Question

What is the trigonometric form of `(2-4i)*(3-2i)`?

Answer 1

See the explanation below.

Explanation:

First, expand the expression.

`(2-4i) * (3-2i)`
= `6 - 4i - 12i +8i^2`
= `6 - 4i - 12i -8`
= `-2 - 16i`

To convert this to trigonometric form, you need to know the values of `r` and `theta`.

You can use the following equations:
`r^2 = x^2+y^2` and `tan theta = (y)/(x)`

`r^2 = x^2+y^2`
`r = sqrt(x^2+y^2)`
`r = sqrt((-2)^2+(-16)^2)`
`r = 2sqrt65`

`tan theta = (y)/(x)`
`tan theta = (-16)/(-2)`
`tan theta = 8`
`theta = tan^-1(8)`
`theta ~~ 1.45`

So, the answer is `2sqrt65` `cis` `1.45` or `2sqrt65 (cos 1.45 + i sin 1.45)`.