Why Does (e^(1+i*2*pi))^(1+i*2*pi) Not Equal e^((1+i*2*pi)(1+i*2*pi))?

[eltodesukane]

(e^{1+i*2*pi})^{1+i*2*pi}=e^{(1+i*2*pi)(1+i*2*pi)}
why is this false (according to Wolfram Alpha) ?

 

[pbuk]

Because the exponent laws are only true for real numbers.

 

[DrewD]

$\begin{align} (e^{1+i2\pi})^{1+i2\pi}&=(e\cdot e^{i2\pi})\cdot(e\cdot e^{i2\pi})^{i2\pi}\\ &=(e\cdot1)(e\cdot1)^{i2\pi}=e \end{align}$

but

$\begin{align} e^{(1+i2\pi)(1+i2\pi)}&=e^{1-4\pi^2+i4\pi}\\ &=e^{1-4\pi^2}\cdot1\neq e \end{align}$

The complex exponential is defined (I recall) so that $e^{z_1+z_2}=e^{z_1}e^{z_2}$. Beyond that the algebra needs to be derived. This exponent "law" is only a law for real numbers.

 

[WWGD]

Because you have to choose a branch of log to define $z^a$ when z is complex with non-zero imaginary part and a is not a rational number. Take, e.g.,

(e^iπ/4))^iπ/4 .

Let z =e^iπ/4 . Then

(e^iπ/4))^iπ/4=z^iπ/4:=e^zlog(iπ/4) . Using the

$Logz$ branch, i.e., the main branch (because the argument of iπ/4 is precisely π/4 in the main branch Logz), we have

Log(iπ/4)=ln(1)+iπ/4=iπ/4 and then it works out:

(e^iπ/4)^iπ/4=(cos(π/4)+isin(π/4))^iπ/4= $( \sqrt 2/2+isin \sqrt 2/2)^{i\pi/4}= ( \sqrt 2/2+isin \sqrt 2/2)^{ln(1)+i\pi/4} =e^{-\pi^2/16}$

. But if you want to use $2 \pi$ as argument, then you need to work in a branch where $2\pi$ makes sense.