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

- x3_drummerchick

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- x3_drummerchick

@zepdrix

- x3_drummerchick

@mathmale

- zepdrix

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

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## More answers

- x3_drummerchick

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)

- zepdrix

No exponential? :)
Aw ok fine fine.

- x3_drummerchick

i havent learned the one you've posted

- x3_drummerchick

sorry :( lol

- zepdrix

\[\large\rm z=-256i\]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.

- x3_drummerchick

okay that makes sense

- zepdrix

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

- x3_drummerchick

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

- zepdrix

So I guess our angle is going to be 3pi/2 or -pi/2.
Can you visualize that?|dw:1450024718250:dw|

- x3_drummerchick

yes, because those are θ when the point is (0,-1)

- x3_drummerchick

im sorry, those are the radians*

- zepdrix

So our complex number,\[\large\rm z^{1/4}=4\left[~~\color{orangered}{0}~+i\color{royalblue}{(-1)}~~\right]^{1/4}\]can be written this way,\[\large\rm z^{1/4}=4\left[~~\color{orangered}{\cos\left(-\frac{\pi}{2}\right)}~+i\color{royalblue}{\sin\left(-\frac{\pi}{2}\right)}~~\right]^{1/4}\]

- x3_drummerchick

okay

- zepdrix

Since we're taking a root, we have to do a little bit of extra work.
This is probably going to be the trickiest of all the steps,
because it's going to make our expression look even crazier.
We have to consider all of the other angles which are `co-terminal` with angle -pi/2.

- zepdrix

So for example we could also say that the angle\[\large\rm \cos\left(-\frac{\pi}{2}+2\pi\right)=0\]If we spin around the circle an entire time, and land in the same spot, then this still holds true, ya?

- x3_drummerchick

+2pi k

- zepdrix

good good.

- x3_drummerchick

yes

- zepdrix

\[\large\rm z^{1/4}=4\left[\cos\left(-\frac{\pi}{2}+2k \pi\right)+i\sin\left(-\frac{\pi}{2}+2k \pi\right)\right]^{1/4}\]Good, so we're adding this to the end of our angle, to allow for rotations.

- zepdrix

Let's get a common denominator before we proceed.\[\large\rm z^{1/4}=4\left[\cos\left(-\frac{\pi}{2}+\frac{4k \pi}{2}\right)+i\sin\left(-\frac{\pi}{2}+\frac{4k \pi}{2}\right)\right]^{1/4}\]

- zepdrix

And NOWWWW, finally!
We're at a point where we can apply De'Moivre's Theorem.

- zepdrix

Uh oh, the silence.
Brain esplode? :(

- x3_drummerchick

sorry, i was taking notes :) i see where this is going now

- x3_drummerchick

now we can do the "n-1" and plug it in- values 3,2,1 and 0?

- zepdrix

I'm not sure what you mean by "n-1",
but yes, we will only consider the first four positive integer values of k,
if we considered the one after that, k=4,
it would end up giving us the same root that k=0 gave us.
So k=0,1,2,3 will give us our 4 unique 4th roots.

- x3_drummerchick

gotcha

- zepdrix

So applying De'Moivre's Theorem, bringing the exponent into the angle,\[\large\rm z^{1/4}=4\left[\cos\left(-\frac{\pi}{8}+\frac{4k \pi}{8}\right)+i\sin\left(-\frac{\pi}{8}+\frac{4k \pi}{8}\right)\right]\]I brought the 1/4 into the angle ^

- zepdrix

Hopefully I'm not stealing too much of the fun here :)
It's just difficult to ask you to do a step when it's this big messy thing lol

- x3_drummerchick

im just taking notes and predicting it as i go along lol, but i see where its headed

- zepdrix

\[\large\rm z_k^{1/4}=4\left[\cos\left(-\frac{\pi}{8}+\frac{4k \pi}{8}\right)+i\sin\left(-\frac{\pi}{8}+\frac{4k \pi}{8}\right)\right],~~k=0,1,2,3\]
And then we just plug in some k values to find our roots, ya?\[\large\rm z_0^{1/4}=4\left[\cos\left(-\frac{\pi}{8}+0\right)+i\sin\left(-\frac{\pi}{8}+0\right)\right]\]Do we want decimal approximations for our final answer?
Otherwise it's going to be pretty messy.

- x3_drummerchick

he wants them in z = r (cosθ + i sinθ)

- x3_drummerchick

he just wanted to make sure we understand the b asics

- x3_drummerchick

this helped alot, thanks!

- zepdrix

oo that's no fun :) lol
fair enough though.
Think you can find the other three roots?

- x3_drummerchick

yes, thank you! :)

- zepdrix

cool, yay team \c:/
np

- x3_drummerchick

:)

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