Answer
$$z=\sqrt{13} \left ( \cos 2.16 +i \sin 2.16 \right )$$
Work Step by Step
We plot the complex number $z=-2+3i$ the same way we plot $(-2,3)$ in the rectangular coordinate system.
Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi $.
So, for the complex number $z=-2+3i$, we have$$r=\sqrt{(-2)^2+3^2}=\sqrt{13}, \\ \tan \theta =\frac{3}{-2}=-\frac{3}{2}.$$Since $\cos \theta =\frac{a}{r} = \frac{-2}{\sqrt{13}} \lt 0$, the argument, $\theta$, must lie in quadrant $II$. So$$\theta =\tan^{-1} (-\frac{3}{2}) \approx 2.16$$Thus, the complex number has the polar form$$z=\sqrt{13} \left ( \cos 2.16 +i \sin 2.16 \right ).$$
