r = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}
\theta = \tan^{-1} \left( \frac{5}{-3} \right) + \pi = \tan^{-1} \left( \frac{-5}{3} \right) + \pi = \tan^{-1} \left( -\frac{5}{3} \right) + \pi
Thus, the trigonometric form is:
\sqrt{34} \left( \cos \left( \tan^{-1} \left( -\frac{5}{3} \right) + \pi \right) + i \sin \left( \tan^{-1} \left( -\frac{5}{3} \right) + \pi \right) \right)