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Math Help - Equations with i

  1. #1
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    Equations with i

    Hello i have some problems with solving two things connected with complex numbers:


    1)
    |i^{\hat{z}}|=e
    I tried with logarythming both sides, so i get:
    \hat{z}=\frac{1}{ln i}=\frac{1}{i ( \frac{ \pi}{2}+ 2k \pi)}
    and here i stuck...

    2)
    arg( i^{ \hat{z}})= \frac{ \pi}{2}
    Here i even do not have idea how to start
    Last edited by Vermax; October 18th 2009 at 04:49 AM.
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  2. #2
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    Quote Originally Posted by Vermax View Post
    Hello i have some problems with solving two things connected with complex numbers:


    1)
    |i^{\hat{z}}|=e
    I tried with logarythming both sides, so i get:
    \hat{z}=\frac{e}{ln i}=\frac{e}{i ( \frac{ \pi}{2}+ 2k \pi)}
    and here i stuck...

    2)
    arg( i^{ \hat{z}})= \frac{ \pi}{2}
    Here i even do not have idea how to start
    Please post the question exactly as it's worded in wherever you got it from.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Please post the question exactly as it's worded in wherever you got it from.
    Yes sir. So the question is: "Solve the equation". In other words, "find z".
    And sorry for posting this thread in Algebra.
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  4. #4
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    There is something a bit off about this question.
    I have settled on a solution if we restrict i^{\overline{z}} its principle value.
    Consider these: \left| {\exp (z)} \right| = e^{\text{Re}(z)} and  i^{\overline{z}}  = \exp (\overline{z}\log (i)) = \exp (\overline{z}(i(\pi /2))).

    If \left| {i^{\overline{z}}} \right| = e\; \Rightarrow \;\text{Im}(z) = \frac{2}{\pi} and \text{Re}(z) is any real number

    If  \arg (i^{\overline{z}} ) = \frac{\pi}{2} then \text{Re}(z)=(4k+1)~\&~ \text{Im}(z) = \frac{2}{\pi}
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  5. #5
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    Plato used "\overline{z}" rather than "\hat{z}" to get \overline{z}, the standard notation for "complex conjugate".
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  6. #6
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    Thank you Plato, however what is very weird for me that the answer for 2) is:
    z=\frac{1-4l}{1+4k}+iy
    ,where y is a real number, and l,k are integers.
    Does it have any sense?

    According to the first equation, the solution is just simple:
    \overline{z}=\frac{1}{i ( \frac{ \pi}{2}+ 2k \pi)}
    so:
    z=x-\frac{1}{i ( \frac{ \pi}{2}+ 2k \pi)}
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