What's the modulus of $(\sqrt{11}-i)^{1000}$?

Question

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?

Answer 1

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, $\lvert z\rvert = \sqrt{12}$.