2}} \right| < \frac{5}{2} \cr} $$

$${z_2} = \frac{{2{z_1} + {z_3}}}{{2 + 1}}$$
$$ \Rightarrow {z_2}$$  divides the line segment joining $${z_1},{z_3}$$  in the ratio $$2 : 1$$  internally.
Note : If $${z_1},{z_2},{z_3}$$   are related by $$a{z_1} + b{z_2} + c{z_3} = 0,$$     where $$a + b \pm c = 0,$$    then $${z_1},{z_2},{z_3}$$   will be collinear points.
ALTERNATE SOLUTION
$$\eqalign{ & {\text{Given, }}2{z_1} + {z_3} = 3{z_2} \cr & {z_2} = \frac{{2{z_1} + {z_3}}}{{2 + 1}} \cr} $$
So, $${z_2}$$ divided the line joining the point $${z_1},\,{z_3}$$  are in the ratio $$1 : 2$$
So, $${z_1},{z_2},{z_3}$$   are collinear.

$$\eqalign{ & {\left( {x - 1} \right)^3} + 8 = 0 \cr & \Rightarrow \,\,\left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{\frac{1}{3}}} \cr & \Rightarrow \,\,x - 1 = - 2\,\,{\text{or }} - 2\omega \,\,{\text{or }} - 2{\omega ^2} \cr & {\text{or }}x = - 1\,\,{\text{or 1}} - 2\omega \,\,{\text{or 1}} - 2{\omega ^2}. \cr} $$

Let $$z = a + bi$$

$$\eqalign{ & \Rightarrow \bar z = a - bi \cr & \therefore i\bar z - iz = i\left[ {\left( {a - bi} \right) - \left( {a + bi} \right)} \right] = 5 \cr & \Rightarrow i\left[ { - 2bi} \right] = 5 \cr & \Rightarrow b = \frac{5}{2} \cr} $$
So from figure it is clear that
$$\eqalign{ & x = 1,y = \frac{5}{2} + \frac{3}{2} = 4 \cr & {z_2} = 1 + 4i \cr} $$

$$\eqalign{ & {z^7} = - 1 \cr & \therefore \,\,{\text{expression}} = {\left( {{z^7}} \right)^{12}} \cdot {z^2} + {\left( {{z^7}} \right)^{25}} + {\left( {{z^7}} \right)^{41}} \cdot {z^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( { - 1} \right)^{12}} \cdot {z^2} + {\left( { - 1} \right)^{25}} + {\left( { - 1} \right)^{41}} \cdot {z^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {z^2} - 1 - {z^2} = - 1. \cr} $$

$$\eqalign{ & \ln \left( {\frac{{a - ib}}{{a + ib}}} \right) = \ln \left| {\frac{{a - ib}}{{a + ib}}} \right| + i\left[ {2n\pi + \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right] \cr & = i\left[ {2n\pi + \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right]{\text{ Since, }}\left| {\frac{{a - ib}}{{a + ib}}} \right| = 1 \cr & \therefore {\text{Arg}}\left[ {i\ln \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right] \cr & = {\text{Arg}}\left[ { - 2n\pi - \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)} \right] = 0\,\,{\text{or }}\pi \cr & {\text{As }}2n\pi + \arg \left( {\frac{{a - ib}}{{a + ib}}} \right)\,\,{\text{is a real number}}{\text{.}} \cr} $$