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Prove that $\frac{1}{x}=\bar{x}$ if and only if $1=|x|$.

Here's what I have so far:

$\bar{x}$ is the conjugate of $x$. Multiplying both sides of $\frac{1}{x}=\bar{x}$ by $x$ yields $1=x \cdot \bar{x} = -x^2$.

Taking the square root of both sides $1=-x^2$ can be written as $1=|x|i$.

Am I on the right track? If yes, what should I consider from here? If not, what should I reconsider? Thank you!

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    No. $x\bar{x}=|x|^2$. – 2017-01-27
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    What is the connection with the title ? – 2017-01-27

4 Answers 4

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Note that $x\cdot\overline{x}=|x|^2\neq -x^2$, so you can instead proceed from here by just taking square roots. You can then prove this in the opposite direction by reversing your previous steps

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    Taking square roots isn't required here, and anyway taking square roots of complec numbers is a minefield of errors... – 2017-01-27
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    @5xum the modulus is a real number.... – 2017-01-27
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    Yeah, but $x$ and $\overline x$ aren't... – 2017-01-27
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    @5xum I only took a square root of the modulus? – 2017-01-27
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    So, you only took the square root of one side of an equation? How does that work? – 2017-01-27
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    $\sqrt{|x|^2}=\sqrt{1}$. I don`t see the problem? @5xum – 2017-01-27
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For $x\ne 0$,

$$\frac1x=\bar x\iff1=x\bar x\iff1=|x|^2\iff1=|x|$$ (last equivalence because $|x|$ is known to be positive).

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We have $$\frac {1}{x} = \bar x $$ $$\Rightarrow \frac {1}{x}(x) = \bar x (x) $$ $$\Rightarrow 1 = x \bar x = |x|^2$$

May I let you conclude? Hope it helps.

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$\Rightarrow$ Assume that $\frac{1}{x}=\overline{x}$. Then $1=x\overline{x}$. Thus, $$|x|^2=x\overline{x}=1.$$ This implies that $|x|=1$.

$\Leftarrow$ Assume that $|x|=1$. Clearly, $x\neq 0$. Then $$x\overline{x}=|x|^2=1,$$ and we get $$\frac{1}{x}=\overline{x}.$$