Question

Argument and modulus of $\frac{1+i}{1-i}$ are respectively
[RPET 1984; MP PET 1987; Karnataka CET 2001]

• 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

$\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{(1+i{)}^2}{2}$
Now 1 + i = r (cos $\theta$ + i sin $\theta$) $\Rightarrow$ r ~cos $\theta$ = 1, r sin $\theta$ = 1
$\Rightarrow r=\sqrt{2},\theta =\pi /4$
$\therefore 1+i=\sqrt{2}(cos\frac{\pi }{4}+isin\frac{\pi }{4})$
$\Rightarrow \frac{1}{2}(1+i{)}^2=\frac{1}{2}.2{(cos\frac{\pi }{2}+isin\frac{\pi }{4})}^2$
By De Moivre's Theorem, $(cos\frac{\pi }{2}+isin\frac{\pi }{2})$
Hence the amplitude is $\frac{\pi }{2}$ and modulus is 1.

Trick: $arg\left(\frac{1+i}{1-i}\right)=arg(1+i)-arg(1-i)$
$={45}^o-(-{45}^o)={90}^o$
$\left|\frac{1+i}{1-i}\right|=\left|\frac{1+i}{1-i}\right|=\frac{\sqrt{2}}{\sqrt{2}}=1.$