Under what conditions does $\log a^b=b \log a$ for complex numbers $a,b$ Use the branch of $\log$ with $-\pi< \theta <\pi$.
For solve the problem I need understand the meaning of brach, and the definition of $\log z$ for $z\in \mathbb{C}$.
First of all I understand that given $z\in \mathbb{C} \setminus \lbrace 0 \rbrace $ we define $\log z=\log(|z|)+(arg(z)+2 \pi k)i$ for $k\in \mathbb{Z}$
Too I know that if $a,b\in \mathbb{C}$ we define $a^b=e^{b \log a}$ where $a\neq 0$.
Now for give a correct answer I need know the correct interpretation of the branch of logarithm which I don´t understand.
according to my low skills in complex variable I Imagine that,first I should take the LHS
$$\log a^b=log(e^{b \log a})=b \log a$$ which is true for any $a,b\in \mathbb{C}, a\neq 0$.
Someone can give my a explanation about my mistake with my assumption, give me a step-by step solution or much better give me a intuitive and formally explanation of the meaning of Branch of logarithm
Best Answer
By one definition, $\log$ is an inverse function to $\exp$, i.e. $\log(z)$ should be a number $w$ with $\exp(w) = z$. The trouble is, there are infinitely many $w$ that work: $\exp(2 \pi i) = 1$, so if $\exp(w) = z$ then also $\exp(w + 2 \pi i k) = z$ for any integer $k$. To be able to give a definite meaning to $\log(z)$, we have to make a choice from these infinitely many possibilities. We can do it in such a way that $\log(z)$ is an analytic function on the complex plane with a curve starting at $0$ and going out to complex infinity removed. Such a function is called a branch of the logarithm, and the excluded curve is called a branch cut. If you choose the version of $\log(z)$ with imaginary part in $(-\pi, \pi)$, that's called the principal branch, and its branch cut is the negative real axis.
$a^b$ is defined, for complex numbers, as $\exp(b \log(a))$. Again, in general that's ambiguous, so there are branches to consider. The principal branch uses the principal branch of logarithm.
Now $\log(a^b) = w$ where $\exp(w) = a^b = \exp(b \log(a))$, so $w = b \log(a) + 2 \pi i k$ for some integer $k$. If you're using the principal branch of $\log$ (both for $\log(a^b)$ and $\log(a)$, the imaginary parts of both $w$ and $\log(a)$ must be in $(-\pi, \pi)$. You want $k = 0$ according to your formula, so what you need is that the imaginary part of $b \log(a)$ is between $-\pi$ and $\pi$.