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#redirect [[exponentiation#Complex exponents with complex bases]]
{{db|still no context to this article, so I believe this can be speedied under A1. Also, recreation of earlier speedily deleted article, so please protect}}
{{hangon}}
{{Importance|date=December 2006}}
{{incoherent}}<!--I can't find a better tag to say "This is in jargon; please rewrite in English prose.-->
At the beginning we let take X ,Y ,z as complex numbers :-
:<math> x = r e^{i\theta} \,</math>
:<math>y = k e^{i\varphi}= A + B i</math>
:<math>Z = q e^{i \phi} \,</math>
Then :
:<math> x^{y} = (r e^{i\theta})^{(k e^{i\varphi})} = e^{k[(\cos(\varphi) \ln(r)+ \sin(\varphi) \theta)+ i (\cos(\varphi) \theta - \sin(\varphi)\ln(r))]} \,</math>
==The Proof==
if
:<math> x^{y} = z \,</math>
then x and y are known and we are trying to find z . we change to logarthim form and we get :-
:<math>\log_z(x) = {\ln(x)\over\ln(z)} = y = A + Bi \,</math>
:<math>= {\ln(r)+ i \theta \over\ln(q)+ i \phi} \, </math>
:<math>= {(\ln(q) \ln(r)+ \theta \phi)+i(\phi \ln(r) - \theta \ln(q) ) \over (\ln(r))^{2}+ (\theta)^{2}}</math>
We going to say that <math>g = (\ln(r))^{2}+ (\theta)^{2} </math> and we continue ...
:<math>y = {(\ln(q) \ln(r)+ \theta \phi)+i(\phi \ln(r) - \theta \ln(q) ) \over g } </math>
:<math>= {(\ln(q) \ln(r)+ \theta \phi) \over g} + i {(\phi \ln(r) - \theta \ln(q) )\over g}= A + Bi </math>
Then that means .....
:<math> \ln(q) \ln(r)+ \theta \phi = A g \,</math> ----> 1
:<math> - \theta \ln(q) + \phi \ln(r) = B g \,</math> ------>2
: <math>\delta = \begin{bmatrix}
\ln(r) & \theta \\
\theta & -ln(r) \\
\end{bmatrix} = - (\ln(r)^{2} - (\theta)^{2} = - g</math>
:<math>\delta(\ln(q)) = \begin{bmatrix}
A g & \theta \\
B g & -\ln(r) \\
\end{bmatrix} = -g \begin{bmatrix}
-A & \theta \\
-B & -\ln(r) \\
\end{bmatrix} = -g [ A \ln(r) + B \theta ]\,</math>
: <math>\delta(\theta) = \begin{bmatrix}
\ln(r) & A g \\
\theta & B g \\
\end{bmatrix} = -g \begin{bmatrix}
\ln(r) & -A \\
\theta & -B \\
\end{bmatrix} = -g [ - A \theta + B \ln(r)] </math>
<math>\ln(q) = { \delta(\ln(q)) \over \delta } = A \ln(r) + B \theta </math>
<math> q = e^ {A \ln(r) + B \theta} \,</math>
<math>\theta = {\delta(\theta) \over \delta } = A \theta - \ln(r) B</math>
Then we get....
:<math> x^{y} = (r e^{i\theta})^{(A + b i)} = e^{(A \ln(r)+ B \theta)+ i (A \theta - B\ln(r))} \,</math>
we return y to the poler form by sub <math>A=k \cos(\varphi)</math> , <math>sub B=k \sin(\varphi)</math> . then we get....
:<math> x^{y} = (r e^{i\theta})^{(k e^{i\varphi})} = e^{k[(\cos(\varphi) \ln(r)+ \sin(\varphi) \theta)+ i (\cos(\varphi) \theta - \sin(\varphi)\ln(r))]} \,</math>
==Applications==
'''complex number power by real number:'''
then we use the formula with B=0
:<math> x^{y} = (r e^{i\theta})^{(A)} = e^{(A \ln(r)+ 0)+ i (A \theta - 0)} \,</math>
:<math> = r^{A} e^{i (A \theta)} \,</math>
'''real number power by complex number'''
we use the formula with <math>\theta = 0</math>
:<math> x^{y} = (r)^{(A + b i)} = e^{(A \ln(r)+ 0)+ i (0 - B\ln(r))} \,</math>
:<math> = r^a e^{i (-B\ln(r))} \,</math>
[[Category:Complex numbers]]
[[Category:Exponentials]]
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