# 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?

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Logarithms aren't going to work, unless you are very careful with them and understand complex logarithms. – Thomas Andrews Jul 28 '14 at 13:42
Soo basically I can't cancel powers? @ThomasAndrews – Gummy bears Jul 28 '14 at 13:43
Consider $-1^2 = 1^2$. Can you cancel powers? – Théophile Jul 28 '14 at 13:44
Oops... I see now. @Théophile Thanks. Very stupid mistake by me. – Gummy bears Jul 28 '14 at 13:45
In general, you can't cancel powers, even with real numbers. – Thomas Andrews Jul 28 '14 at 13:45

## 5 Answers

Two points:

1. 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.

2. $$\frac{1+i}{1-i} = i$$

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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 $$\Bigl(\frac{1+i}{1-i}\Bigr)^n=1\quad\Leftrightarrow\quad i^n=1\quad\Leftrightarrow\quad n=4k\ ,\ k\in{\Bbb Z}\ .$$

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Hint: Try writing: $$\frac{1+i}{1-i} = \frac{1+i}{1-i}\cdot\frac{1+i}{1+i}$$

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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 $z=re^{i\theta}$.

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Note that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$, which means $$\frac{1+i}{1-i} = e^{i \pi/2} = i$$

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