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

1. 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

2. zepdrix

Have you learned about the exponential form or no?

3. 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?

4. 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.

5. anonymous

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

6. anonymous

what to do..

7. 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..

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

9. zepdrix

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

10. anonymous

ok...

11. zepdrix

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

12. zepdrix

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

13. 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.

14. 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?

15. 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

16. zepdrix

Good, distribute! :)

17. zepdrix

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

18. zepdrix

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

19. anonymous

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

20. anonymous

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

21. zepdrix

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

22. 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?

23. zepdrix

Noooo you silly billy!

24. zepdrix

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

25. zepdrix

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

26. 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..

27. 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.

28. anonymous

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

29. 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.

30. 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 :)

31. 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.

32. 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!

33. zepdrix

yay team \c:/