Question

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

Answer 1

`-13-i`

Explanation:

This is an interesting questions, but a solution based on the idea of `e^(itheta) = costheta + isintheta`

Consider `-1-4i`, the modulus is `root2 ( 1^2 + 4^2 )` = `root2 17`
also `1-3i`, the modulus is `root2 ( 1^3 + 3^2 )` = `root2 10`

Now consider the arguments, `arg(-1-4i)` = `arctan4 + pi` as easy to see by sketching the argand diagram

and the `arg(1-3i)` = `2pi - arctan3`

hence `(-1-4i)(1-3i)` = `root2 17 * root 2 10 * e^(arctan4 + pi) * e^(2pi - arctan3 )`

= `root2 170 e^((3pi + arctan4 - arctan3)i)`

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

Yielding an interesting result of `cos(3pi + arctan4 - arctan3) = -13/root2 170`

What is linked to `cos(arctan(x)) = (1+x^2)^(-1/2)`