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AmTran_Bus

  • one year ago

Complex numbers

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  1. AmTran_Bus
    • one year ago
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    If Z=x+2iy, then find Re(z*) Where z=x+iy is the formula and x=Re(z) and y=IM(z)

  2. AmTran_Bus
    • one year ago
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    Re= real , IM=imag.

  3. AmTran_Bus
    • one year ago
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    * is the conjugate

  4. ganeshie8
    • one year ago
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    \(z^*\) is obtained by reflecting the complex number over real axis it is called the conjugate of \(z\)

  5. ganeshie8
    • one year ago
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    when you reflect something over real axis, notice that the real component is not changed

  6. AmTran_Bus
    • one year ago
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    Here is how it says to do it: Re(z*) =Re(x-2iy) =x. Can you help me understand it that way?

  7. ganeshie8
    • one year ago
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    just enter the part that is not attached to \(i\)

  8. AmTran_Bus
    • one year ago
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    I thought the same thing until part B that wants Re(z^2). I know the right answer for it but cant get it.

  9. ganeshie8
    • one year ago
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    z = x+2iy z^2 = (x+2iy)^2 = ?

  10. AmTran_Bus
    • one year ago
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    But I thought it only wanted the real part. If you square all of it for this one, why did you only worry about the x in the other and not the imaginary 2iy in the other one?

  11. ganeshie8
    • one year ago
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    because they are asking just the real part

  12. AmTran_Bus
    • one year ago
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    Yea, so why is it not z^2, so z=x, so z^2 = x^2?

  13. ganeshie8
    • one year ago
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    z is not x z is x+2iy

  14. AmTran_Bus
    • one year ago
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    So what does the Re they specifify there mean?

  15. ganeshie8
    • one year ago
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    to find the real part of z^2, first you need to find z^2

  16. AmTran_Bus
    • one year ago
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    Ohh. Ohh. Ohh.

  17. AmTran_Bus
    • one year ago
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    So can you let me work through it real quick?

  18. ganeshie8
    • one year ago
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    sure

  19. AmTran_Bus
    • one year ago
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    So I need (x+2iy)^2 first, which gives x^2+4iy-4y? is that right?

  20. ganeshie8
    • one year ago
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    (x+2iy)^2 = x^2 + 4iy - 4y^2

  21. AmTran_Bus
    • one year ago
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    How is the last part that?

  22. AmTran_Bus
    • one year ago
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    I thought 2iy*2iy was 4i^2*y^2 Or -4y^3 Oh I understand.

  23. ganeshie8
    • one year ago
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    (2iy)^2 = 4i^2y^2 = -4y^2

  24. AmTran_Bus
    • one year ago
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    Oh so x^2-4y^2 is all the non imaginary stuff.

  25. ganeshie8
    • one year ago
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    \(z = x+i2y\) \(z^2 = (x+2iy)^2 = \color{red}{x^2-4y^2} + i\color{purple}{4y}\)

  26. ganeshie8
    • one year ago
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    yes x^2-4y^2 is the real part

  27. AmTran_Bus
    • one year ago
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    THANK YOU x1,000,000,000

  28. ganeshie8
    • one year ago
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    np:)

  29. AmTran_Bus
    • one year ago
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    Wait, why does the book sayx^2+4ixy-4y^2? Where did the x in the 4ixy come from? @ganeshie8

  30. ganeshie8
    • one year ago
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    right, \(z = x+i2y\) \(z^2 = (x+2iy)^2 = \color{red}{x^2-4y^2} + i\color{purple}{4xy}\)

  31. ganeshie8
    • one year ago
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    we're just using the identity \[\large (\heartsuit + \spadesuit)^2~~ =~~ \heartsuit^2 + 2\heartsuit\spadesuit + \spadesuit^2 \]

  32. AmTran_Bus
    • one year ago
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    |dw:1440213734204:dw| Sorry I'm so slow

  33. ganeshie8
    • one year ago
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    |dw:1440213819660:dw|

  34. AmTran_Bus
    • one year ago
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    yeppers. You came through again.

  35. ganeshie8
    • one year ago
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    np

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