Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\text{Given }}S = \left\{ {\left( {x,y} \right):y = x + 1\,{\text{and }}0 < x < 2} \right\} \cr & \because \,\,x \ne x + 1\,\,{\text{for any }}x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S \cr & \therefore \,\,S\,\,{\text{is not reflexive}}{\text{.}} \cr} $$
Hence $$S$$ in not an equivalence relation.
$$\eqalign{ & {\text{Also }}T = \left\{ {\left( {x,y} \right):x - y\,\,{\text{is an integer}}} \right\} \cr & \because \,\,x - x = 0\,\,{\text{is an integer }}\forall \,x \in R \cr & \therefore \,\,T\,\,{\text{is reflexive}}{\text{.}} \cr} $$
If $$x - y$$  is an integer then $$y - x$$  is also an integer
∴ $$T$$ is symmetric.
If $$x - y$$  is an integer and $$y - z$$  is an integer then $$(x - y) + (y - z) = x - z$$      is also an integer.
∴ $$T$$ is transitive
Hence $$T$$ is an equivalence relation.

$${\text{Given }}\left| z \right| = 1\,{\text{and }}z \ne \pm 1$$
To find locus of $$\omega = \frac{z}{{1 - {z^2}}}$$
We have $$\omega = \frac{z}{{1 - {z^2}}} = \frac{z}{{z\overline z - {z^2}}}\,\,\,\,\,\,\,\,\left[ {\because \,\left| z \right| = 1\,\, \Rightarrow {{\left| z \right|}^2} = z\overline z = 1} \right]$$
$$\,\,\,\,\, = \frac{1}{{\overline z - z}}$$
= purely imaginary number
∴ $$\omega $$ must lie on $$y$$ - axis.

$$\frac{{1 + i}}{{1 - i}} = \frac{{{{\left( {1 + i} \right)}^2}}}{{\left( {1 - i} \right)\left( {1 + i} \right)}} = \frac{{1 - 1 + 2i}}{2} = i$$
Now $${i^n} = 1$$
⇒ the smallest positive integral value of $$n$$ should be 4.

$$\eqalign{ & {\left| {{x_1}{z_1} - {y_1}{z_2}} \right|^2} + {\left| {{y_1}{z_1} - {x_1}{z_2}} \right|^2} \cr & = {\left| {{x_1}{z_1}} \right|^2} + {\left| {{y_1}{z_2}} \right|^2} - 2\operatorname{Re} \left( {{x_1}{y_1}{z_1}{z_2}} \right) + {\left| {{y_1}{z_1}} \right|^2} + {\left| {{x_1}{z_2}} \right|^2} + 2\operatorname{Re} \left( {{x_1}{y_1}{z_1}{z_2}} \right) \cr & = {x_1}^2{\left| {{z_1}} \right|^2} + {y_1}^2{\left| {{z_2}} \right|^2} + {y_1}^2{\left| {{z_1}} \right|^2} + {x_1}^2{\left| {{z_2}} \right|^2} \cr & = 2\left( {{x_1}^2 + {y_1}^2} \right)\left( {{4^2}} \right) = 32\left( {{x_1}^2 + {y_1}^2} \right) \cr} $$
Alternate Solution
$$\eqalign{ & {\text{Given, }}\left| {{z_1}} \right| = \left| {{z_2}} \right| = 4......\left( {\text{i}} \right) \cr & {\text{Now, }}{\left| {{x_1}{z_1} - {y_1}{z_2}} \right|^2} + {\left| {{y_1}{z_1} - {x_1}{z_2}} \right|^2} \cr & {\text{ = }}{\left| {{x_1}{z_1}} \right|^2} + {\left| {{y_1}{z_2}} \right|^2} - 2{x_1}{z_1}.{y_1}{z_2} + {\left| {{y_1}{z_1}} \right|^2} + {\left| {{x_1}{z_2}} \right|^2} + 2{y_1}{z_1}{x_1}{z_2} \cr & {\text{ = }}{\left| {{x_1}} \right|^2}{\left| {{z_1}} \right|^2} + {\left| {{y_1}} \right|^2}{\left| {{z_2}} \right|^2} + {\left| {{y_1}} \right|^2}{\left| {{z_1}} \right|^2} + {\left| {{x_1}} \right|^2}{\left| {{z_2}} \right|^2} \cr & = x_1^2{\left( 4 \right)^2} + y_1^2{\left( 4 \right)^2} + y_1^2{\left( 4 \right)^2} + x_1^2{\left( 4 \right)^2}\,\,\,\,\,\,\left( {\because \,{{\left| z \right|}^2} = {z^2}} \right) \cr & {\text{ = 2}}{\text{.}}{\left( 4 \right)^2}.x_1^2 + 2.{\left( 4 \right)^2}.y_1^2 \cr & {\text{ = 32}}\left( {x_1^2 + y_1^2} \right) \cr} $$

