# Thread: forgot how to solve nth root

1. ## forgot how to solve nth root

(a)
how do you find the fifth root of -16 ( 2) ^ (1/2) ( 1+i)

i found that the mod of z^5 is 4. and the argument of ( 1+i ) = (Pi /2)

from there, i said that the arg of z is ( Pi / 10 + 2k Pi / 5)

thus the answer for the fifth root of z would be of the form,

z = 4 (cos ( Pi / 10 + 2k Pi / 5) + i sin (Pi / 10 + 2k Pi / 5) )

am i right to say that?

(b)
also is there a fixed method that i can do to compute questions like ( 1-i) ^ ( -3) and ( -1 + i) ^16?

becos now im just doing them via expansion and im wondering if there is a shorter method that i can do to solve it..

thanks!

2. Originally Posted by alexandrabel90
(a)
how do you find the fifth root of -16 ( 2) ^ (1/2) ( 1+i)

What did you write here? Is it $-16\sqrt{2}(1+i)$?

i found that the mod of z^5 is 4. and the argument of ( 1+i ) = (Pi /2)

This is wrong if the number is what I wrote above: if $z=-16\sqrt{2}(1+i)=-16\sqrt{2}-16\sqrt{2}i$ , then $|z|=\sqrt{16^2\cdot 2+16^2\cdot2}=\sqrt{4\cdot 16^2}=2\cdot 16=32$
Also, its argument is $Arg(z)=\arctan\left(\frac{Im(z)}{Re(z)}\right)=\ar ctan 1=\frac{\pi}{4}+k\pi\,,\,\,k\in\mathbb{Z}$

So try now to solve your question with the above info...and check how come you thought it was something else!

Tonio

from there, i said that the arg of z is ( Pi / 10 + 2k Pi / 5)

thus the answer for the fifth root of z would be of the form,

z = 4 (cos ( Pi / 10 + 2k Pi / 5) + i sin (Pi / 10 + 2k Pi / 5) )

am i right to say that?

(b)
also is there a fixed method that i can do to compute questions like ( 1-i) ^ ( -3) and ( -1 + i) ^16?

becos

Oh, and please: it is "because"...common, a little university level's basic writing.

now im just doing them via expansion and im wondering if there is a shorter method that i can do to solve it..

thanks!
.