Question

How do you change `10z^12` into polar coordinates where `z = 2 +4i`?

Answer 1

`2+4i`

`theta = arctan(b/a)`

`theta = arctan(2)`

`r = sqrt(2^2+4^2) = sqrt(20) = 2sqrt(5)`

polar form of `2+4i` `=>` `2sqrt(5)(cos(arctan(2))+isin(arctan(2)))`

`(2sqrt(5)(cos(arctan(2))+isin(arctan(2))))^12`

using the moivre formula

`(2sqrt(5))^12(cos(arctan(2))+isin(arctan(2)))^12`

`6,4*10^7(cos(12arctan(2))+isin(12arctan(2))) = z^12`

`6,4*10^8(cos(12arctan(2))+isin(12arctan(2))) = 10z^12`