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Thread: Complex number

  1. #1
    Junior Member
    Aug 2009

    Complex number

    Hi everybody,

    How to show that: \forall a \in \mathbb{C}*, \exists z_0 \in \mathbb{C} such as: a=e^{z_0}, and if e^z=a then \exists k \in \mathbb{Z}: z=z_0+2k\pi, i don't know how to show that, can you help me please???

    And thanks anyway.
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  2. #2
    Newbie driegert's Avatar
    Feb 2010
    Kingston, Ontario
    Hey there.

    So just as a reminder we know that you can represent a complex number z as:

    z = x + iy in rectangular coordinates or
    z = re^{i \theta} where  r = \sqrt{x^{2} + y^{2}} and \theta = tan^{-1}(y/x) in polar coordinates (you have to adjust theta for the quadrant depending on the signs on x and y).


    e^{z} = e^{x + iy} = e^{x}e^{iy} By Euler's formula we can change the second e like this:

    e^{x}e^{iy} = e^{x}(cos(y) + isin(y))

    So actually, now that I've done that.. it's not going to work out at all unless you actually meant:

     z = z_{0} + 2ik \pi

    So you need an 'i' term with the 2k \pi term.

    So yeah... If there wasn't a typo, perhaps some of my rambling will jog your memory.
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