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Math Help - Complex Powers of Complex Numbers

  1. #1
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    Complex Powers of Complex Numbers

    Hey,
    I have the complex number (-(root 3)+i)^i. and using complex number mathematics I have put the complex number into exponential form: 2^i * e^(-5pi/6).
    However, this is not in exponential form as e^(-5pi/6) is not an imaginary number. Therefore the 2^i is the imaginary number. How do I get 2^i into exponential form using natural logs?

    Thanks
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  2. #2
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    Quote Originally Posted by littlea86 View Post
    I have the complex number (-(root 3)+i)^i. and using complex number mathematics However, this is not in exponential form as e^(-5pi/6) is not an imaginary number.
    Here is the rule for complex exponentiation: z^w=\exp(w\log(z)).
    So (-\sqrt{3}+i)^i=\exp(i\log(-\sqrt{3}+i).
     \log(-\sqrt{3}+i)=ln(2)+ i \left( {\frac{{ - 5\pi }}{6} + 2n\pi }).
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  3. #3
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    2^i= e^{ln(2^i)}= e^{i ln(2)}

    (-\sqrt{3}+ i)^i= re^{i\theta} with r= e^{-5\pi/6}, as you say, and \theta= ln(2).
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    Thank you so much for all your help
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  5. #5
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    Quote Originally Posted by littlea86 View Post
    Hey,
    How do I find (-(root3)+i)^i, expressing the answer in trigonometric form using exact values?

    I have aready got the equation into the form 2cis5pi/6 but now i have to put it to the power of i (2cis5pi/6)^i, what do i have to do now?

    Thanks
    \displaystyle z^{\alpha}=e^{\alpha \text{Ln} z}=e^{i \text{Ln} (-\sqrt{3}+i)}

    \text{Ln}=\ln|z|+i\text{Arg}(z)

    e^{-\text{Arg}(z)}\left(\cos(\ln(z))+i\sin(\ln(z))\rig  ht)

    \displystyle |z|=\sqrt{\left(\sqrt{3}\right)^2+1^2}=2, \ \ \text{Arg}(z)=\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{5\pi}{6}
    Last edited by dwsmith; March 15th 2011 at 12:41 PM. Reason: Added more steps and fixed a mistake
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