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