Is there any formula to calculate $2^i$ for example? What about $x^z$? I was surfing through different pages and I couldn't seem to find a formula like de Moivre's with $z^x$.
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1You have to read about multi-valued complex functions. There have been many questions about that on MathSE if you look around. http://mathworld.wolfram.com/ComplexExponentiation.html – jdods Jul 14 '16 at 14:06
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1This question occurs very frequently. See my answer to a similar question, for example. – MPW Jul 14 '16 at 14:54
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By definition, for non-rational exponents, $$ x^z=e^{z\log(x)} $$ This definition is fine as far as it goes, but the limitation is on the values of $\log(x)$ for $x\in\mathbb{C}$. Since $e^{2\pi i}=1$, logarithms, as inverses of the exponential function, are unique up to an integer multiple of $2\pi i$.
Usually, when the base is a positive real number, we use the real value of the logarithm, so $$ 2^i=e^{i\log(2)}=\cos(\log(2))+i\sin(\log(2)) $$ However, if $2$ is viewed as a complex number, we might equally well say $$ 2^i=e^{i\log(2)-2k\pi}=e^{-2k\pi}\cos(\log(2))+ie^{-2k\pi}\sin(\log(2)) $$ for any $k\in\mathbb{Z}$.
robjohn
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@GEdgar: Yes. At first, I did not include, "for non-rational exponents", so the "pretty much" was there to recognize there is another definition for rational exponents. When I added "for non-rational exponents", I should have removed the "pretty much". I have edited the answer. – robjohn Jul 14 '16 at 14:54
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As with the reals, you want to define $a^b$ as $e^{b\log a}$. The logarithm has issues in the complex plane (you cannot make it continuous) but these difficulties are not seen by the exponential.
The key is the identity $$ e^{it}=\cos t+i\,\sin t.$$ This allows you to define the exponential of any $z=s+it$ via $$ e^z=e^{s+it}=e^se^{it}=e^s\cos t+i\,e^s\sin t. $$
In your concrete example, you have $$ 2^i=e^{i\,\log 2}=\cos(\log 2)+i\,\sin(\log 2). $$
Martin Argerami
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