Question

will give medals! - if you an explain how to solve this:
find all fourth roots of -256i (this is my example problem)

Answer 1

@zepdrix

Answer 2

@mathmale

Answer 3

So umm,
are you familiar with Euler's Identity?
You've seen complex numbers in both trig and exponential form, yes?\[\large\rm e^{i \theta}=\cos \theta+i \sin \theta\]

Answer 4

no, but i was given demoires theorum: if z=r( (cosθ + i sinθ), then z^n= r^n( cos n θ + i sin n θ)

Answer 5

No exponential? :)
Aw ok fine fine.

Answer 6

i havent learned the one you've posted

Answer 7

sorry :( lol

Answer 8

\[\large\rm z=-256i\]Taking our 4th root gives us,\[\large\rm z^{1/4}=\left(-256i\right)^{1/4}\]\[\large\rm z^{1/4}=256^{1/4}\left(-i\right)^{1/4}\]\[\large\rm z^{1/4}=4(-i)^{1/4}\]Let's take it one step further, and write our complex number like this:\[\large\rm z^{1/4}=4(0-i)^{1/4}\]Check those steps out a sec :)
Lemme know if confusion sets in and rots your brain.

Answer 9

okay that makes sense

Answer 10

What we have in the brackets is some complex number which we can write in trig form.
\[\large\rm z^{1/4}=4(~~\color{orangered}{0}~+\color{royalblue}{-i}~~)^{1/4}\]\[\large\rm z^{1/4}=4(~~\color{orangered}{\cos \theta}~+\color{royalblue}{i \sin \theta}~~)^{1/4}\]
Can you figure out which angle... when you take the
cosine ... gives you zero
sine ... gives you -1

Answer 11

no no, this makes sense, youre showing me how the cosine and sine are derived

Answer 12

So I guess our angle is going to be 3pi/2 or -pi/2.
Can you visualize that?