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

1 Answer
Oct 29, 2017

-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)