Complex $ \sqrt[n]{\cdot} $

Question

Define $ b^{z}:=e^{\log\left(b\right)\cdot z} $, as a function of 2 complex variables $ (b,z) \in D(1,1)\times \mathbb{C} $

Where $log $ here is defined only on the disk $ D(1,1) $ and given by $$ \log\left(z\right)=\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\left(z-1\right)^{k} $$

Now, given a positive integer $ n $, I want to prove that there exist a unique function $ f=\sqrt[n]{\cdot}:D\left(1,1\right)\to D\left(1,1\right) $

such that $$ \left(f\left(z\right)\right)^{n}=z,\thinspace\thinspace\thinspace and\thinspace\thinspace\thinspace\thinspace\thinspace f\left(1\right)=1 $$

Here's what I have tried:

Define $$ f\left(z\right)=e^{\frac{1}{n}\log\left(z\right)} $$

This will be our function. And indeed $ f\left(1\right)=e^{\frac{1}{n}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k}\left(1-1\right)^{k}}=e^0=1 $

Next, I want to prove that indeed $(f(z))^n=z$.

But Im not sure why would it even be true.

Those are the steps Im not sure about:

$ \left(f\left(z\right)\right)^{n}=\left(e^{\frac{1}{n}\log\left(z\right)}\right)^{n}\underset{?}{=}e^{\frac{n}{n}\log\left(z\right)}\underset{?}{=}z $

This power rule is true in this complex case? also, I know in general $ e^{\log\left(z\right)}\neq z $, but in this particular case where we defined $ log $ as the power series over the disk $(1,1) $ , is it true?

If indeed those steps are true, how can I show that this function is unique?

Thanks in advance, any help would be appreciated.

Answer 1

The function $$ \log(z)=\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\left(z-1\right)^{k} $$ is holomorphic in $D(1, 1)$ as the value of a power series with convergence radius $1$. It satisfies $\log(1) = 0$ and $\log(z)' = 1/z$. It follows that $e^{\log(z)} = z$ in $D(1, 1)$, as demonstrated in Primitive function of $ 1/z $, complex analysis.

The complex exponential function satisfies $e^{z+w} = e^z e^w$ for all complex numbers $z, w$, and consequently $e^{nz} = (e^z)^n$ for $z \in \Bbb C$ and positive integers $n$.

So your calculation is correct: $f(z)=e^{\frac{1}{n}\log(z)}$ is holomorphic in $D(1,1)$, it satisfies $f(1) = 1$ and $$ f(z)^n = \left( e^{\frac{1}{n}\log(z)}\right)^n = e^{\log(z)} = z \, . $$

Uniqueness: $f$ is the only holomorphic function in $D(1, 1)$ satisfying $f(z)^n = z$ and $f(1) = 1$: Assume that $g$ is another such function. Then $h=f/g$ is holomorphic in $D(1, 1)$ and satisfies $$ \tag{*} 0 = h(z)^n -1 = (h(z)-1)(h(z)-\omega)\cdots(h(z)-\omega^{n-1}) $$ with the primitive $n$-th root of unity $\omega = e^{2\pi i/n}$. It follows that the zeros of at least one of the factors on the right-hand side of $(*)$ must have an accumulation point in $D(1,1)$, and then the identity theorem for holomorphic functions implies that that factor is identically zero.

So we have shown that $f(z)/g(z) = \omega^k$ for some $k$, and since $f(1)=g(1)=1$ it follows that $f(z) = g(z)$.

Alternatively one can show that $f(z)^n = z$ implies $$ f'(z) = \frac{f(z)}{nz} $$ and use this to prove that if $g$ is another such function then the derivative of $h=f/g$ is zero.

Remark: One can also show that $f$ is the only continuous function in $D(1, 1)$ satisfying $f(z)^n = z$ and $f(1) = 1$.