Question

Argument and modulus of ${\frac{1+i}{1-i}}^{2013}$ are respectively

• A. $\frac{-\pi }{2}$ and $1$
• B. $\frac{\pi }{2}$ and $\sqrt{2}$
• C. $0$ and $\sqrt{2}$
• D. $\frac{\pi }{2}$ and $1$

Answer 1

The correct option is D $\frac{\pi }{2}$ and $1$

$\frac{1+i^{2013}}{1-i}$

$\Rightarrow \frac{1+\left(i^{2012}\right).i}{1-i}[\because (i^4{)}^{503}=1^{503}=1]$

$\Rightarrow \frac{1+i}{1-i}\frac{(1+i)}{(1+i)}$

$=\frac{(1+i{)}^2}{1^2-i^2}$

$=\frac{1+i^2-2i}{2}=-i$

$2=0+i$

$z=x+iy$

Modulus of $z=\sqrt{x^2+y^2}(\because x=0,y=1)$

$z=\sqrt{0+1}$

$z=1$

$z=0+i=r\cos \theta +i\lambda \sin \theta$

$\cos \theta =0,\sin \theta =1(\because r=1)$

Thus, $\theta$ lies in $1^{st}$ Quadrant

$Argument={90}^o$

$=90\times \frac{\pi }{180}$

$=\frac{\pi }{2}$

$\therefore$ Argument & Modulus $\Rightarrow \frac{\pi }{2}\propto 1$

Option $D$ is correct