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Why is $|z^n| = |z|^n, z \in \mathbb{C}$.

I have up to:

$$|z|^n = ((a^2+b^2)^{1/2})^n = ((a^2+b^2)^n)^{1/2}$$ and $$|z^n| = |(a+bi)^n| = |\sum_{k=0}^{n}{{n}\choose{k}}a^k(bi)^{n-k}| = \sqrt{\sum_{k=0}^{n}({{n}\choose{k}}a^k(bi)^{n-k})^2} = \sqrt{\sum_{k=0}^{n}{{n}\choose{k}}^2(a^{k^2})((bi)^{n-k})^2} = \sqrt{\sum_{k=0}^{n}{{n}\choose{k}}^2(a^{k^2})((bi)^{n-k})^2} = \sqrt{\sum_{k=0}^{n}{{n}\choose{k}}^2a^{2^k}b^{2^{n-k}}}(-1)^{n-k}$$

I can see $$\sqrt{(a^2+b^2)^n}$$ coming from the last equality, but is this track possible?

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    $\begingroup$ Just prove it for two complex numbers, i.e, $|zw| = |z||w|$. Then apply induction. Why go through this mess? $\endgroup$ – Shalop Nov 10 '15 at 23:08
  • $\begingroup$ Oh wow. Thank you. Yes that is perfect. $\endgroup$ – user3517214 Nov 10 '15 at 23:09
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By properties of absolute values we have $$|z_1\cdot z_2\cdots z_n|=|z_1|\cdot|z_2|\cdots|z_n|$$ So if $z_1=z_2=\dots=z_n$, then $$|z\cdot z\cdots z|=|z^n|=|z|\cdot|z|\cdots|z|=|z|^n$$

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  • $\begingroup$ Works, like Shalops answer, for me. Thank you. $\endgroup$ – user3517214 Nov 10 '15 at 23:19
  • $\begingroup$ I don't know, this feels like begging the question… I think you should explicitly prove the more general result before invoking it in the context of an exercise like this. $\endgroup$ – Aaron Golden Nov 10 '15 at 23:36
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It's easier if you represent $z$ in polar, rather than cartesian form. $z = r e^{i\theta}$, so $z^n = r^n e^{i n \theta}$. $\left|z^n\right|=\left|r^n e^{i n \theta}\right| = r^n = \left|r e^{i \theta}\right|^n = \left|z\right|^n$.

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