# How do you multiply  (-1-4i)(1-3i)  in trigonometric form?

Oct 29, 2017

$- 13 - i$

#### Explanation:

This is an interesting questions, but a solution based on the idea of ${e}^{i \theta} = \cos \theta + i \sin \theta$

Consider $- 1 - 4 i$, the modulus is $\sqrt[2]{{1}^{2} + {4}^{2}}$ = $\sqrt[2]{17}$
also $1 - 3 i$, the modulus is $\sqrt[2]{{1}^{3} + {3}^{2}}$ = $\sqrt[2]{10}$

Now consider the arguments, $a r g \left(- 1 - 4 i\right)$ = $\arctan 4 + \pi$ as easy to see by sketching the argand diagram

and the $a r g \left(1 - 3 i\right)$ = $2 \pi - \arctan 3$

hence $\left(- 1 - 4 i\right) \left(1 - 3 i\right)$ = $\sqrt[2]{17} \cdot \sqrt[2]{10} \cdot {e}^{\arctan 4 + \pi} \cdot {e}^{2 \pi - \arctan 3}$

= $\sqrt[2]{170} {e}^{\left(3 \pi + \arctan 4 - \arctan 3\right) i}$

Hence when computed via calculator we get; $- 13 - i$

Yielding an interesting result of $\cos \left(3 \pi + \arctan 4 - \arctan 3\right) = - \frac{13}{\sqrt[2]{170}}$

What is linked to $\cos \left(\arctan \left(x\right)\right) = {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}$