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.

 

[CellCoree]

This statement is false because according to Wolfram Alpha, the left side of the equation simplifies to 1, while the right side simplifies to e^(-4*pi^2). These two values are not equal, therefore the equation is not true. Additionally, the laws of exponents do not allow for the exponent to be multiplied by another exponent in this way. The correct way to express this equation would be (e^(1+i*2*pi))^(1+i*2*pi) = e^(1+i*2*pi)^2.