Question

How do you multiply `(9-i)(7-3i)` in trigonometric form?

Answer 1

`2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))`, where `alpha=tan^(-1)(-1/9)` and `beta=tan^(-1)(-3/7)`

Explanation:

A complex number of form `a+bi` can be written as `r(cos theta +i sin theta)`, where `r=sqrt(a^2+b^2)` and `theta=tan^(-1)(b/a)`.

Using this `(9-i)=sqrt82(cos alpha+i sin alpha)`, and `alpha=tan^(-1)(-1/9)`

Similarly, `(7-3i)=sqrt58(cos beta+i sin beta)`, and `beta=tan^(-1)(-3/7)`

Multiplication of `(sqrt82(cos alpha+i sin alpha)` and `sqrt58(cos beta+i sin beta)` is given by

`2sqrt(1189)(cos (alpha+beta)+isin(alpha+beta))`, where `alpha=tan^(-1)(-1/9)` and `beta=tan^(-1)(-3/7)`, as `sqrt(82*58)=2sqrt(1189)`.