Question

How do you multiply `(3+5i)(6-6i)` in trigonometric form?

Answer 1

`6(8+2i)`

Explanation:

Given that

`(3+5i)(6-6i)`

`=\sqrt34(\cos(\tan^{-1}(5/3))+i\sin(\tan^{-1}(5/3)))\cdot 6\sqrt2(\cos(-\pi/4)+i\sin(-\pi/4))`

`=12\sqrt17(\cos(\tan^{-1}(5/3)-\pi/4)+i\sin(\tan^{-1}(5/3)-\pi/4))`

`=12\sqrt17(\cos(\tan^{-1}(5/3))\cos(\pi/4)+\sin(\tan^{-1}(5/3))\sin(\pi/4)+i(\sin(\tan^{-1}(5/3))\cos(\pi/4)-\cos(\tan^{-1}(5/3))\sin(\pi/4)))`

`=12\sqrt17(\frac{3}{\sqrt34}1/\sqrt2+5/\sqrt34 1/\sqrt2+i(\frac{5}{\sqrt34}1/\sqrt2-3/\sqrt34 1/\sqrt2))`

`={12\sqrt17}/{2\sqrt17}(8+2i)`

`=6(8+2i)`