Complex Number MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\text{Let }}z = {\left( 1 \right)^{\frac{1}{n}}} = {\left( {\cos 2k\pi + i\sin 2k\pi } \right)^{\frac{1}{n}}} \cr & \,\,z = \cos \frac{{2k\pi }}{n} + i\sin \frac{{2k\pi }}{n},k = 0,1,2,.....,n - 1. \cr & Let\,\,\,{z_1} = \cos \left( {\frac{{2{k_1}\pi }}{n}} \right) + i\sin \left( {\frac{{2{k_1}\pi }}{n}} \right) \cr & {\text{and }}{z_2} = \cos \left( {\frac{{2{k_2}\pi }}{n}} \right) + i\sin \frac{{2{k_2}\pi }}{n} \cr} $$
be the two values of $$z.$$ s.t. they subtend $$\angle $$ of 90° at origin.
$$\eqalign{ & \therefore \,\,\frac{{2{k_1}\pi }}{n} - \frac{{2{k_2}\pi }}{n} = \pm \frac{\pi }{2} \cr & \Rightarrow \,\,4\left( {{k_1} - {k_2}} \right) = \pm n \cr} $$
As $${k_1}$$ and $${k_2}$$ are integers and $${k_1} \ne {k_2}.$$
$$\therefore \,\,n = 4k,k \in {\text{I}}$$

Given $$z = x + iy$$   where $$x$$ and $$y$$ are integer
Also $$z{\overline z ^3} + \overline z {z^3} = 350$$
$$\eqalign{ & \Rightarrow \,\,{\left| z \right|^2}\left( {{{\overline z }^2} + {z^2}} \right) = 350 \cr & \Rightarrow \,\,\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 175 \cr & \Rightarrow \,\,\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 25 \times 7\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{or}}\,\,\,\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 35 \times 5\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & \because \,\,\,x\,{\text{and}}\,y\,{\text{are}}\,\operatorname{integers} \cr & \therefore \,\,\,{x^2} + {y^2} = 25\,\,\,{\text{and}}\,{x^2} - {y^2} = 7\,\,\,\,\,\,\,\,\,\left[ {{\text{From}}\,{\text{eq}}\,\left( {\text{i}} \right)} \right] \cr & \Rightarrow \,\,{x^2} = 16\,{\text{and}}\,{y^2} = 9 \cr & \Rightarrow \,\,x = \pm 4\,{\text{and}}\,y = \pm 3 \cr & \therefore \,\,{\text{Vertices}}\,{\text{of}}\,{\text{rectangle}}\,{\text{are}} \cr & \,\,\left( {4,3} \right),\left( {4, - 3} \right),\left( { - 4, - 3} \right),\left( { - 4,3} \right). \cr & {\text{So, area of rectangle}} = 8 \times 6 = 48\,{\text{sq}}{\text{. units}} \cr & {\text{Now from eq}}{\text{.}}\left( {{\text{ii}}} \right) \cr & {\text{or }}{x^2} + {y^2} = 35\,\,{\text{and }}{x^2} - {y^2} = 5 \cr} $$
$$ \Rightarrow \,\,{x^2} = 20,$$   which is not possible for any integral value of $$x$$

