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Find a value for $2^{-4i}$? I have no idea what to do or how to find the value. My thoughts are that I should use logarithm. Can someone please show me how to solve this?

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  • $\begingroup$ Some context would be helpful here in order to present an answer which would most likely agree with your background. Was this a question which came up in a class? If so, what subject? $\endgroup$ – process91 Dec 3 '14 at 1:15
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    $\begingroup$ Hint: For complex numbers $a$ and $b$, we define $a^b = \exp(b\log a)$. $\endgroup$ – Tacet Dec 3 '14 at 1:18
  • $\begingroup$ You can obtain one value with this formula: $\cos(4 \log 2) - \sin(4 \log 2) i$. $\endgroup$ – James47 Dec 3 '14 at 22:50
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I'll outline one way of approaching this problem. You should first be aware of Euler's formula: $$ e^{ic}=\cos c + i\sin c. $$ Now note that we may write $$ e^{b+ic} = e^be^{ic}=e^b(\cos c + i\sin c). $$ What may be of use to you is that for a real number $a$ you can define $a^{b+ic}$ by letting $a=e^{\ln a}$; doing this will give you the following: \begin{align} a^{b+ic} &= e^{\ln a(b+ic)}\\[0.5em] &= e^{b\ln a + i(c\ln a)}\\[0.5em] &= e^{b\ln a}(\cos(c\ln a)+i\sin(c\ln a))\\[0.5em] &= a^b(\cos(c\ln a)+i\sin(c\ln a)) \end{align} Now apply this to your problem where $a=2, b=0, c=-4$. Then you get that $$ 2^0(\cos(-4\ln 2)+i\sin(-4\ln 2))=\cos(-4\ln 2)+i\sin(-4\ln 2). $$

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Use the equation $e^{ix}=\cos{x}+i\sin{x}$:

$2^{-4i}=(e^{\ln2})^{-4i}=e^{(-4\ln2)i}=\cos{(-4\ln2)}+i\sin{(-4\ln2)}$

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Simply writing,

$2^{-4i} = e^{-4i*\log(2)} = \cos(4\log2) - i\sin(4\log2)$

You can get the required solution.

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