$\lim_{n\to \infty} n(\sqrt[n]{z} - 1)$

Question

Let $z\in \mathbb C $ with $\vert arg(z)\vert < \pi$. Let $\sqrt[n]{z}$ be the principal n-th root of z. How can one calculate this limit?

$$\lim_{n\to \infty} n(\sqrt[n]{z} - 1)$$

I'm sure it also depends on $\vert z \vert$..I've tried expressing it into a power series but I don't think I was doing it correctly or I haven't really gotten anywhere...

Thank you!

Answer 1

For $|\arg(z)| < \pi$, we have :

$$\begin{array}{rcl} z^{\epsilon} & = & \displaystyle\exp\big(\epsilon \log(z)\big) \\ \\ & = & \displaystyle 1+\epsilon \log(z) + \frac{(\epsilon \log(z))^2}{2!}+\cdots\end{array}$$

Then :

$$\dfrac{z^{\epsilon} - 1}{\epsilon} = \log(z) + \epsilon \frac{\log(z)^2}{2!}+\cdots$$

It's clear now that :

$$\lim_{n\to \infty} n(\sqrt[n]{z} - 1) = \lim_{\epsilon\to 0} \dfrac{z^{\epsilon} - 1}{\epsilon} = \log(z)$$

Answer 2

If $z=\rho(\cos\theta+i\sin\theta)$, then$$\sqrt[n]z=\sqrt[n]\rho\left(\cos\left(\frac\pi n\right)+i\sin\left(\frac\pi n\right)\right)$$and therefore$$n\left(\sqrt[n]z-1\right)=n\left(\sqrt[n]\rho\cos\left(\frac\pi n\right)-1\right)+in\sqrt[n]\rho\sin\left(\frac\pi n\right).$$So, you can solve the problem computing the limits$$\lim_{n\to\infty}\frac{\rho^{1/n}\cos\left(\frac\pi n\right)-1}{\frac1n}\text{ and }\lim_{n\to\infty}\frac{\rho^{1/n}\sin\left(\frac\pi n\right)}{\frac1n},$$each of which is simply the derivative at $0$ of an appropriate function.

Answer 3

For the correct branch of $\log$ and using the equivalence of $\exp(\cdots) - 1$: $$ \lim_{n\to\infty}n(\sqrt[n]{z} - 1) = \lim_{n\to\infty}n(e^{(\log{z})/n} - 1) = \lim_{n\to\infty}n(\log{z})/n = \log z. $$