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AmTran_Bus

  • one year ago

Really confused all I need is explanation and maybe some steps.

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  1. AmTran_Bus
    • one year ago
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    Express 6i in the form re^(i theta).

  2. AmTran_Bus
    • one year ago
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    @ParthKohli @paki @Nnesha

  3. AmTran_Bus
    • one year ago
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    I have the steps. My book says 6i = r cos theta + ir sin theta Then r=6. How?

  4. AmTran_Bus
    • one year ago
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    I understand why we use Euler's formula.

  5. ParthKohli
    • one year ago
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    Yeah well, each complex number is an addition of its "real"-part and its "imaginary"-part. Whenever you compare two complex numbers, you compare their real and imaginary parts separately. You know what the best way to find out the polar-form of a complex number is? Graphing it. |dw:1440264084863:dw|

  6. ParthKohli
    • one year ago
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    |dw:1440264204510:dw|

  7. AmTran_Bus
    • one year ago
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    That is what @phi and @IrishBoy123 said, too, but I still need to understand the actual math.

  8. AmTran_Bus
    • one year ago
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    Like, I see and think I get the graph. My book has it.

  9. phi
    • one year ago
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    find the magnitude by multiplying by the complex conjugate: \[ (r \cos x + i r \sin x ) (r \cos x- i r \sin x )= r^2 \cos^2 x + r^2 \sin^2 x \\ = r^2 ( \cos^2 x + \sin^2 x) \\= r^2 \] and the magnitude is r

  10. phi
    • one year ago
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    thus if you start with 0 + 6 i and find the magnitude, you will get 6 that means r = 6

  11. AmTran_Bus
    • one year ago
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    How did you get the 0 in 0+6i?

  12. phi
    • one year ago
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    6i is 0 + 6i (if we insist on showing the real component is 0)

  13. AmTran_Bus
    • one year ago
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    Oh. The z=x+iy formula. Ok.

  14. AmTran_Bus
    • one year ago
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    So now I need tan-1, which is 6/0?

  15. ParthKohli
    • one year ago
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    This is the actual math. If you want to look at it the boring way, here you go:\[\tan \theta=\frac{y}{x} = \frac{\Im(z)}{\Re(z) } = \frac{6}{0}\]\[\Rightarrow \theta = \pi/2\]\[r = \sqrt{x^2 + y^2} = \sqrt{\left(\Re(z)\right)^2 + \left(\Im(z)\right)^2} = 6\]

  16. phi
    • one year ago
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    yes, the angle can be found using inverse tan imag/real

  17. AmTran_Bus
    • one year ago
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    Thanks! So to express it in that certain form, It must be 6e^(i pi/2)

  18. phi
    • one year ago
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    yes, you are using Euler's \[ e^{ix}= \cos x + i \sin x \]

  19. IrishBoy123
    • one year ago
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    |dw:1440264786581:dw|

  20. AmTran_Bus
    • one year ago
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    Why does my calc say tan 6/0 is undef and yall get pi/2?

  21. IrishBoy123
    • one year ago
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    because 6/0 is undefined

  22. IrishBoy123
    • one year ago
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    ask it for \(arctan \frac{6}{0.00000001}\)

  23. AmTran_Bus
    • one year ago
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    got 1.57

  24. AmTran_Bus
    • one year ago
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    ok tks

  25. IrishBoy123
    • one year ago
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    my calc blows up too. its software just crashes on a singularity, it doesn't ask which function is coming next....

  26. AmTran_Bus
    • one year ago
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    Can yall help me with one more? 4-sqrt(2)i?

  27. AmTran_Bus
    • one year ago
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    Do I need to multiply it by 4+sqrt(2)i?

  28. phi
    • one year ago
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    you can do sum of squares (then take the square root)

  29. AmTran_Bus
    • one year ago
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    Ok. Let me get that ans for ya!

  30. AmTran_Bus
    • one year ago
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    Its just 16+2 right?

  31. phi
    • one year ago
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    yes if you plot it, you see it's pythagoras

  32. AmTran_Bus
    • one year ago
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    Why does the book say sqrt(16+2) and then 3 sqrt(2)?

  33. phi
    • one year ago
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    |dw:1440265488511:dw|

  34. AmTran_Bus
    • one year ago
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    If I took a picture of the book, would you help me make sense of what its saying?

  35. phi
    • one year ago
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    do you get that the length (distance from the origin) of the point 4 + sqr(2) I is sqr(18)?

  36. AmTran_Bus
    • one year ago
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    Yes. But this book does not do it that way.

  37. AmTran_Bus
    • one year ago
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    It starts under the 6i

  38. phi
    • one year ago
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    they are doing the same thing, but left out the simplifying step. sqrt(16+2) = sqr(18)= sqr(2*9)= sqr(2)*sqr(9) = sqr(2)*3 or \[ 3 \sqrt{2}\]

  39. AmTran_Bus
    • one year ago
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    Got ya. Thanks. Its just hard for me to see sqrt(16+2), because I thought 4- sqrt(2)i was not under a root.

  40. AmTran_Bus
    • one year ago
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    oh you take sqrt to get mag.

  41. phi
    • one year ago
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    if you think pythagoras with sides 4 (along the x-axis ) and -sqr(2) (along the y or "i" axis) and c^2 = a^2 + b^2 that should help you remember that x^2 + y^2 is the magnitude squared thus 16+2 = 18 is the mag squared and we want sqr(18) for the actual magnitude.

  42. AmTran_Bus
    • one year ago
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    Yea, I see where pythagoras helps understand that now.

  43. AmTran_Bus
    • one year ago
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    Thanks for that.

  44. IrishBoy123
    • one year ago
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    this is from the Gospel of St Mary [Boas, Maths for Physical Sciences] notice how the Argand diagram features prominently

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  45. AmTran_Bus
    • one year ago
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    Wow that is really nice to have. Thanks!

  46. AmTran_Bus
    • one year ago
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    Just solved 3 on my own got em all right

  47. IrishBoy123
    • one year ago
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    https://gyazo.com/a85e151f5d058e0a0a631873c5c59849

  48. IrishBoy123
    • one year ago
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    you have 30 seconds!

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