Question

polar form of (1+I)^2/ 1-i​

Answer 1

Step-by-step explanation:

We have,

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

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

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

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

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

$= i + {i}^{2}$

$= - 1 + i$

So,

$z = - 1 + i$

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

$z = \sqrt{2} ( \cos( \frac{3\pi}{4} ) + i \sin( \frac{3\pi}{4} ))$