Expression $$ = {\left( {1 + i} \right)^{{n_1}}} + {\left( {1 - i} \right)^{{n_1}}} + {\left( {1 + i} \right)^{{n_2}}} + {\left( {1 - i} \right)^{{n_2}}}$$
$$\eqalign{ & = {2^{\frac{{{n_1}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} + i\frac{1}{{\sqrt 2 }}} \right)^{{n_1}}} + {2^{\frac{{{n_1}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} - i\frac{1}{{\sqrt 2 }}} \right)^{{n_1}}} + {2^{\frac{{{n_2}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} + i\frac{1}{{\sqrt 2 }}} \right)^{{n_2}}} + {2^{\frac{{{n_2}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} - i\frac{1}{{\sqrt 2 }}} \right)^{{n_2}}} \cr & = {2^{\frac{{{n_1}}}{2}}}\left\{ {{{\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)}^{{n_1}}} + {{\left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right)}^{{n_1}}}} \right\} + {2^{\frac{{{n_2}}}{2}}}\left\{ {{{\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)}^{{n_2}}} + {{\left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right)}^{{n_2}}}} \right\} \cr & = {2^{\frac{{{n_1}}}{2}}}\left\{ {\cos \frac{{{n_1}\pi }}{4} + i\sin \frac{{{n_1}\pi }}{4} + \cos \frac{{{n_1}\pi }}{4} - i\sin \frac{{{n_1}\pi }}{4}} \right\} + {2^{\frac{{{n_2}}}{2}}}\left\{ {\cos \frac{{{n_2}\pi }}{4} + i\sin \frac{{{n_2}\pi }}{4} + \cos \frac{{{n_2}\pi }}{4} - i\sin \frac{{{n_2}\pi }}{4}} \right\},\,\,{\text{if }}{n_1},{n_2}\,{\text{are integers}} \cr & = {2^{\frac{{{n_1}}}{2}}} \cdot 2\cos \frac{{{n_1}\pi }}{4} + {2^{\frac{{{n_2}}}{2}}} \cdot 2\cos \frac{{{n_2}\pi }}{4} = {\text{real}}{\text{.}} \cr} $$

If $$z = x + iy,\sqrt {{{\left( {x + 1} \right)}^2} + {y^2}} = x + 2\,{\text{and }}0 = y + 2.$$

$$\eqalign{ & E = 4 + 5{\left( \omega \right)^{334}} + 3{\left( \omega \right)^{365}} = 4 + 5\omega + 3{\omega ^2} \cr & \,\,\,\,\,\, = 1 + 2\omega + 3\left( {1 + \omega + {\omega ^2}} \right) = 1 + \left( { - 1 + i\sqrt 3 } \right) \cr & \,\,\,\,\,\, = \,\,i\sqrt 3 \cr} $$

$$\eqalign{ & z = \cos \theta + i\sin \theta \cr & \Rightarrow \,\,{z^{2m - 1}} = {\left( {\cos \theta + i\sin \theta } \right)^{2m - 1}} \cr} $$
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos \left( {2m - 1} \right)\theta + i\sin \left( {2m - 1} \right)\theta \,\,\,\,\,\,\,\,\left[ \begin{array}{l} {\rm{using\,\, De\,\, Moiver's\,\, theorem}}\\ {\left( {\cos \theta + i\sin \theta } \right)^n} = \cos n\theta + i\sin n\theta \end{array} \right]\]
$$\eqalign{ & \therefore \,\,{\text{Im}}\left( {{z^{2m - 1}}} \right) = \sin \left( {2m - 1} \right)\theta \cr & \therefore \,\,\sum\limits_{m = 1}^{15} {{\text{Im}}\left( {{z^{2m - 1}}} \right)} = \sum\limits_{m = 1}^{15} {\sin \left( {2m - 1} \right)\theta } \cr & = \sin \theta + \sin 3\theta + \sin 5\theta + ..... + {\text{upto 15 terms}} \cr & {\text{ = }}\frac{{\sin \left[ {15\left( {\frac{{2\theta }}{2}} \right)} \right].\sin \left[ {\theta + 14 \times \theta } \right]}}{{\sin \theta }} \cr} $$
\[\left[ \begin{array}{l} {\rm{Using\,\, sin}}\alpha + \sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha + 2\beta } \right) + ..... + n\,{\rm{terms}}\\ {\rm{ = }}\frac{{\sin \left( {\frac{{n\beta }}{2}} \right).\sin \left[ {\alpha + \frac{{\left( {n - 1} \right)\beta }}{2}} \right]}}{{\sin \left( {\frac{\beta }{2}} \right)}} \end{array} \right]\]
$$\eqalign{ & = \frac{{\sin 15\theta .\sin 15\theta }}{{\sin \theta }} \cr & = \frac{{\sin {{30}^ \circ }.\sin {{30}^ \circ }}}{{\sin {2^ \circ }}} \cr & = \frac{1}{{4\sin {2^ \circ }}} \cr} $$

