Find the modulus and argument of the complex number 1+3i/1+2i

Q : 13 Find the modulus and argument of the complex number $\small \frac{1+2i}{1-3i}$ .

Answers (1)

Let
$z = \small \frac{1+2i}{1-3i}$
Now, multiply the numerator and denominator by $(1+3i)$

$\Rightarrow z=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}= \frac{1+3i+2i+6i^2}{1^2-(3i)^2}= \frac{1+5i-6}{1-9i^2}= \frac{-5+5i}{10}$ $= -\frac{1}{2}+i\frac{1}{2}$
Therefore,
$r\cos \theta= -\frac{1}{2} \ \ and \ \ r\sin \theta =\frac{1}{2}$
Square and add both the sides
$r^2\cos^2\theta +r^2\sin^2\theta= \left ( -\frac{1}{2} \right )^2+\left ( \frac{1}{2} \right )^2$
$r^2(\cos^2\theta +\sin^2\theta)= \left ( \frac{1}{4} \right )+\left ( \frac{1}{4} \right )$
$r^2= \frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \sin^2\theta +\cos^2\theta = 1)$
$r = \frac{1}{\sqrt2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Therefore, the modulus is $\frac{1}{\sqrt2}$
Now,
$\frac{1}{\sqrt2} \cos\theta = -\frac{1}{2} \ \ and \ \ \frac{1}{\sqrt2} \sin\theta = \frac{1}{2}$
$\cos\theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin\theta = \frac{1}{\sqrt2}$
Since the value of $\cos\theta$ is negative and the value of $\sin\theta$ is positive and we know that it is the case in II quadrant
Therefore,
Argument $=\left ( \pi-\frac{\pi}{4} \right )= \frac{3\pi}{4}$

Therefore, Argument and modulus are $\frac{3\pi}{4} \ \ and \ \ \frac{1}{\sqrt2}$ respectively