[SOLVED] Solve z^4 = (-4i)^2 in trig form, and express on complex plane

[Marija]

Hi everyone, can anyone solve this
z^4=(-4i)^2 (complex number equation)
I need to solve it and express on complex plane.
Oh and i need it in trigonometric form!
Sorry for my English im from Serbia.
P.s. i have my exam tomorrow
Wish me luck ))

 

[Marija]

I try to solve it but i am not sure is it right.
I got
Z=2(cos((π+2kπ)/4)+isin((π+2kπ)/4))
Where k is 0,1,2,3
Can anyone just check iz out?

 

[Marija]

And i got this

 

[Romsek]

I try to solve it but i am not sure is it right.
I got
Z=2(cos((π+2kπ)/4)+isin((π+2kπ)/4))
Where k is 0,1,2,3
Can anyone just check iz out?

This is correct.

 

[HallsofIvy]

Whoever gave you this problem clearly expects you to know that \(\displaystyle r (\cos(\theta)+ i\sin(theta))^n= r^n(\cos(n\theta)+ i\sin(n\theta))\) where "n" does not have to be an integer.
Also (-4i)^2= (-4)^2(i)^2= -16.

So you want to find r and \(\displaystyle \theta\) such that \(\displaystyle r^4 (cos(4\theta)+ isin(4\theta))= -16= 16((-1)+ i(0)\). That is, you want to find r (a positive real number) and \(\displaystyle \theta\) (between 0 and \(\displaystyle 2\pi\)) such that \(\displaystyle r^4= 16\), \(\displaystyle cos(4\theta)= -1\), and \(\displaystyle sin(4\theta)= 0\).