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Thread: Complex Numbers

  1. July 8th 2011, 12:04 AM #1
    Blizzardy
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    Complex Numbers

    Hi guys, need help with a question:

    Given that z = 2 + 2i, express z* in the exponential form. Hence find the set of the positive interger n for which (z*)^n is purely imaginary.

    z* = 2^(3/2) e^[i(-pi/4)]
    (z*)^n = 2^(3n/2) e^[i(-npi/4)]

    (i) Given that (z*)^n is purely imaginary,
    -npi/4 = pi/2, 3pi/2, 5pi/2, ...
    -npi/4 = (2m + 1)(pi/2) where m 'is an element of all intergers, and m > than or = 0'
    n = -(4m+2)
    (But my ans was wrong)

    Anyway, if the question wants to find the set of intergers for n for which (z*)^n is real, can I get the answer simply by letting '-npi/4 = 0, pi, 2pi, 3pi,...' ?

    Thanks in advance!
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  2. July 8th 2011, 12:51 AM #2
    Sudharaka
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    Re: Complex Numbers

    Quote Originally Posted by Blizzardy View Post
    Hi guys, need help with a question:

    Given that z = 2 + 2i, express z* in the exponential form. Hence find the set of the positive interger n for which (z*)^n is purely imaginary.

    z* = 2^(3/2) e^[i(-pi/4)]
    (z*)^n = 2^(3n/2) e^[i(-npi/4)]

    (i) Given that (z*)^n is purely imaginary,
    -npi/4 = pi/2, 3pi/2, 5pi/2, ...
    -npi/4 = (2m + 1)(pi/2) where m 'is an element of all intergers, and m > than or = 0'

    This is incorrect. It should be, -\frac{n\pi}{4}=2k\pi\pm\frac{\pi}{2}\mbox{ where }k\in Z. I have shown how to obtain this below.

    n = -(4m+2)
    (But my ans was wrong)

    Anyway, if the question wants to find the set of intergers for n for which (z*)^n is real, can I get the answer simply by letting '-npi/4 = 0, pi, 2pi, 3pi,...' ?

    Thanks in advance!
    \displaystyle\overline{z}^n = 2^{\frac{3n}{2}}e^{i\left(-\frac{n\pi}{4}}\right)}=2^{\frac{3n}{2}}\left(\cos \frac{n\pi}{4}-i\sin\frac{n\pi}{4}\right)

    Since \displaystyle\overline{z}^n is purely imaginary,

    \cos \frac{n\pi}{4}=0\Rightarrow -\frac{n\pi}{4}=2k\pi\pm\frac{\pi}{2}\mbox{ where }k\in Z

    Hope you understood.
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