Write 2i in polar form.

Let, $z=2 i$

Let $0=r \cos \theta$ and $2=r \sin \theta$

By squaring and adding, we get

$(0)^{2}+(2)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$

$\Rightarrow 0+4=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$\Rightarrow 4=r^{2}$

$\Rightarrow r=2$

$\therefore \cos \theta=0$ and $\sin \theta=1$

Since, θ lies in first quadrant, we have

$\theta=\frac{\pi}{2}$

Thus, the required polar form is $2\left(\cos \left(\frac{\pi}{2}\right)+i \sin \left(\frac{\pi}{2}\right)\right)$