Distributing powers on complex numbers Can I not distribute powers on complex numbers as I do with real numbers?
For example:
Consider
$$\left(\frac{1 + i}{1-i}\right)^n = 1$$
Distributing powers as in real numbers:
$$(1+i)^n = (1 - i)^n$$
Taking log both sides and eliminating $n$ I am left with
$2i = 0$ which is obviously incorrect. So I am lead to believe that I can't distribute powers in complex numbers?
 A: Distributing powers is fine, as long as the exponent is an integer.  Your problem is with the logarithms.
In fact the easiest way (I'm assuming that what you want is to solve for $n$) is to simplify the bit in brackets first:
$$\frac{1+i}{1-i}=\frac{1+i}{1-i}\frac{1+i}{1+i}=\frac{2i}{2}=i\ .$$
Therefore
$$\Bigl(\frac{1+i}{1-i}\Bigr)^n=1\quad\Leftrightarrow\quad
  i^n=1\quad\Leftrightarrow\quad
  n=4k\ ,\ k\in{\Bbb Z}\ .$$
A: Hint: Try writing: $$\frac{1+i}{1-i} = \frac{1+i}{1-i}\cdot\frac{1+i}{1+i}$$
A: Two points:


*

*The complex $\log$ function is much more complicated than the real $\log$. That is because $a^x=a^y$ does not imply that $x=y$ in the complex numbers.

*$$\frac{1+i}{1-i} = i$$
A: You can distribute powers. The problem in your reasoning is that you worked like if you were dealing with real numbers when taking logarithms.
As you can see here: http://en.wikipedia.org/wiki/Complex_logarithm the principal value of the complex logarithm is defined as:
$$log(z)=log(r)+i\theta$$
If $z=re^{i\theta}$.
A: Note that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$, which means $$\frac{1+i}{1-i} = e^{i \pi/2} = i$$
