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Check if an $n \in \mathbb{Z_+}$ exists that satisfies: $$(1+2i)^n = (1-2i)^n$$ Without using the unique prime factorization ring $\mathbb{Z}[i]$.

I know that $(1-2i)^n = \overline{(1+2i)^n}$ so everything comes to prove that $(1+2i)^n \neq \overline{(1+2i)^n}$ for any $n \in \mathbb{Z_+}$

Let say $z = 1+2i$ then I tried to write it as $z^n - \overline{z}^n = 4i(z^{n-1} + z^{n-2}*\overline{z} + z^{n-3}*\overline{z^2} + \cdots + z^2*\overline{z^{n-3}} + z*\overline{z^{n-2}} + \overline{z^{n-1}})$

But this doesn't get me anywhere. Induction probably doesn't fit. The trigonometric form also doesn't help me.

VereX
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1 Answers1

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Think again when you say that the polar form doesn't help.

$1+2i$ has $r = \sqrt5$ and $\theta = \arctan 2$, so $(1+2i)^n$ has $r = \sqrt{5}^n$ and $\theta = (\arctan 2)n$.

$1-2i$ has $r = \sqrt5$ and $\theta = -\arctan 2$, so $(1+2i)^n$ has $r = \sqrt{5}^n$ and $\theta = -(\arctan 2)n$.

Since in both numbers $r$ is the same, then we just need $\theta$ to be the same. However, since by this question, $n \arctan 2$ does not divide $2\pi$ radians, the only integer possible is $n=0$.

Toby Mak
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