Question

How do you multiply `(-2-9i)(-3-4i)` in trigonometric form?

Answer 1

`(-30+35i)`

Explanation:

Any complex equation in the form of `a+bi` i.e. (vector form) can be written as `re^(thetai)` (rectangular form)
Here,
`r rarr` magnitude of the vector, and
`theta rarr` the angle between the vector form and the components

now, `re^(thetai)` is equivalent to `r(costheta +isintheta)`

this tells us,
`a=rcostheta`
and `b=rsintheta`
thus, by solving the 2 above equations, `r=sqrt(a^2+b^2)`
and `theta=tan^-1(b/a)`

So, solving for `(-2-9i)`,
`r_1 = sqrt85`
`theta_1=77.47` degrees

And,solving for `(-3-4i)`,
`r_2 = 5`
`theta_2=53.13` degrees

so, we get,
`(−2−9i)(−3-4i)=sqrt85 e^(77.47i)` x `5e^(53.13i)= 5sqrt85 e^(130.6i)`

`:. (−2−9i)(−3-4i)=5sqrt85(cos130.6+isin130.6)`

`:. (−2−9i)(−3-4i)=5sqrt85(-0.651+0.759i)`

`:. (−2−9i)(−3-4i)=(-30+35i)`