TITLE: What's the modulus of $(\sqrt{11}-i)^{1000}$?
QUESTION [6 upvotes]: The modulus of $\sqrt{11}-i$ is $\sqrt{11+1} = \sqrt{12}$ and the modulus of that squared is $\sqrt{144}$ so is the answer $12^{500}$? Or does the pattern change in some kind of way?

REPLY [2 votes]: Take $\newcommand{\e}{\mathrm e}\newcommand{\i}{\mathrm i} z= \sqrt{11}-\i = r\e^{\theta\i}$, where $r=\lvert z\rvert$. Then
$$\begin{align}
z^{1000} &= (r\e^{\theta\i})^{1000} \\
&= r^{1000}\e^{1000\theta\i} \\
\end{align}$$
The modulus of $\e^{1000\theta\i}$ is $1$, so $\lvert z^{1000}\rvert=r^{1000}=\vert z\rvert^{1000}$. As you said, $\lvert z\rvert = \sqrt{12}$.