Question

find the modulus and argument of the complex numbers.
$\frac{2+14i}{{\left(2-i\right)}^2}$

Answer 1

Consider the given complex expression ,

$\frac{2+14i}{{(2-i)}^2}$

$\frac{2+14i}{{(2-i)}^2}=\frac{2+14i}{(4+i^2-2i)}$

$\because i^2=-1$

$=\frac{2+14i}{4-1-4i}$

$=\frac{2+14i}{3-4i}$

Rationalize

$\frac{2+14i}{3-4i}\times \frac{3+4i}{3+4i}$

Now ,$\frac{2+14i}{3-4i}\times \frac{3+4i}{3+4i}=\frac{6+50i+56i^2}{{\left(3\right)}^2-{\left(4i\right)}^2}$

$\frac{6+50i+56i^2}{{\left(3\right)}^2-{\left(4i\right)}^2}=\frac{6+50i+56i^2}{3^2-16i^2}$

$\frac{6+50i+56(-1)}{9-16(-1)}=\frac{6+50i-56}{9+16}$

$\frac{50i-50}{25}=i-1=-1+i$

Now,

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

$-1=r\cos \theta$ ….(2)

$1=r\sin \theta$ ……(3)

Now taking square and add equations (2) and (3), we get

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

$2=r^2\left(1\right)$

$r=\sqrt{2}$

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

$\frac{\sin \theta }{\cos \theta }=-1$

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

$\tan \theta =\tan (\pi -\frac{\pi }{4})=\tan \frac{3\pi }{4}$

$\tan \theta =\tan \frac{3\pi }{4}$

$\theta =\frac{3\pi }{4}$

Put the value of $r$ and $\theta$ in equation (1), we get

$-1+i=\sqrt{2}\left(\cos \frac{3\pi }{4}i\sin \frac{3\pi }{4}\right)$

Hence, this is the required question.