Complex numbers exponential

 ??? 

At the beginning we let take X ,Y as complex numbers :-

${\displaystyle x=re^{i\theta }\,}$

${\displaystyle y=ke^{i\varphi }=A+Bi}$

${\displaystyle Z=qe^{i\phi }\,}$

Then :

${\displaystyle x^{y}=(re^{i\theta })^{(ke^{i\varphi })}=e^{k[(\cos(\varphi )\ln(r)+\sin(\varphi )\theta )+i(\cos(\varphi )\theta -\sin(\varphi )\ln(r))]}\,}$

The Proof

if

${\displaystyle x^{y}=z\,}$ 

then x and y are known and we are trying to find z . we change to logarthim form and we get :-

${\displaystyle \log _{z}(x)={\ln(x) \over \ln(z)}=y=A+Bi\,}$ 

${\displaystyle ={\ln(r)+i\theta \over \ln(q)+i\phi }\,}$ 

${\displaystyle ={(\ln(q)\ln(r)+\theta \phi )+i(\phi \ln(r)-\theta \ln(q)) \over (\ln(r))^{2}+(\theta )^{2}}}$ 

We going to say that ${\displaystyle g=(\ln(r))^{2}+(\theta )^{2}}$  and we continue ...

${\displaystyle y={(\ln(q)\ln(r)+\theta \phi )+i(\phi \ln(r)-\theta \ln(q)) \over g}}$ 

${\displaystyle ={(\ln(q)\ln(r)+\theta \phi ) \over g}+i{(\phi \ln(r)-\theta \ln(q)) \over g}=A+Bi}$ 

Then that means .....

${\displaystyle \ln(q)\ln(r)+\theta \phi =Ag\,}$  ----> 1

${\displaystyle -\theta \ln(q)+\phi \ln(r)=Bg\,}$  ------>2

${\displaystyle \delta ={\begin{bmatrix}\ln(r)&\theta \\\theta &-ln(r)\\\end{bmatrix}}=-(\ln(r)^{2}-(\theta )^{2}=-g}$ 

${\displaystyle \delta (\ln(q))={\begin{bmatrix}Ag&\theta \\Bg&-\ln(r)\\\end{bmatrix}}=-g{\begin{bmatrix}-A&\theta \\-B&-\ln(r)\\\end{bmatrix}}=-g[A\ln(r)+B\theta ]\,}$ 

${\displaystyle \delta (\theta )={\begin{bmatrix}\ln(r)&Ag\\\theta &Bg\\\end{bmatrix}}=-g{\begin{bmatrix}\ln(r)&-A\\\theta &-B\\\end{bmatrix}}=-g[-A\theta +B\ln(r)]}$ 

${\displaystyle \ln(q)={\delta (\ln(q)) \over \delta }=A\ln(r)+B\theta }$ 

${\displaystyle q=e^{A\ln(r)+B\theta }\,}$ 

${\displaystyle \theta ={\delta (\theta ) \over \delta }=A\theta -\ln(r)B}$ 

Then we get....

${\displaystyle x^{y}=(re^{i\theta })^{(A+bi)}=e^{(A\ln(r)+B\theta )+i(A\theta -B\ln(r))}\,}$ 

we return y to the poler form by sub ${\displaystyle A=k\cos(\varphi )}$  , ${\displaystyle subB=k\sin(\varphi )}$  . then we get....

${\displaystyle x^{y}=(re^{i\theta })^{(ke^{i\varphi })}=e^{k[(\cos(\varphi )\ln(r)+\sin(\varphi )\theta )+i(\cos(\varphi )\theta -\sin(\varphi )\ln(r))]}\,}$ 

Applications

complex number power by real number:

then we use the formula with B=0

${\displaystyle x^{y}=(re^{i\theta })^{(A)}=e^{(A\ln(r)+0)+i(A\theta -0)}\,}$ 

${\displaystyle =r^{A}e^{i(A\theta )}\,}$ 

real number power by complex number

we use the formula with ${\displaystyle \theta =0}$ 

${\displaystyle x^{y}=(r)^{(A+bi)}=e^{(A\ln(r)+0)+i(0-B\ln(r))}\,}$ 

${\displaystyle =r^{a}e^{i(-B\ln(r))}\,}$