Find roots of (-16i)^(1/4) using polar form.
Can someone guide me in solving this?
I need to understand this concept.
Thanks.

- anonymous

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- zepdrix

\[\Large\rm -16i=16\color{orangered}{(-i)}=16\color{orangered}{e^{i\frac{3\pi}{2}}}\]\[\large\rm =16\color{orangered}{\left(\cos\frac{3\pi}{2}+i \sin\frac{3\pi}{2}\right)}\]I would start with something like this maybe :o hmm

- zepdrix

Have you learned about the exponential form or no?

- anonymous

I'm still learning it now.
yes that's what I did.
then we have to apply this: z^n=r^n(cos nΘ + isin nΘ), right?

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## More answers

- zepdrix

De'Moivre's Theorem? Yes, good.
But we have to be careful. Since we're taking roots, we'll have to have a 2k pi uhhh somewhere in there.

- anonymous

can you explain that part? I'm still not sure

- anonymous

what to do..

- zepdrix

This is the same as our equation we've established above.\[\large\rm =16\left[\cos\left(\frac{3\pi}{2}+2\pi k\right)+i \sin\left(\frac{3\pi}{2}+2\pi k\right)\right]\]It's easier to explain in exponential form, one sec..

- zepdrix

If I have a point in the complex plane let's say: \(\LARGE\rm z=e^{i\frac{2pi}{5}}\)
The representation of this is that you extend out radially 1 along the real axis, then rotate an angle of 2pi/5.
But we could also get to that exact same point by adding 2pi to the angle, yes?
It's just an extra spin around the circle.\(\LARGE\rm z=e^{i\frac{2pi}{5}+2\pi}\)

- zepdrix

But then you have to realize, we could spin any number of times around the circle to get to that same point, \(\LARGE\rm z=e^{i\frac{2pi}{5}+2k\pi}\)

- anonymous

ok...

- zepdrix

So we can get to -i by rotating that exact amount, or by rotating that amount + any integer multiple of 2pi.

- zepdrix

So maybe this is a step I should have written in between:\[\Large\rm 16\color{orangered}{e^{i\frac{3\pi}{2}}}=16\color{orangered}{e^{i\frac{3\pi}{2}+2k \pi}}\]

- zepdrix

What's going to happen is...
When we apply De'Moivre's thing,
we're going to get some roots.. 4 unique roots since it is a 4th root.
After that, the roots will start to repeat.
So we're only interested in the first 4 values of k, starting from 0.

- zepdrix

\[\large\rm (-16i)^{1/4}=16^{1/4}\left[\cos\left(\frac{3\pi}{2}+2\pi k\right)+i \sin\left(\frac{3\pi}{2}+2\pi k\right)\right]^{1/4}\]So we apply De'Moivre's to this :o
Too confusing?

- anonymous

Yeah.. Give me some moment to understand it...
So from that,
It will be like:
\(\large\rm (-16i)^{1/4}=16^{1/4}\left[\cos\left
(\frac{1}{4})(\frac{3\pi}{2}+2\pi k\right)+i \sin\left(\frac{1}{4})(\frac{3\pi}{2}+2\pi k\right)\right]^{1/4}\)
then... what will happen to 2pik? just leave it like that? do i have to distribute 1/4 on that part also?
sorry, i'm still learning this, so I will be slow

- zepdrix

Good, distribute! :)

- zepdrix

Not to steal all of the fun,
but it gives us something like this, ya?\[\large\rm =2\left[\cos\left(\frac{3\pi}{8}+\frac{2\pi k}{4}\right)+i \sin\left(\frac{3\pi}{8}+\frac{2\pi k}{4}\right)\right]\]

- zepdrix

Oh common denominator would be very helpful actually.\[\large\rm =2\left[\cos\left(\frac{3\pi}{8}+\frac{4\pi k}{8}\right)+i \sin\left(\frac{3\pi}{8}+\frac{4\pi k}{8}\right)\right]\]

- anonymous

lol yes. I'm getting the same thing.
give me a sec.

- anonymous

\(\large\rm =2\left[\cos\left(\frac{7\pi k}{8}\right)+i \sin\left(\frac{7\pi k}{8}\right)\right]\)

- zepdrix

Woops! The first term doesn't have a k!
We won't be able to combine them that way :O

- anonymous

oh alright. so it will be:
\(\large\rm =\left[\cos\left(\frac{3\pi}{4}+\frac{4\pi k}{4}\right)+i \sin\left(\frac{3\pi}{4}+\frac{4\pi k}{4}\right)\right]\) ? can I do that? I cancelled 2 from the outside?

- zepdrix

Noooo you silly billy!

- zepdrix

In general: \(\large\rm 2\cos(\theta)\ne\cos(2\theta)\)

- zepdrix

Can't just bring the multiplier inside willy nilly :)

- anonymous

okay... so what else can we do then?
For multiplying to complex numbers in polar form, is it okay just to leave it like that?
Just asking..

- zepdrix

So to find our roots, we want to plug in values for k.
We start by plugging in k=0, that gives us our first, or principal, root.\[\large\rm =2\left[\cos\left(\frac{3\pi}{8}+0\right)+i \sin\left(\frac{3\pi}{8}+0\right)\right]\]So there is the first of our four roots that we're looking for.

- anonymous

any values for k? so the solutions will be too many?

- zepdrix

The values for k are going to be k=0,1,2 and 3.
Since this is a `4th root` we're looking for, we only want the first `four` integer values for k, starting from 0.
We could keep counting beyond that, but what you'll come to find out is that, your k=5 gives you the same result as k=0!
We only get a unique root, for the first 4 values of k.

- anonymous

ok... I think I can manage it from here. Thank you so much!
I'll try to do more problems until I get it :)

- zepdrix

It looks a lot nicer in exponential form: \(\large\rm 2e^{i\frac{3\pi}{8}},~2e^{i\frac{7\pi}{8}}, \) blah blah blah...
But if you like the sines and cosines form, you can stick with that.
After you get your 4 solutions, you might have to adjust them a tiny bit depending on what interval you want for your angle.
Is this for a class?
Traditionally we make the interval from -pi to pi,
but it's quite possible your teacher would prefer your angle be between 0 and 2pi.

- anonymous

I haven't learned the exponential form. I think it will be next.
I'm just learning it by myself for now, so that I will be ready for our next lecture.
I will ask my prof about that once he discuss this topic.
Thanks for the headstart about exponential form!

- zepdrix

yay team \c:/

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