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Ahsome

  • one year ago

Complex Number Equation

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  1. ahsome
    • one year ago
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    \[z=x+yi\]\[|z|=1\]Prove that \(w\) is a real number, when \(w=\dfrac{z}{1+z^2}\) Note: \(y\ne0\) and \(1+z^2\ne0\)

  2. ahsome
    • one year ago
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    @ganeshie8

  3. ganeshie8
    • one year ago
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    Let \(z=x+iy = re^{i\theta}\) then \[w=\dfrac{re^{i\theta}}{1+r^2e^{i2\theta}}\] multiply top and bottom by \(e^{i(-\theta)}\) and get \[w = \dfrac{r}{e^{i(-\theta)}+r^2e^{i\theta}}\]

  4. ganeshie8
    • one year ago
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    For \(w\) to be a real number, it must be the case that the denominator is real, so set the imaginary part of denominator equal to 0 : \[\sin(-\theta) + r^2\sin(\theta) = 0\]

  5. ahsome
    • one year ago
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    Ok, I have oficially confirmed that we haven't learnt the info needed to solve that question

  6. ahsome
    • one year ago
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    I'm not sure with \(e^{i\theta}\) etc means

  7. ganeshie8
    • one year ago
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    you haven't learned polar form ?

  8. ganeshie8
    • one year ago
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    ohk then forget about that, lets work it in rectangular form

  9. ahsome
    • one year ago
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    We have. Its given in \(z=r\times\text{cis}\times\theta\)

  10. ahsome
    • one year ago
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    But ok, I prefer rectangular

  11. ganeshie8
    • one year ago
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    Yes, \(re^{i\theta}\) is euler form for \(r*cis(\theta)\)

  12. ganeshie8
    • one year ago
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    Okay lets try and work it in rectangular

  13. ahsome
    • one year ago
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    Ok then :)

  14. ganeshie8
    • one year ago
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    maybe start by plugging in \(z=x+iy\)

  15. ahsome
    • one year ago
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    Into the fraction?

  16. ganeshie8
    • one year ago
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    \[w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} = \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\]

  17. ganeshie8
    • one year ago
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    next multiply top and bottom by the conjugate of bottom

  18. ganeshie8
    • one year ago
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    so that the denominator becomes real

  19. ahsome
    • one year ago
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    Would we take the \(i\) as common, to make it easier?

  20. ganeshie8
    • one year ago
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    what is the conjugate of denominator ?

  21. ahsome
    • one year ago
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    \((x^2-y^2+1)-2xyi\)?

  22. ganeshie8
    • one year ago
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    yes multiply that top and bottom

  23. ganeshie8
    • one year ago
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    \[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}y(1+x^2-y^2-2x^2)}{(1+x^2-y^2)^2+(2xy)^2} \end{align}\] set the imaginary part equal to 0

  24. ahsome
    • one year ago
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    Why?

  25. ganeshie8
    • one year ago
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    because we're looking for a conditaion that makes w real

  26. ganeshie8
    • one year ago
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    w is real means, its imaginary part is 0, yes ?

  27. ahsome
    • one year ago
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    Ahh, ok

  28. ahsome
    • one year ago
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    So, wherever the is an imaginary, its equal to 0?

  29. ganeshie8
    • one year ago
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    \[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}} \end{align}\] thats the imaginary part, set that equal to 0

  30. ganeshie8
    • one year ago
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    *that blue thing is the imaginary part

  31. ahsome
    • one year ago
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    Why is the bottom imaginary? I don't see an i

  32. ganeshie8
    • one year ago
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    first write that in standard form

  33. ganeshie8
    • one year ago
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    standard form of complex number : \(a+ib\)

  34. ahsome
    • one year ago
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    Yup

  35. ganeshie8
    • one year ago
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    \[\begin{align}w=\dfrac{z}{1+z^2} = \dfrac{x+iy}{1+(x+iy)^2} &= \dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\\~\\ &=\dfrac{x+\color{red}{i}y}{1+x^2-y^2 + 2xy\color{red}{i}}\times \dfrac{1+x^2-y^2 - 2xy\color{red}{i}}{1+x^2-y^2 - 2xy\color{red}{i}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2) +\color{red}{i}\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}}\\~\\ &=\dfrac{x(1+x^2-y^2+2xy^2)}{\color{black}{(1+x^2-y^2)^2+(2xy)^2}} +\color{red}{i}\dfrac{\color{blue}{y(1+x^2-y^2-2x^2)}}{\color{blue}{(1+x^2-y^2)^2+(2xy)^2}} \\~\\ \end{align}\]

  36. ahsome
    • one year ago
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    \[\dfrac{x(1+x^2−y^2+2xy^2)}{(1+x^2−y^2)^2+(2xy)^2}+\dfrac{iy(1+x2−y2−2x2)}{(1+x^2−y^2)^2+(2xy)^2}\]

  37. ahsome
    • one year ago
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    Don't we just cancel the right fraction?

  38. ganeshie8
    • one year ago
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    set that equal to 0

  39. ganeshie8
    • one year ago
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    we want to know when the imaginary part of \(w\) equals \(0\)

  40. ahsome
    • one year ago
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    Ok. That essentially just removes the right fraction, right?

  41. ganeshie8
    • one year ago
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    idk what you mean by just removing the right fraction

  42. ganeshie8
    • one year ago
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    we can't just remove right fraction

  43. ahsome
    • one year ago
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    Nevermind. Lets continue

  44. ganeshie8
    • one year ago
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    set the imaginary part equal to 0

  45. ahsome
    • one year ago
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    Ok. So its just\[0=\dfrac{iy(1+x^2−y^2−2x^2) } {(1+x^2−y^2)^2+(2xy)^2}\]

  46. ganeshie8
    • one year ago
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    remove "i", imaginary part of w is \[\dfrac{y(1+x^2−y^2−2x^2) } {(1+x^2−y^2)^2+(2xy)^2}\]

  47. ganeshie8
    • one year ago
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    set that equal to 0

  48. ahsome
    • one year ago
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    \[{0=\frac{y(1+x^2−y^2−2x^2)}{(1+x^2−y^2)^2+(2xy)^2}}\]?

  49. ganeshie8
    • one year ago
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    Yep

  50. ahsome
    • one year ago
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    \[0=y(1+x^2−y^2−2x^2)\]?

  51. ganeshie8
    • one year ago
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    Yes

  52. ahsome
    • one year ago
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    \[0=y+x^2y-y^3-2x^2y\] \[0=y-x^2y-y^3\]

  53. ganeshie8
    • one year ago
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    factor out y

  54. ahsome
    • one year ago
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    \[0=y(1-x^2-y^2)\]\[0=1-x^2-y^2\]

  55. ahsome
    • one year ago
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    \[x^2+y^2=1\]

  56. ganeshie8
    • one year ago
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    Looks good!

  57. ganeshie8
    • one year ago
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    from the hypothesis \(y\ne 0\) therefore \(x^2+y^2=1\)

  58. ganeshie8
    • one year ago
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    which is same as \(|z|=1\)

  59. ahsome
    • one year ago
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    And that's how we are meant to solve it? AWESOME :D

  60. ganeshie8
    • one year ago
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    Yes if we're allowed to use only rectangular form

  61. ahsome
    • one year ago
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    Great

  62. ahsome
    • one year ago
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    Thank you SO MUCH @ganeshie8. I really do wonder how you solve like half of these questions ;)

  63. ganeshie8
    • one year ago
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    np:)

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