x3_drummerchick
  • x3_drummerchick
will give medals! - if you an explain how to solve this: find all fourth roots of -256i (this is my example problem)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
x3_drummerchick
  • x3_drummerchick
@zepdrix
x3_drummerchick
  • x3_drummerchick
@mathmale
zepdrix
  • 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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x3_drummerchick
  • 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
  • zepdrix
No exponential? :) Aw ok fine fine.
x3_drummerchick
  • x3_drummerchick
i havent learned the one you've posted
x3_drummerchick
  • x3_drummerchick
sorry :( lol
zepdrix
  • 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
  • x3_drummerchick
okay that makes sense
zepdrix
  • 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
  • x3_drummerchick
no no, this makes sense, youre showing me how the cosine and sine are derived
zepdrix
  • zepdrix
So I guess our angle is going to be 3pi/2 or -pi/2. Can you visualize that?|dw:1450024718250:dw|
x3_drummerchick
  • x3_drummerchick
yes, because those are θ when the point is (0,-1)
x3_drummerchick
  • x3_drummerchick
im sorry, those are the radians*
zepdrix
  • 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
  • x3_drummerchick
okay
zepdrix
  • 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
  • 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
  • x3_drummerchick
+2pi k
zepdrix
  • zepdrix
good good.
x3_drummerchick
  • x3_drummerchick
yes
zepdrix
  • 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
  • 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
  • zepdrix
And NOWWWW, finally! We're at a point where we can apply De'Moivre's Theorem.
zepdrix
  • zepdrix
Uh oh, the silence. Brain esplode? :(
x3_drummerchick
  • x3_drummerchick
sorry, i was taking notes :) i see where this is going now
x3_drummerchick
  • x3_drummerchick
now we can do the "n-1" and plug it in- values 3,2,1 and 0?
zepdrix
  • 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
  • x3_drummerchick
gotcha
zepdrix
  • 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
  • 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
  • x3_drummerchick
im just taking notes and predicting it as i go along lol, but i see where its headed
zepdrix
  • 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
  • x3_drummerchick
he wants them in z = r (cosθ + i sinθ)
x3_drummerchick
  • x3_drummerchick
he just wanted to make sure we understand the b asics
x3_drummerchick
  • x3_drummerchick
this helped alot, thanks!
zepdrix
  • zepdrix
oo that's no fun :) lol fair enough though. Think you can find the other three roots?
x3_drummerchick
  • x3_drummerchick
yes, thank you! :)
zepdrix
  • zepdrix
cool, yay team \c:/ np
x3_drummerchick
  • x3_drummerchick
:)

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