Question

Find roots of (-16i)^(1/4) using polar form.

Can someone guide me in solving this?
I need to understand this concept.
Thanks.

Answer 1

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

Answer 2

Have you learned about the exponential form or no?

Answer 3

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?

Answer 4

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.

Answer 5

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

Answer 6

what to do..

Answer 7

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..

Answer 8

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.\(\LARGE\rm z=e^{i\frac{2pi}{5}+2\pi}\)

Answer 9

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

Answer 10

ok...

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

Good, distribute! :)

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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?

Answer 23

Noooo you silly billy!

Answer 24

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

Answer 25

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

Answer 26

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..

Answer 27

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.

Answer 28

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

Answer 29

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.

Answer 30

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

Answer 31

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.

Answer 32

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!

Answer 33

yay team \c:/