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Ahsome

  • one year ago

Question with Complex Numbers

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  1. ahsome
    • one year ago
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    If \(z = x + iy\) where \(y ≠ 0\) and \(1+ z^2 ≠ 0\), show that the number \(w=\dfrac{z}{1+z^2}\)is real only if \(|z| = 1\)

  2. anonymous
    • one year ago
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    would it help if i prove the converse?

  3. ahsome
    • one year ago
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    What is the converse, @divu.mkr?

  4. UsukiDoll
    • one year ago
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    it's the reversed version of a statement

  5. anonymous
    • one year ago
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    that if |z| = 1 then 'w' is real?

  6. UsukiDoll
    • one year ago
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    for example. if A then B is the original statement the converse would be If B then A and I'm going in and out of the question, my openstudy is real bad

  7. ahsome
    • one year ago
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    Sure, if that works @divu.mkr

  8. anonymous
    • one year ago
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    I dont know how to use latex... so try to understand just use this (z)(z bar) = |z|^2 so z=1/zbar and if a complex number is real then w= w bar take bar on both the sides replace 'z bar' by z they will become same

  9. ahsome
    • one year ago
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    So, \(z\times \overline{z}=|z|^2\) Therefore, \(z=\dfrac{1}{\overline{z}}\) I don't get what you mean if a complex number is real, then w = w bar

  10. anonymous
    • one year ago
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    when we take a bar over a complex number the sign of the complex quantity reverses so if the complex number is purely real then there will be no change in the complex number as the complex quantity is absent

  11. UsukiDoll
    • one year ago
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    \[\bar{z} \]

  12. UsukiDoll
    • one year ago
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    latex is \bar{insertsomethinghere} for the bar over the area

  13. ahsome
    • one year ago
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    Ahh, I see. So, \(w\) is a complex number (cause you are dividing a complex by a complex), but since we have established its real, its conjugate should be the same..?

  14. ahsome
    • one year ago
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    Cause there is no imaginary component?

  15. anonymous
    • one year ago
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    yep, exactly

  16. ahsome
    • one year ago
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    Great :) So, we turn: \[\huge{w=\frac{z}{1+z^2}}\] To \[\huge{\overline{w}=\frac{\overline{z}}{\overline{1+z^2}}}\]

  17. ahsome
    • one year ago
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    Right..?

  18. anonymous
    • one year ago
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    yep now just one of them

  19. ahsome
    • one year ago
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    \[\huge{w}=\frac{\frac{1}{z}}{1+(\frac{1}{z})^2}\]?

  20. anonymous
    • one year ago
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    yeah doing good

  21. anonymous
    • one year ago
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    but it will be w bar

  22. ahsome
    • one year ago
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    Didn'w we say w = w bar tho?

  23. ahsome
    • one year ago
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    @divu.mkr?

  24. anonymous
    • one year ago
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    you substituted in wrong eqn ...just saying that

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