Convert the following in the polar form: (ii) (1+3i)/ (1-2i)

Q: 5 Convert the following in the polar form:

(ii) $\small \frac{1+3i}{1-2i}$

Answers (1)

Let
$z =\frac{1+3i}{1-2i}$

Now, multiply the numerator and denominator by $1+2i$
$\Rightarrow z =\frac{1+3i}{1-2i} \times \frac{1+2i}{i+2i}= \frac{1+2i+3i-6}{1+4}= \frac{-5+5i}{5}=-1+i$
Now,
let
$r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1$
On squaring both and then add
$r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2$
$r^2=2$
$r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Now,
$\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}$
Since the value of $\cos \theta$ is negative and   $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
$z = r\cos \theta + ir\sin \theta$
   $=\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}$
   $=\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$
Therefore, the required polar form is