Question

If `z=(a+i b)^5+(b+i a)^5` , then prove that `R e(z)=I m(z),w h e r ea ,b in Rdot`

Answer 1

Let a = r cos `theta`, b = r `sin theta`
Then `z = (a+ib)^(5) + (b + ia)^(5)`
`r^(5){(cos theta + isin theta)^(5) + (sin theta + cos theta)^(5) + (sin theta + icos theta)^(5)}`
`=r^(5)[(cos 5theta + isin 5theta)+{cos((pi)/(2)-theta)+isin((pi)/(2) -theta)}^(5)]`
`r^(5)[(cos 5theta + isin 5theta)+cos 5((pi)/(2) - theta)+isin5((pi)/(2) - theta)]`
`=r^(5) [(cos 5theta + sin 5theta)(1+i)]`
Clearly Re(z) = Im(z)` = r^(5) (cos 5theta + sin 5theta)`