Question

Using De Moivre's theorem, find the least positive integer $n$ such that ${\left(\frac{2i}{1+i}\right)}^n$ is a positive integer

Answer 1

We have, $\frac{2i}{1+i}=\frac{2i}{1+i}\times \frac{1-i}{1-i}$
$=\frac{2(1+i)}{2}=1+i$

Now, $1+i=r\cos \theta +ir\sin \theta$
$r\cos \theta =1,r\sin \theta =1$

$\therefore r^2({\cos }^2\theta +{\sin }^2\theta )=(1{)}^2+(1{)}^2$
$r^2=2\Rightarrow r=\sqrt{2}$
and

$\tan \theta =\frac{1}{1}$

$\tan \theta =\tan \left(\frac{\pi }{4}\right)$

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

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

$\therefore {\left(\frac{2i}{1+i}\right)}^n={\left[\sqrt{2}(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})\right]}^n$

$=2^{n/2}(\cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4})$
which is a positive integer

If $\frac{n\pi }{4}=0,2\pi ,4\pi ,6\pi ,.....$
$\Rightarrow n=0,8,16,24,.....$
$\Rightarrow$ The least positive integer of $n$ is $8$.