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Ahsome

  • one year ago

How to find conjugate of complex number

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  1. ahsome
    • one year ago
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    Given \(w=2\text{cis}\left(-\dfrac{\pi}{5}\right)\) Find \(\overline{w}\) Is there anyway of solving this without turning into \(x+yi\), then finding conjugate, then back to \(\text{cis}\) form?

  2. ahsome
    • one year ago
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    @ganeshie8?

  3. amistre64
    • one year ago
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    what does W multiplied by it conjugate result in?

  4. ahsome
    • one year ago
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    \(x^2+y^2\)?

  5. ahsome
    • one year ago
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    Also, imaginary part = 0

  6. amistre64
    • one year ago
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    seems fair to me what is the 2 part in W ?

  7. ahsome
    • one year ago
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    The magnitude

  8. amistre64
    • one year ago
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    for 2 complex numbers (if memory serves) (n,a)(m,b) = (nm,a+b) n,m being magnitudes, and a,b being angles

  9. ahsome
    • one year ago
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    Yup :)

  10. amistre64
    • one year ago
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    cos(0) + sin(0) = 1 so a+b = 0 right? or am i going down a rabbit hole?

  11. anonymous
    • one year ago
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    I think you can get the conjugate by using the negative theta rcis(-theta)

  12. ahsome
    • one year ago
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    Yes, I understand that

  13. ahsome
    • one year ago
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    And @Harindu, reallly?

  14. ahsome
    • one year ago
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    Well, i guess that would make sense

  15. ahsome
    • one year ago
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    Or would it be thetha - \(\pi\)?

  16. ahsome
    • one year ago
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    nvm

  17. anonymous
    • one year ago
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    sorry I haven't touched complex numbers for sometime. But a + bi = r cis(theta) Where: a = rCos(theta) b = rSin(theta ) cos only b gets negative if theta is negative it works out

  18. ahsome
    • one year ago
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    Yup, I get that. Thanks @Harindu :)

  19. anonymous
    • one year ago
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    You are welcome

  20. ahsome
    • one year ago
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    How would you solve \[\frac{1}{w}\] tho?

  21. ahsome
    • one year ago
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    @Harindu?

  22. amistre64
    • one year ago
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    (n,a)^k = (n^k, ka)

  23. ahsome
    • one year ago
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    In this case, would it be -1?

  24. amistre64
    • one year ago
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    after all (n,a)(n,a) = (nn,a+a) yes, in this case k=-1

  25. anonymous
    • one year ago
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    hmm turn into a complex number first . I mean get the denominator to be real so I think you have to use x + yi form . I dnt remember sorry

  26. ahsome
    • one year ago
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    Wait, isn't 1 = \(1\times cis(\theta)\)?

  27. amistre64
    • one year ago
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    w1/w2 (n,a) (m^(-1), -b) = (n/m, a-b)

  28. ahsome
    • one year ago
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    Then, you can minus them. \(\text{Let }1=z = 1\times cis(\theta)\) \(\text{Let }w = 4\times cis\left(\dfrac{-\pi}{5}\right)\) \[\frac{1}{w}\]\[\dfrac{4}{1}\times cis\left(\dfrac{-\pi}{5}-1\right)\]Does that work?

  29. ahsome
    • one year ago
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    Althought it would be 1/4 and \(1-\frac{-\pi}{5}\)

  30. amistre64
    • one year ago
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    \[\frac1{4cis(-\frac{\pi}{5})}\] \[\frac{1}{4}cis(\frac{\pi}{5})\]

  31. ahsome
    • one year ago
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    :D

  32. ahsome
    • one year ago
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    THANK YOU :)

  33. amistre64
    • one year ago
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    \[(4,-\frac{\pi}{5})^{-1}\implies(4^{-1},-(-\frac{\pi}{5}))\]

  34. amistre64
    • one year ago
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    youre welcome

  35. ahsome
    • one year ago
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    No problem. For future reference, if it ever occured that: \[\frac{x}{w}\] We would simply say that \[\frac{x}{w}=\frac{x\times cis(0)}{w}\] Then do \[\frac{x}{|w|}\times cis\left(0-\arg(w)\right)\]?

  36. ahsome
    • one year ago
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    If x is a real number and w is a complex number?

  37. amistre64
    • one year ago
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    yep, seems fair to me

  38. ahsome
    • one year ago
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    Thank you :)

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