I'm having a hard time rewriting the following complex number to the exponential and polar form .
$$8ie^{i\frac{\pi}{6}}$$
I need almost the same form but without the imaginary number near to 8 .
I'm really stuck here, I tried to square both sides but I didn't get anywhere .
I'm not looking for direct answer instead I'm interested in a beginner friendly approach and explanation .
We know that $$e^{i\theta} = \cos \theta + i\sin \theta \tag {1}$$ Thus, $$ie^{i\theta} = i (\cos \theta + i\sin \theta) =i\cos \theta- \sin \theta \tag {2} $$ as $i^2=-1$.
You can also use the fact that $i = e^{i\frac {\pi}{2}}$, giving us, $ie^{i\frac {\pi}{6}} = e^{i\frac {\pi}{2}} e^{i\frac {\pi}{6}} = e^{i\frac {\pi}{2}+\frac {\pi}{6}} = e^{i\frac {2\pi}{3}} $ and then use $(1)$.
Hope you can take it from here.
$ie^{i {π} /6}=$ $i(\cos \frac{π}{6}+i\sin \frac{π}{6})$ =i$\cos \frac{π}{6}-\sin \frac{π}{6}$ and $\cos(\frac{π}{2} -x)=\sin x$