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tdabboud

  • 2 years ago

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

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

  2. EarthCitizen
    • 2 years ago
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    nope

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

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

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

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

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

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

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

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

  11. JamesJ
    • 2 years ago
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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 \( \theta \in [0,2\pi) \).

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

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

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

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

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

  17. JamesJ
    • 2 years ago
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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.

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

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

  20. tdabboud
    • 2 years ago
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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)

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

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

  23. tdabboud
    • 2 years ago
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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

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

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

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

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

  28. EarthCitizen
    • 2 years ago
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    or 1.849 rads

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

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

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

  32. tdabboud
    • 2 years ago
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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

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

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

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

  36. tdabboud
    • 2 years ago
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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

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

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

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

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

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

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