Question

Find the modulus and the principal value of the argument of the following numbers:
(a) $1-i$
(b) $-1-\sqrt{3i}$
(c) $1+\sqrt{2}+i$

Answer 1

(a) Let $1=r\cos \theta$ $-1=r\sin \theta$
Squaring and adding these relations, we get
$r^2({\cos }^2\theta +{\sin }^2\theta )=1^2+{(-1)}^2=2$
or $r^2=2$ i.e. $r=\sqrt{2}$
Then $\cos \theta =1╱\sqrt{2}$, $\sin \theta =-1╱\sqrt{2}$,
The value of $\theta$ between $-\pi$ and $\pi$ which satisfies these equations is $-\left(\pi ╱4\right),$
Thus $|1-i|=r\sqrt{2}$ and $\arg (1-i)=-\left(\pi ╱4\right)$.
(b) Ans. $|-1-\sqrt{3i}|=2,$
$(-1-\sqrt{3i})=-\left(2\pi ╱3\right)$
$\because \arg (1+\sqrt{3i})=\pi ╱3$
$\because \arg (1+\sqrt{3i})=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
or $\arg (1+\sqrt{3i})=\frac{2\pi }{3}$
(c) Let $r\cos \theta =1+\sqrt{2}$, $r\sin \theta =1$.
Then $r^2={(1+\sqrt{2})}^2+1=4+2\sqrt{2}$,
$\therefore r=\sqrt{4+2\sqrt{\left(2\right)}}$.
On dividing , $\tan \theta =\frac{1}{\sqrt{\left(2\right)+1}}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$
Hence $\theta =\pi ╱8$ (Standard result) as proved below.
$\alpha =\frac{\pi }{4}\therefore \frac{\alpha }{2}=\frac{\pi }{8}\therefore \tan \frac{\pi }{4}=\frac{2t}{1-t^2}=1$
when $t=\tan \left(\alpha ╱2\right)=\tan \left(\pi ╱8\right)$ or $1-t^2=2t$$1-t^2+2t+1=1+1$
or ${(t+1)}^2={\left(\sqrt{2}\right)}^2$
$\therefore =t=\pm \sqrt{2}-1=\sqrt{2}-1$
The other value $-\sqrt{2}-1$ is being rejected as $\tan \pi ╱8$ is + ive.
Hence $|1+\sqrt{2}+i|=\sqrt{4+2\sqrt{\left(2\right)}}$ and arg $(1+\sqrt{2}+i)=\pi ╱8$