How to calculate $\displaystyle\lim_{n\to\infty}\sqrt[n]{|1-z^n|}$ with $z\in\mathbb{C}$ and $|z|\ne 1$?

Question

As stated in the title: How does one calculate $$\displaystyle\lim_{n\to\infty}\sqrt[n]{|1-z^n|}$$ with $z\in\mathbb{C}$ and $|z|\ne 1$?

Answer 1

Hint:

For any real $\;x>0\;$ we have

$$\lim_{n\to\infty}\sqrt[n]x=1$$

and from here:

$$\left(a_n>0\;\;and\;\;\lim_{n\to\infty}a_n=L>0\;,\;\;L\in\Bbb R\right)\implies \lim_{n\to\infty}\sqrt[n]{a_n}=1$$

I think the last line's true also when $\;L=0\;$ , but it definitely is not true if the limit is $\;\infty\;$ , and that's why the restriction $\;L\in\Bbb R\;$ .

Answer 2

Since $\lvert z\rvert\ne1$ so $z=re^{i\theta}$ where $r\ne1$ and then $$\sqrt[n]{|1-z^n|}=\left((1-r^n\cos n\theta)^2+r^{2n}\sin^2n\theta\right)^{1/2n}=\left(1+r^{2n} -2r^n\cos n\theta\right)^{1/2n}$$ and then we have two cases

• if $r=|z|>1$ then $1+r^{2n} -2r^n\cos n\theta\sim_\infty r^{2n}$ and then

$$\boxed{\displaystyle\lim_{n\to\infty}\sqrt[n]{|1-z^n|}=r=|z|\;\;}$$

• if $r=|z|<1$ then $$\log\left(1+r^{2n} -2r^n\cos n\theta\right)\sim_\infty -2r^n\cos n\theta$$ and then $$\boxed{\displaystyle\lim_{n\to\infty}\sqrt[n]{|1-z^n|}=1}$$

Answer 3

The case where $|z|<1$ is essentially answered by @DonAntonio. You can also prove it directly using the squeeze theorem and triangle inequality. (Try to do it before looking under the spoiler box!)

$1 - r^n <|1-z^n| < 1+ r^n$, where $r = |z|$.

Since $\sqrt[n]{ }$ is an increasing function,

$\sqrt[n]{1-r^n} < \sqrt[n]{|1-z^n|} < \sqrt[n]{1+r^n}$. Taking the limit as $n$ to $\infty$, along with the squeeze theorem, we get that the limit is equal to 1.

For $|z| > 1$, you should get a different limit than in the case of $|z|<1$. This makes sense intuitively: if $|z|>1$, then $z^n$ gets very large, and so $|1 - z^n|$ looks a lot more like $|z|^n$. (On the other hand, if $|z|<1$, then $z^n$ gets very small, so $|1 - z^n|$ is a lot like 1, which matches with DonAntonio's answer.)

See if you can figure out what to do for $|z|>1$ before you look at my answer which is hidden in the spoiler box:

$\sqrt[n]{|1-z^n|} = |z| \sqrt[n]{|1-(\frac{1}{z})^n|}$. Since $|1/z|<1$ in this case, you should be able to compute the limit based on the case $|z|<1$.