Question

Find the modulus and the argument of the complex number $z=\frac{(1+2i)}{(1-3i)}$

Answer 1

We have,

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

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

$=\frac{1(1+3i)+2i(1+3i)}{1²-\left(3i\right)²}$

$=\frac{(1+3i+2i+6i²)}{(1+9)}$

$=\frac{(-5+5i)}{10}$

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

Now, modulus of the complex number is $=|\frac{-1}{2}+\frac{1}{2i}|$

$=\sqrt{(-1/2)²+\left(1/2\right)²}$

$=\sqrt{1/4+1/4}$

$=\frac{1}{\sqrt{2}}$

Now,

$\tan \varphi =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|$

$=\left|\frac{\frac{1}{2}}{\frac{-1}{2}}\right|=1$

$\tan \varphi =\tan \frac{\pi }{4}$

$\varphi =\frac{\pi }{4}$

$\varphi$ lies on $2^{nd}$ quadrant so,

$arg\left(z\right)=\pi -\varphi$

$arg\left(z\right)=\pi -\frac{\pi }{4}$

$arg\left(z\right)=\frac{3\pi }{4}$

Hence, this is the answer.