As $$\alpha $$ lies on the circle $${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$$
$$\therefore \,\,{\left| {\alpha - {z_0}} \right|^2} = {r^2}$$
$$\eqalign{ & \Rightarrow \,\,\left( {\alpha - {z_0}} \right)\left( {\overline a - {{\overline z }_0}} \right) = {r^2} \cr & \Rightarrow \,\,\alpha \overline \alpha - \alpha {\overline z _0} - \overline \alpha {z_0} + {z_0}{\overline z _0} = {r^2} \cr & \Rightarrow \,\,{\left| \alpha \right|^2} + {\left| {{z_0}} \right|^2} - \alpha {\overline z _0} - \overline \alpha {z_0} = {r^2}\,\,\,\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
Also $$\frac{1}{{\overline \alpha }}$$ lies on the circle $${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = 4{r^2}$$
$$\eqalign{ & \therefore \,\,{\left| {\frac{1}{{\overline \alpha }} - {z_0}} \right|^2} = 4{r^2} \cr & \Rightarrow \,\,\left( {\frac{1}{{\overline \alpha }} - {z_0}} \right)\left( {\frac{1}{\alpha } - {{\overline z }_0}} \right) = 4{r^2} \cr & \Rightarrow \,\,\frac{1}{{\alpha \overline \alpha }} - \frac{{{z_0}}}{\alpha } - \frac{{{{\overline z }_0}}}{{\overline \alpha }} + {z_0}{\overline z _0} = 4{r^2} \cr & \Rightarrow \,\,\frac{1}{{{{\left| \alpha \right|}^2}}} - \frac{{{z_0}\overline \alpha }}{{{{\left| \alpha \right|}^2}}} - \frac{{{{\overline z }_0}\alpha }}{{{{\left| \alpha \right|}^2}}} + {\left| {{z_0}} \right|^2} = 4{r^2} \cr & \Rightarrow \,\,1 + {\left| \alpha \right|^2}{\left| {{z_0}} \right|^2} - {z_0}\overline \alpha - {\overline z _0}\alpha = 4{r^2}{\left| \alpha \right|^2}\,\,\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Subtracting $${\text{e}}{{\text{q}}^n}$$ (i) from (i) we get
$$\eqalign{ & 1 - {\left| \alpha \right|^2} + {\left| {{z_0}} \right|^2}\left( {{{\left| \alpha \right|}^2} - 1} \right) = {r^2}\left( {4{{\left| \alpha \right|}^2} - 1} \right) \cr & {\text{or }}\,\,\left( {{{\left| \alpha \right|}^2} - 1} \right)\left( {{{\left| {{z_0}} \right|}^2} - 1} \right) = {r^2}\left( {4{{\left| \alpha \right|}^2} - 1} \right) \cr & {\text{Using }}{\left| {{z_0}} \right|^2} = \frac{{{r^2} + 2}}{2}{\text{ we get}} \cr & \left( {{{\left| \alpha \right|}^2} - 1} \right)\frac{{{r^2}}}{r} = {r^2}\left( {4{{\left| \alpha \right|}^2} - 1} \right) \cr & \Rightarrow \,\,{\left| \alpha \right|^2} - 1 = 8{\left| \alpha \right|^2} - 2 \cr & \Rightarrow \,\,\left| \alpha \right| = \frac{1}{{\sqrt 7 }} \cr} $$

