Question

Find the modulus and argument of the complex numbers.
$\frac{5-i}{2-3i}$

Answer 1

Consider the given complex number,$\frac{5-i}{2-3i}$ .

By rationalisation of given number .

$\frac{5-i}{2-3i}=\frac{5-i}{2-3i}\times \frac{2+3i}{2+3i}=\frac{10-13i-3i^2}{4-9i^2}$

$\because i^2=-1$

$\frac{10-13i-3i^2}{4-9i^2}=\frac{10-13i-3(-1)}{4-9(-1)}=\frac{13-13i}{13}=1-i$

$\frac{5-i}{2-3i}=1-i$

Now,

$1-i=r(\cos \theta -i\sin \theta )$ ……(1)

$1=\cos \theta$ …..(2)

$1=\sin \theta$ …..(3)

Taking square and add equation (1) and (2),we get

$r^2{\cos }^2\theta +r^2{\sin }^2\theta =1^2+1^2$

$r^2({\cos }^2\theta +{\sin }^2\theta )=2$

$r^2=2$

$r=\sqrt{2}$

Equation (3) divided by equation (2), we get

$\tan \theta =1$

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

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

The value of $\theta$ and $r$ put in equation (1), we get

$1-i=\sqrt{2}(\cos \theta -i\sin \theta )$

This is the required answer.