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mathslover

  • 3 years ago

If \(z_1\) and \(z_2\) are two complex numbers such that \(|\frac{z_1 - z_2}{z_1 + z_2}| =1 \) , prove that, \(\frac{iz_1}{z_2}=k\) is a real number. Find the angle between the lines from the origin to the points \(z_1 + z_2 \) and \(z_1 - z_2\) in terms of k.

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  1. mathslover
    • 3 years ago
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    @AravindG @amistre64 @ghazi @hartnn

  2. AravindG
    • 3 years ago
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    oh too small latex ..my eye hurts

  3. amistre64
    • 3 years ago
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    isnt a general complex number: a + bi ? or do we need to define it in trig terms?

  4. mathslover
    • 3 years ago
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    If \(\large{z_1}\) and \(\large{z_2}\) are two complex numbers such that \(\large{|\frac{z_1 - z_2}{z_1 + z_2}| = 1 }\) , prove that , \(\large{\frac{i z_1}{z_2} = k }\) , where k is a real number. Find the angle between the lines from the origin to the points \(\large{z_1 + z_2}\) and \(\large{z_1 - z_2}\) in terms of k... It did hurt me too :)

  5. mathslover
    • 3 years ago
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    Both conditions can be applied... @amistre64

  6. mathslover
    • 3 years ago
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    how @AravindG ?

  7. mathslover
    • 3 years ago
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    well I started like this : \[\large{| \frac{\frac{z_1}{z_2} -1}{\frac{z_1}{z_2} +1}| = 1}\] \[\large{\textbf{or} \space |\frac{z_1}{z_2} -1| = |\frac{z_1}{z_2}+1| }\]

  8. mathslover
    • 3 years ago
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    I am thinking of squaring both sides, should I go for it ?

  9. amistre64
    • 3 years ago
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    \[\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}=1\] \[(a+bi)-(n+mi)=(a+bi)+(n+mi)\] \[-(n+mi)=(n+mi)\] \[0=2(n+mi)\] just a thought ....

  10. AravindG
    • 3 years ago
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    i think I made a mistake there ..wait lemme think

  11. mathslover
    • 3 years ago
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    But amistre where did modulus go ? In LHS ...

  12. AravindG
    • 3 years ago
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    oh god i was right :P

  13. mathslover
    • 3 years ago
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    \[\large{|\frac{z_1-z_2}{z_1+z_2}| = 1 \ne \frac{z_1 -z_2}{z_1+z_2} =1 }\]

  14. mathslover
    • 3 years ago
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    ^ not necessary that : (a+bi)-(n+mi) / (a+bi)+(n+mi) = 1 ...

  15. shubhamsrg
    • 3 years ago
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    You can do it easily by interpretting is geometrically.

  16. shubhamsrg
    • 3 years ago
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    it*

  17. amistre64
    • 3 years ago
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    im just not that attuned to complex calculations :)

  18. AravindG
    • 3 years ago
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    |z1+z2| =|z1-z2| this means lenth of the vectors z1+z2 and z1-z2 are equal Now think of z1 and z2 as two vectors z1+z2 is one diagonal of the parallelogram ,z1-z2 is the other diagonal Given the diagonal lengths are equal thus the parallelogram is a rectangle So angle between z1 and z2 is \(\dfrac{\pi}{2}\)

  19. AravindG
    • 3 years ago
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    ;) geometry is wonderful

  20. mathslover
    • 3 years ago
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    wait...

  21. mathslover
    • 3 years ago
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    can you draw that parallelogram please @AravindG

  22. mathslover
    • 3 years ago
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    I am confused that whether that parallelogram is possible or not with two diagonals as : z1 + z2 and z1-z2

  23. hartnn
    • 3 years ago
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    \(\huge |\frac{(a+bi)-(n+mi)}{(a+bi)+(n+mi)}|=1 \\ (a-n)^2+(b-m)^2=(a+n)^2+(b+n)^2 \\ \implies an+bm=0\) \(\huge \frac{i(a+bi)}{(n+mi)}=\frac{i(a+bi)}{(n+mi)}\dfrac{(n-mi)}{(n-mi)}= \\ \huge =\dfrac{i(an+bm-inb+ami)}{(n^2+m^2)}=\dfrac{i^2(0+am+nb)}{n^2+m^2}= \\ =k\)

  24. mathslover
    • 3 years ago
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    well right :) I got it ..

  25. mathslover
    • 3 years ago
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    But what about the second one..

  26. mathslover
    • 3 years ago
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    I did the first one like this : \[\large{|\frac{z_1}{z_2} - 1| = | \frac{z_1}{z_2} + 1|}\] Squaring both sides : \[\large{|\frac{z_1}{z_2}|^2 + 1 - 2Re (\frac{z_1}{z_2}) = |\frac{z_1}{z_2}|^2 + 1 + 2 Re(\frac{z_1}{z_2})}\] \[\large{4Re(\frac{z_1}{z_2}) = 0}\] \[\large{Re(\frac{z_1}{z_2}) = 0}\] therefore (z_1)/z_2 is purely imaginary number. Therefore \[\large{\frac{z_1}{z_2} = i \frac{z_1}{z_2} = k}\] where k is a real number

  27. AravindG
    • 3 years ago
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    @mathslover where do you have confusion in my working ?

  28. mathslover
    • 3 years ago
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    I am confused whether that parallelogram is possible or not?

  29. mathslover
    • 3 years ago
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    can you draw that for me please?

  30. AravindG
    • 3 years ago
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    just consider z1 and z2 as two vectors .... P.S. I am bad at drawing

  31. mathslover
    • 3 years ago
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    |dw:1361019499183:dw| is it right diagram ?

  32. mathslover
    • 3 years ago
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    instead of drawing can you tell me that what are co ordinates of the parallelogram : like A(z_1) B(z_2) C(z_(-1)) D(z_(-2)) ...

  33. AravindG
    • 3 years ago
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    i would say this : |dw:1361019666403:dw|

  34. AravindG
    • 3 years ago
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    gosh u made me draw that ...:P

  35. mathslover
    • 3 years ago
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    |dw:1361019850236:dw|

  36. mathslover
    • 3 years ago
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    is C (z_1 - z_2)

  37. AravindG
    • 3 years ago
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    c is z1+z2

  38. mathslover
    • 3 years ago
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    oh got it now :)

  39. AravindG
    • 3 years ago
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    :)

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