Convert each of the complex numbers in the polar form: (3) 1

Convert each of the complex numbers in the polar form:

Q : 3 $1-i$

Answers (1)

Given problem is
$z=1-i$
Now, let
$r\cos \theta = 1 \ \ \ and \ \ \ r\sin \theta = -1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (1)^2+(-1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$     $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = 1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1$
$\cos \theta = \frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}$
Since values of $\sin \theta$ is negative and value $\cos \theta$ is positive and we know that this is the case in the IV quadrant
Therefore,
$\theta = -\frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ IV \ quadrant)$
Therefore,
$1-i= r\cos \theta +ir\sin \theta$
   $= \sqrt2\cos \left ( -\frac{\pi}{4} \right ) +i\sqrt2\sin \left ( -\frac{\pi}{4} \right )$
   $= \sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )$

Therefore, the required polar form is $\sqrt2\left ( \cos \left ( -\frac{\pi}{4} \right ) +i\sin \left ( -\frac{\pi}{4} \right ) \right )$