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?
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2No, it doesn't change. The answer is indeed $12^{500}$. – José Carlos Santos Jun 12 '19 at 22:35
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1Welcome to Maths SX! It is perfectly correct, since the modulus is multiplicative. – Bernard Jun 12 '19 at 22:35
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Hello and welcome to MSE! Please write mathematical expressions using mathjax in the future. I have edited your question to be formatted correctly. – JMJ Jun 12 '19 at 22:41
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MathJax tutorial – gen-ℤ ready to perish Jun 13 '19 at 00:57
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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, $\lvert z\rvert = \sqrt{12}$.
gen-ℤ ready to perish
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