Question

Argument and modules of $[\frac{1+i}{1-i}{]}^{2\pi i}$ 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 A $\frac{-\pi }{2}$ and $1$

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

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

Rationalizing the above, we get

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

$z=0+i=1$

here x =0 ,y = 1

modules of $2=\sqrt{x^2+y^2}=\sqrt{0+1^2}=1$

modules of z = 1

$z=i,$ As per the question, $(i{)}^{2\pi i}$

& $z=r(cos\theta +sin\theta )$

$\Rightarrow \left(r\right(cos\theta +1sin\theta ){)}^{2\pi i}$

$=\left[r\right(cos\left(2\pi i\right))+\theta +isin(2\pi i\left)\theta \right]$

$(i{)}^{2\pi i}=(r{)}^{2\pi i}$

$\Rightarrow r=i$

$r^{2}co{s}^2\theta =0;r^{2}si{n}^2\theta =1$

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

$0+1=r^{2}co{s}^2\theta +rsi{n}^2\theta$

$1=r^2(co{s}^2\theta +si{n}^2\theta )$

$1=r^2\Rightarrow r=1$

argument of$z=\frac{\pi }{2}$

Option (D) is correct.