Hello everyone,

Need a bit of help.

(-2-2i)^5

I want to turn this into Cartesian form using the modulus.

Into the form
z^n=r^n(cos n0 +i sin n0)

So far I have r=2sqrt(2)

The angle ive worked out as pi/4 or 45 degrees (im not 100% confident here).

Now im a bit unsure how I put it together.

Thanks for any help!

3 answers

yes r^2 = 4 + 4 = 8
so
r = sqrt 8 = 2 sqrt 2

- 2 - 2 i is in quadrant 3, 45 degrees below -x axis

Theta = pi + pi/4 = 5 pi/4
or 180 + 45 = 225 degress
so

x = 8^.5 e^(5 pi i/4)

now raise to power five
raise (8^.5)^5 = 8^2.5
multiply angle by 5
25 pi/4 = 24 pi/4 + pi/4
which is
6 pi + pi/4
which is just pi/4 because every 2 pi is all the way around
so
8^2.5 (cos 45 + i sin 45

8^2.5 (sqrt2 /2 + i sqrt 2/2)
2^7.5 (2^.5 + i 2^.5)/2
(2^8 + i 2^8)/2
2^7 + i 2^7
or
128 + 128 i

Well, that was entertaining :)

Thanks Damon, really helped.