# How is this possible?

1. Dec 10, 2008

### naggy

How can this be true(z is complex)

z^a = exp(a * ln(z))

but in general ln(z^a) is not equal to a * ln(z)

2. Dec 10, 2008

### jostpuur

Are you mainly interested in the case where $a\in\mathbb{R}$? (Answer would be simpler then)

3. Dec 10, 2008

### jostpuur

Well, anyway, I'll assume that $a\in\mathbb{R}$. Arbitrary $a\in\mathbb{C}$ would result in more calculating, but I don't believe there would be anything highly different.

This equation is usually true, because it is the definition of the left side. First you must decide what argument function $\textrm{arg}:\mathbb{C}\to\mathbb{R}$ you are using. This choice will fix the branch of the logarithm,

$$\log(z) = \log |z| + i\textrm{arg}(z),$$

and then we denote

$$z^a := e^{a\log (z)}.$$

The equation is not necessarily true, if you start using other branches of logarithm later, of course.

This is not true always because of the branch cuts.

$$\log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})$$

$$a\log(z) = a\log |z| + ia\textrm{arg}(z)$$

The equation

$$a\log |z| = \log(e^{a\log |z|})$$

is always true, but

$$a\textrm{arg}(z) = \textrm{arg}(e^{ia\textrm{arg}(z)}) + 2\pi n$$

may need some $n\in\mathbb{Z}$ depending on how the argument function happens to behave with these $a,z$.

Last edited: Dec 10, 2008
4. Dec 10, 2008

### naggy

So... basically it's not true due to the extra 2*pi*n factor at the end?

No that can´t be true. Then the equation would be true for the principal value?

No wait. It's the extra iarg factor in this

$$\log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})$$

If there is no iarg, then there is only log(e^(a*log|z|)) which is |z|^a and this is true for the real numbers since |z| is really replaced by x

Last edited: Dec 10, 2008