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Math Help - Solve e^z - 2 = 2i.

  1. #1
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    Solve e^z - 2 = 2i.

    Hi, could someone help me with this one?

    Find all solutions of e^{z}-2 = 2i



    Thanks.
    Last edited by mr fantastic; March 23rd 2011 at 03:40 AM. Reason: Re-titled.
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  2. #2
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    Define \displaystyle z = x + iy.


    \displaystyle e^z - 2 = 2i

    \displaystyle e^z = 2 + 2i.


    Working on the RHS:

    \displaystyle 2 + 2i = 2(1 + i)

    \displaystyle = 2\sqrt{2}\,\textrm{cis}\,{\left(\frac{\pi}{4} + 2\pi n\right)}, where \displaystyle n \in \mathbf{Z}.


    Working on the LHS:

    \displaystyle e^z = e^{x + iy} = e^x\,\textrm{cis}\,y.


    So \displaystyle e^x\,\textrm{cis}\,y = 2\sqrt{2}\,\textrm{cis}\,{\left(\frac{\pi}{4} + 2\pi n\right)}.

    Clearly \displaystyle e^x = 2\sqrt{2} \implies x = \ln{(2\sqrt{2})} and \displaystyle y = \frac{\pi}{4} + 2\pi n.


    Therefore, \displaystyle z = \ln{(2\sqrt{2})} + i\left(\frac{\pi}{4} + 2\pi n\right).



    Alternatively, just use the fact that \displaystyle \log{Z} = \ln{|Z|} + i\arg{Z}.

    If \displaystyle e^z = 2(1 + i)

    Then \displaystyle z = \log{\left[2(1 + i)\right]}

    \displaystyle = \ln{|2(1 + i)|} + i\arg{[2(1 + i)]}

    \displaystyle = \ln{(2\sqrt{2})} + i\left(\frac{\pi}{4} + 2\pi n\right)} where \displaystyle n \in \mathbf{Z}.
    Last edited by Prove It; March 22nd 2011 at 04:41 PM.
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  3. #3
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    To start with, do the same thing you would if it were a "real numbers" problem. Add 2 to both sides to get
    e^z= 2+ 2i
    and then z= ln(2+ 2i)

    Now, if a complex number, z, is written in "polar form", z= re^{i\theta} then ln(z)= ln(re^{i\theta)}= ln(r)+ i\theta

    So just write 2+ 2i in "polar form". Don't forget that whatever \theta is, adding 2\pi to it doesn't change z but will give different logarithms. That the reason for the "all solutions".
    Last edited by HallsofIvy; March 23rd 2011 at 05:20 AM.
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  4. #4
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    Thanks so much, appreciate it!
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