Answer
Ans. $\color{blue}{\dfrac{1+i}{1-i} =e^{i\pi/2}}$
Work Step by Step
Given: $\frac{1+i}{1-i}$
Rationalizing the denominator term
= $\frac{1+i}{1-i}$×$\frac{1+i}{1+i}$
= $\frac{1+2i+i^2}{1^2-i^2}$
= $\frac{2i}{i}$
= $i$
= $0 +i$
= $x+iy$, where $x=0$ and $y=1$
= $re^{i\theta}$, where
$\qquad \vert r\vert = \sqrt{x^2+y^2} = \sqrt{0^2+1^2} = \sqrt{0+1} = \sqrt{1} = 1\implies \color{blue}{r = 1} \ge 0$ and
$\qquad \tan(\theta) = y/x = 1/0 = 0,\theta\in[-\pi,\pi] \implies \color{blue}{\theta = \pi/2}$.
Thus,
$\dfrac{1+i}{1-i}$ = $i$ = $e^{i\pi/2}$
Ans. $\color{blue}{\dfrac{1+i}{1-i} = e^{i\pi/2}}$
