How do you convert $9i$ to polar form?

Hint: We will use the polar representation of a complex number to solve this question. The polar representation of complex number $a+ib$ is given as $z=r\left( \cos \theta +i\sin \theta \right)$, where r is the distance between the point from origin and is expressed as $r=\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $\theta $ is the angle which is expressed as $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ .

Complete step by step answer:
We have been given a complex number $9i$.
We have to convert it into the polar form.
Now, we know that the polar form of a complex number $a+ib$ is given as $z=r\left( \cos \theta +i\sin \theta \right)$, where r is the distance between the point from origin and is expressed as $r=\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $\theta $ is the angle which is expressed as $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ .
Now, we have a complex number $9i$, we can write it in standard form as
$\Rightarrow 0+9i$
Now, we get $a=0,b=9$
Now, let us find the value of r, substituting the values in the formula we will get
$\Rightarrow r=\sqrt{{{0}^{2}}+{{9}^{2}}}$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow r=\sqrt{{{9}^{2}}} \\
 & \Rightarrow r=9 \\
\end{align}$
Now, $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$
Substituting the values we will get
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{9}{0} \right)$
Now, simplifying the above obtained expression we will get
$\begin{align}
  & \Rightarrow \theta ={{\tan }^{-1}}\left( \infty \right) \\
 & \Rightarrow \theta =\dfrac{\pi }{2} \\
\end{align}$
Now, the polar form of the given complex number will be
$z=9\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)$

Hence above is the required polar form of $9i$.

Note: In this particular question the point to be noted is that tangent function has undefined or infinite value on $\dfrac{\pi }{2}$ and $\dfrac{5\pi }{2}$. But the given value $9i$ is positive it means $\theta $ must have positive value and lies in the first and second quadrant so we will take the value of $\theta $ as $\dfrac{\pi }{2}$.