We have, $$\left| {z + 1} \right| = \left| {z + 4 - 3} \right|\,\,\,.....\left( {\text{i}} \right)$$
Now, $$\left| {z + 4 - 3} \right| \leqslant \left| {z + 4} \right| + \left| { - 3} \right| \leqslant 3 + 3 = 6$$
$$\left[ {{\text{Given }}\left| {z + 4} \right| \leqslant 3\,\,\& \,\,\left| { - 3} \right| = 3} \right]\,\,\therefore \left| {z + 1} \right| \leqslant 6$$
Again $$\left| {z + 1} \right| \geqslant 0$$   [modulus is always non-negative]
∴ Least value of $$\left| {z + 1} \right|$$  may be zero, which occurs
when $$z = - 1,$$  For $$z = - 1,\left| {z + 4} \right| = \left| { - 1 + 4} \right| = 3$$
Which satisfies the given condition that $${\left| {z + 4} \right| \geqslant 3}$$
Hence, the least and the greatest values of $$\left| {z + 1} \right|$$  are 0 and 6.

$$\left( {\frac{1}{{i - 1}}} \right) = \frac{1}{{ - i - 1}} = \frac{{ - 1}}{{i + 1}}$$

Suppose $$z = \frac{{3 + 2i\sin \theta }}{{1 - 2i\,{{\sin }}\theta }}$$
Since, $$z$$ is purely imaginary, then $$z + \overline z = 0$$
$$\eqalign{ & \Rightarrow \,\,\frac{{3 + 2i\sin \theta }}{{1 - 2i\,{{\sin }}\theta }} + \frac{{3 - 2i\sin \theta }}{{1 + 2i\,{{\sin }}\theta }} = 0 \cr & \Rightarrow \,\,\frac{{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right) + \left( {3 - 2i\sin \theta } \right)\left( {1 - 2i\sin \theta } \right)}}{{1 + 4\,{{\sin }^2}\theta }} = 0 \cr & \Rightarrow \,\,{\sin ^2}\theta = \frac{3}{4}\sin \cr & \Rightarrow \,\,\sin \theta = \frac{{\sqrt 3 }}{2} \cr & \Rightarrow \,\,\theta = - \frac{\pi }{3},\frac{\pi }{3},\frac{{2\pi }}{3} \cr} $$
Now, the sum of elements in $$A$$
$$ = - \frac{\pi }{3} + \frac{\pi }{3} + \frac{{2\pi }}{3} = \frac{{2\pi }}{3}$$

$$\eqalign{ & {\text{Here, }}\left( {1 + z} \right)\left( {1 + {z^3}} \right) = 0 \cr & {\text{or }}{\left( {1 + z} \right)^2}\left( {{z^2} - z + 1} \right) = 0 \cr & \Rightarrow \,\,z = - 1, - 1,\frac{{1 + \sqrt {3}i }}{2}. \cr} $$
The distinct points corresponding to these are
$$\eqalign{ & z = - 1,\frac{{1 + \sqrt {3}i }}{2},\frac{{1 - \sqrt {3}i }}{2} = - 1, - {\omega ^2}, - \omega \cr & AB = \left| {1 - {\omega ^2}} \right| = \left| {{\omega ^2}\left( {\omega - 1} \right)} \right| \cr & AB = \left| {{\omega ^2}} \right|\left| {\omega - 1} \right| = \left| {\omega - 1} \right| \cr & BC = \left| {{\omega ^2} - \omega } \right| = \left| \omega \right|\left| {\omega - 1} \right| = \left| {\omega - 1} \right| \cr & CA = \left| {\omega - 1} \right| \cr & \therefore \,\,AB = BC = CA. \cr} $$