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Thread: For z in the field of complex numbers

  1. October 25th 2012, 01:14 PM #1
    patrickmanning
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    For z in the field of complex numbers

    solve for z^2 = 2i
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  2. October 25th 2012, 01:18 PM #2
    richard1234
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    Re: For z in the field of complex numbers

    Hint: 2i = 2(\cos {\frac{\pi}{2}} + i \sin {\frac{\pi}{2}}) = 2e^{i \frac{\pi}{2}}.
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