$$\eqalign{ & \frac{{z + 4}}{{2z - i}} = \frac{{\left( {x + 4} \right) + iy}}{{2x + i\left( {2y - 1} \right)}} = \frac{{\left\{ {\left( {x + 4} \right) + iy} \right\}\left\{ {2x - i\left( {2y - 1} \right)} \right\}}}{{\left\{ {2x + i\left( {2y - 1} \right)} \right\}\left\{ {2x - i\left( {2y - 1} \right)} \right\}}} \cr & \frac{{z + 4}}{{2z - i}} = \frac{{2x\left( {x + 4} \right) + y\left( {2y - 1} \right) + i\left\{ {2xy - \left( {x + 4} \right)\left( {2y - 1} \right)} \right\}}}{{4{x^2} + {{\left( {2y - 1} \right)}^2}}} \cr & \operatorname{Re} \left( {\frac{{z + 4}}{{2z - i}}} \right) = \frac{1}{2} \cr & \Rightarrow \,\,\frac{{2x\left( {x + 4} \right) + y\left( {2y - 1} \right)}}{{4{x^2} + {{\left( {2y - 1} \right)}^2}}} = \frac{1}{2} \cr} $$
$$ \Rightarrow {\kern 1pt} \,\,16x + 2y = 1,$$    which represents a straight line.

$$\eqalign{ & \left| {\frac{{{z_1} - 2{z_2}}}{{2 - {z_1}{{\overline z }_2}}}} \right| = 1 \cr & \Rightarrow \,{\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}{{\overline z }_2}} \right|^2} \cr & \Rightarrow \,\left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1} - 2{z_2}} } \right) = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {\overline {2 - {z_1}{{\overline z }_2}} } \right) \cr & \Rightarrow \,\left( {{z_1} - 2{z_2}} \right)\left( {{{\overline z }_1} - 2{{\overline z }_2}} \right) = \left( {2 - {z_1}{{\overline z }_2}} \right)\left( {2 - {{\overline z }_1}{z_2}} \right) \cr & \Rightarrow \left( {{z_1}{{\overline z }_1}} \right) - 2{z_1}{\overline z _2} - 2{\overline z _1}{z_2} + 4{z_2}{\overline z _2} = \,4 - 2{\overline z _1}{z_2} - 2{z_1}{\overline z _2} + {z_1}{\overline z _1}{z_2}{\overline z _2} \cr & \Rightarrow \,{\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} = 4 + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} \cr & \Rightarrow \,{\left| {{z_1}} \right|^2} + 4{\left| {{z_2}} \right|^2} - 4 - {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} = 0 \cr & \left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0 \cr & \because \,\left| {{z_2}} \right| \ne 1 \cr & \therefore \,{\left| {{z_1}} \right|^2} = 4 \cr & \Rightarrow \,\left| {{z_1}} \right| = 2 \cr} $$
⇒ Point $${{z_1}}$$ lies on circle of radius 2.

$$\eqalign{ & {\left( {\frac{{\sqrt 3 + i}}{{\sqrt 3 - i}}} \right)^n} = {\left\{ {\frac{1}{i} \cdot \frac{{ - 1 + \sqrt {3}i }}{{ - 1 - \sqrt {3}i }} \cdot \left( { - i} \right)} \right\}^n} \cr & {\left( {\frac{{\sqrt 3 + i}}{{\sqrt 3 - i}}} \right)^n} = {\left\{ {\frac{{\frac{{ - 1 + \sqrt {3}i }}{2}}}{{\frac{{ - 1 - \sqrt {3}i }}{2}}}} \right\}^n} = {\left( { - \frac{1}{\omega }} \right)^n} = 1,\,\,{\text{when }}n = 0,6.....\,\,. \cr} $$

$$\eqalign{ & {\text{Given, }}\frac{{\left( {\cos x + i\sin x} \right)\left( {\cos y + i\sin y} \right)}}{{\left( {\cot u + i} \right)\left( {1 + i\tan v} \right)}} \cr & = \frac{{\left( {\cos x + i\sin x} \right)\left( {\cos y + i\sin y} \right)}}{{\left( {\cos u + i\sin u} \right)\left( {\cos v + i\sin v} \right)}} \cr & = \sin u\cos v\left[ {\cos \left( {x + y - u - v} \right) + i\sin \left( {x + y - u - v} \right)} \right] \cr} $$

$$z\left( {\overline z + \overline \alpha } \right) + \overline z z + \alpha \overline z = 0$$
or, $$z\overline z + \frac{{\overline \alpha }}{2}z + \frac{\alpha }{2}\overline z = 0$$     which is a circle.