Question

Fill in the blanks.
(i) $\frac{\left(1-i\right)}{1-i^{3}}$ equal to P.
(ii) If $\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}$ , then the locus of $z$ is Q.
(iii) The sum of the series $i +i^{2} + i^{3} + ...$ upto $1000$ terms is R .
(iv) If $|z|=4$ and arg $(z)=\frac{5\pi}{6}$, then z = S.

	$P$	$Q$	$R$	$S$
(a)	$2$ $\,$	Parabola $\,$	$-1$ $\,$	$2+2\sqrt{3}\,i$ $\,$
(b)	$-2$ $\,$	Circle $\,$	$0$ $\,$	$-2\sqrt{3}+2i$ $\,$
(c)	$2$ $\,$	Circle $\,$	$1$ $\,$	$2-\sqrt{3}\,i$ $\,$
(d)	$-2$ $\,$	Parabola $\,$	$2$ $\,$	$2-2\sqrt{3}\,i$ $\,$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

$\left(i\right)$ $\frac{\left(1-i\right)^{3}}{1-i^{3}}=\frac{\left(1-i\right)^{3}}{\left(1-i\right)\left(1+i+i^{2}\right)}$
$=\frac{\left(1-i\right)^{2}}{i}=\frac{1+i^{2}-2i}{i}=-2 \, \left(\because\, i^{2}=-1\right)$
$\left(ii\right)$ We have, $\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}$
$\Rightarrow \frac{\left|x+iy-2\right|}{\left|x+iy+2\right|}=\frac{\pi}{6}$
$\Rightarrow \, 6\sqrt{\left(x-2\right)^{2}+y^{2}}=\pi\sqrt{\left(x+2\right)^{2}+y^{2}}$
$\Rightarrow \, 36\left[x^{2}+4-4x+y^{2}\right]=\pi^{2}\left[x^{2}+4x+4+y^{2}\right]$
$\Rightarrow \, \left(36-\pi^{2}\right)x^{2}+\left(36-\pi^{2}\right)y^{2}-\left(144+4\pi^{2}\right)x+144-4\pi^{2}=0$,
which is an equation of a circle.
$\left(iii\right) i+i^{2}+i^{3}+\ldots$ upto $1000$ terms
$=i+i^{2}+i^{3}+i^{4} +\ldots+ i^{1000}=0$
$\left(iv\right)$ We have, $|z| = 4$ and $arg\left(z\right) =\frac{5\pi}{6}$
Let $z = x + iy =r\left(cos\,\theta+isin\,\theta\right)$
$\Rightarrow \, \left|z\right|=r=4$ and $arg \left(z\right)=\theta=\frac{5\pi}{6}$
$\Rightarrow \, z=4\left(cos \frac{5\pi}{6}+i\, sin\frac{5\pi}{6}\right)$
$=4\left[cos\left(\pi-\frac{\pi}{6}\right)+i\, sin\left(\pi-\frac{\pi}{6}\right)\right]$
$=4\left[-cos \frac{\pi}{6}+i\, sin \frac{\pi}{6}\right]=4\left[\frac{-\sqrt{3}}{2}+i\frac{1}{2}\right]$
$=-2 \sqrt{3}+2i$