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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.

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

  2. zepdrix
    • one year ago
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    Have you learned about the exponential form or no?

  3. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    can you explain that part? I'm still not sure

  6. anonymous
    • one year ago
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    what to do..

  7. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    ok...

  11. zepdrix
    • one year ago
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    So we can get to -i by rotating that exact amount, or by rotating that amount + any integer multiple of 2pi.

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

  15. anonymous
    • one year ago
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    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
    • one year ago
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    Good, distribute! :)

  17. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    lol yes. I'm getting the same thing. give me a sec.

  20. anonymous
    • one year ago
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    \(\large\rm =2\left[\cos\left(\frac{7\pi k}{8}\right)+i \sin\left(\frac{7\pi k}{8}\right)\right]\)

  21. zepdrix
    • one year ago
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    Woops! The first term doesn't have a k! We won't be able to combine them that way :O

  22. anonymous
    • one year ago
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    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
    • one year ago
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    Noooo you silly billy!

  24. zepdrix
    • one year ago
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    In general: \(\large\rm 2\cos(\theta)\ne\cos(2\theta)\)

  25. zepdrix
    • one year ago
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    Can't just bring the multiplier inside willy nilly :)

  26. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    any values for k? so the solutions will be too many?

  29. zepdrix
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yay team \c:/

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