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tdabboud

If z= (-2+7i) and w = z^3000 whats the angle theta of w in the region [0,2pi) I want to know how to compute this. I can easily plug it into wolfram but it doesnt explain the steps

  • 2 years ago
  • 2 years ago

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  1. tdabboud
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    would it simply be 3000 * (angle of (-2+7i)) ?

    • 2 years ago
  2. EarthCitizen
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    nope

    • 2 years ago
  3. EarthCitizen
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    \[w=(-2+j7)^{3000}\]

    • 2 years ago
  4. EarthCitizen
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    there's two way's to solve this either directly or use de Moivre's theorem, the choice is yours

    • 2 years ago
  5. tdabboud
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    i havent learned that theorem yet so most likely the direct way

    • 2 years ago
  6. EarthCitizen
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    the direct way is log as you can see the power is 3000

    • 2 years ago
  7. tdabboud
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    yea but we're dealing with complex numbers.. do the same rules apply then

    • 2 years ago
  8. EarthCitizen
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    first thing you need to do here is change z into polar form

    • 2 years ago
  9. EarthCitizen
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    that is in the form of \[r ^{n}<n \theta\]

    • 2 years ago
  10. tdabboud
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    ok then...(7.28,1.85)

    • 2 years ago
  11. JamesJ
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    @EC, that's not polar form for complex numbers. You want to write \[ z = Ae^{i\theta} \] for some positive number \( A \) which is the magnitude of \( z \), and some angle \( \theta \in [0,2\pi) \).

    • 2 years ago
  12. JamesJ
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    Then \[ z^{3000} = A^{3000} e^{3000\theta i} \]

    • 2 years ago
  13. EarthCitizen
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    yh, in polar form \[r<\theta\]

    • 2 years ago
  14. JamesJ
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    (@EC, I see. Not standard notation, but I get it.)

    • 2 years ago
  15. EarthCitizen
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    @ JamesJ is Euler's theorem the only way to solve ?

    • 2 years ago
  16. tdabboud
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    i didnt think it was the only way, there has to be another

    • 2 years ago
  17. JamesJ
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    It's far away the fastest. Unless you use some version of it, you would need to multiply that out explicitly. The whole point of the exercise with such a ridiculous exponent is to make you use some polar form.

    • 2 years ago
  18. JamesJ
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    Do you not have deMoivre's theorem?

    • 2 years ago
  19. EarthCitizen
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    yh couldn't we just say\[(r ^{n}<n \theta)\]

    • 2 years ago
  20. tdabboud
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    yea i do understand that, the coordinates i gave were correct for eulers and polar: w= 7.28 e^j(1.85 rad)

    • 2 years ago
  21. tdabboud
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    using eulers you would then take the log?

    • 2 years ago
  22. EarthCitizen
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    \[e ^{j \theta}= \cos(\theta)+jsin (\theta)\]

    • 2 years ago
  23. tdabboud
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    I am still lost... one of my problems we did they simply took the exponent to the front and multiplied it by the angle the problem was (-3+j)^4 so they did: 4(2.8198 rad) = 11.27 rad... you can then subtract 2pi to get to the correct domain. b but i tried this and its not working correctly

    • 2 years ago
  24. EarthCitizen
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    \[(re ^{j \theta})=r ^{n}e ^{jn \theta}=(7.28)^{3000}(\cos(3000\theta)+jsin(3000\theta)\]

    • 2 years ago
  25. EarthCitizen
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    to convert to radians\[(\theta \times \pi)/180\]

    • 2 years ago
  26. EarthCitizen
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    that's de Moivre's, \[(r<\theta)^{n}=(r ^{n}<n \theta)\]

    • 2 years ago
  27. EarthCitizen
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    \[\theta=105.95\]

    • 2 years ago
  28. EarthCitizen
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    or 1.849 rads

    • 2 years ago
  29. tdabboud
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    the angle you gave 1.85 rad is only for (-2+7j)^1

    • 2 years ago
  30. tdabboud
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    but when raised to the 3000 the angle becomes -.765

    • 2 years ago
  31. EarthCitizen
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    i know there're 2999 more

    • 2 years ago
  32. tdabboud
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    yeah thats the problem I am having i found the angle you gave me easily, but how can you find that angle when raised to the 3000. would it be 3000 times that angle like we have done previously? then just reduce it to 0,2pi

    • 2 years ago
  33. EarthCitizen
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    are you looking for a specific angle ?

    • 2 years ago
  34. tdabboud
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    yeah an angle reduced to the domain between 0 and 2pi. or between 0 and 360 degrees.

    • 2 years ago
  35. EarthCitizen
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    which angle there're 3000 of them ?

    • 2 years ago
  36. tdabboud
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    haha the one between 0 and 360 degrees, there will be only one. the other angles have a greater total degree

    • 2 years ago
  37. EarthCitizen
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    no the angles can't exceed 360 degrees

    • 2 years ago
  38. EarthCitizen
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    all 3000 of them are within the range of 0 to 360

    • 2 years ago
  39. EarthCitizen
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    you can use this formula to find one angle between the range of 0 to 360 \[r ^{1/n}<(\theta+360(k))/n\]

    • 2 years ago
  40. EarthCitizen
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    n= power, k=position of the angle

    • 2 years ago
  41. EarthCitizen
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    http://www.wolframalpha.com/input/?i=%28-2%2Bi7%29^3000 these are all the angles according to wolfy

    • 2 years ago
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