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jazzie Group Title

Find the cube roots of 8(cos 216° + i sin 216°).

  • one year ago
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  1. pradiv Group Title
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    -7.978

    • one year ago
  2. terenzreignz Group Title
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    You again? LOL, let's get down to business :D First off, I'm going to write \[\large r(\cos \theta + i \sin \theta) = r \ cis \theta\] It's shorter that way :P Ready?

    • one year ago
  3. ghazi Group Title
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    8 (cos 216 + i sin 216) =\[8e^{i 216}\] rest you can do ;)

    • one year ago
  4. jazzie Group Title
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    Haha these are tough !! @terenzreignz

    • one year ago
  5. ghazi Group Title
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    \[(8 e^{i 216} ) ^ {1/3}= 2 e^{i 72}\]

    • one year ago
  6. terenzreignz Group Title
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    Okay, since you're awesome (^.^), I'll not use shortcuts... So, if we have \[\large z = r (\cos \theta + i \sin \theta)\] In general, we get \[\large z^p=r^p(\cos \ p\theta + i \sin \ p\theta)\] I say in general, because in the case of fractional exponents (like taking the cube root, for instance), there's a minor catch...

    • one year ago
  7. terenzreignz Group Title
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    Let's look at this part first, for instance: \[\cos \theta + i \sin \theta\] This is equal to \[\large \cos (\theta + 2 \pi) + i \sin (\theta + 2\pi)\] right?

    • one year ago
  8. jazzie Group Title
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    Okay I see

    • one year ago
  9. terenzreignz Group Title
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    In fact, equal to \[\large \cos (\theta + 2k\pi) + i \sin (\theta + 2k\pi)\] for any integer k. Catch me so far?

    • one year ago
  10. terenzreignz Group Title
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    But let's use degrees, it seems the situation calls for it... \[\cos \theta + i \sin \theta = \cos (\theta + 360k^o) + i \sin (\theta + 360k^o\]

    • one year ago
  11. terenzreignz Group Title
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    So back to the question: Cube roots of \[\large 8(\cos \ 216^o + i \sin \ 216^o)\] Now use this rule \[\large z^p=r^p(\cos \ p\theta + i \sin \ p\theta)\] and since we're taking the cube root, take p = 1/3... go ahead now...

    • one year ago
  12. jazzie Group Title
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    Thank you @terenzreignz ! That makes a whole lot more sense now

    • one year ago
  13. terenzreignz Group Title
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    Could you do it from here? Could you post your answers, too?

    • one year ago
  14. jazzie Group Title
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    Yea Ill post the answer I got when I finish !

    • one year ago
  15. terenzreignz Group Title
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    Applying the rule, we get... \[\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)\] \[\large =2(\cos 72^o + i \sin 72^o)\]

    • one year ago
  16. terenzreignz Group Title
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    However, we have to also consider the cube root of \[\large 8[\cos (216+360)^o + i \sin (216+360)^o]\] Since \[\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+360)^o + i \sin (216+360)^o]\]

    • one year ago
  17. terenzreignz Group Title
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    applying the rule again, we get \[\huge 8^{\frac{1}{3}}(\cos \frac{216+360}{3}^o + i \sin \frac{216+360}{3}^o)\] \[\large = 2(\cos \ 192 + i \sin \ 192)\]

    • one year ago
  18. terenzreignz Group Title
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    However still, we have to also consider the cube root of \[\large 8[\cos (216+2(360))^o + i \sin (216+2(360))^o]\]\[=\large 8[\cos (216+720)^o + i \sin (216+720)^o]\] Since \[\large 8[\cos \ 216^o + i \sin \ 216^o]=\large 8[\cos (216+720)^o + i \sin (216+720)^o]\]

    • one year ago
  19. terenzreignz Group Title
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    Applying the rule yet again, we get\[\huge 8^{\frac{1}{3}}(\cos \frac{216+720}{3}^o + i \sin \frac{216+720}{3}^o)\]\[\large = 2(\cos \ 312+ i \sin \ 312)\]

    • one year ago
  20. terenzreignz Group Title
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    Et voila! Your three cube roots. I'll show you a shortcut for this, but first, let's see what happens if continue this process... namely that we take the cube root, only add 360 degrees more to the angle:

    • one year ago
  21. terenzreignz Group Title
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    Consider\[\large 8[\cos (216+3(360))^o + i \sin (216+3(360))^o]\]\[=\large 8[\cos (216+1080)^o + i \sin (216+1080)^o]\] Applying the rule (yet again)...

    • one year ago
  22. terenzreignz Group Title
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    We get \[\huge 8^{\frac{1}{3}}(\cos \frac{216+1080}{3}^o + i \sin \frac{216+1080}{3}^o)\] \[\large = 2(\cos \ 432+ i \sin \ 432)\]

    • one year ago
  23. terenzreignz Group Title
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    But 432 is already past 360, so it's the same if we subtract 360 from it, and we'd get \[\large =2(\cos 72^o + i \sin 72^o)\] Which was the first answer we got, so we really only have 3 cube roots.

    • one year ago
  24. terenzreignz Group Title
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    Anyway, a shortcut for this, when getting the nth roots of a complex number in polar form, you just apply the rule directly, and only once... \[\huge 8^{\frac{1}{3}}(\cos \frac{216}{3}^o + i \sin \frac{216}{3}^o)\] \[\large =2(\cos 72^o + i \sin 72^o)\] And just keep adding \[\frac{360}{n}\] degrees to the angle until you have n distinct answers. In this case, n = 3, so keep addint 360/3 = 120 to the angle until you have three cube roots... The angles are 72 72 + 120 = 192 72 + 120 + 120 = 312

    • one year ago
  25. terenzreignz Group Title
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    And that's it, I have to go now, have fun with Trig :) --.-- Terence out

    • one year ago
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