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There are some ways to define the imaginary axis.Some are obvious, like $\space Re(z)=0 \space$ others not.

I set up a condition that I think defines the imaginary axis.

Let $x$ be a real number, and $z$ a complex number.So,


If $\space z=x+yi$, where $x$ and $y$ are real numbers, ones get

$|x+yi+x|=|x+yi-x| \Leftrightarrow|2x+yi|=|yi| \Leftrightarrow$

$\sqrt{(2x)^2+y^2}=\sqrt{y^2} \Leftrightarrow 4x^2+y^2=y^2 \Leftrightarrow$

$4x^2=0 \Leftrightarrow x=0$

This is correct?Thanks

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Yes, it is. A number which has the same distance from $\pm x$, $x \in \mathbb R$ has to be purely imaginary. – martini May 12 '12 at 19:19
up vote 3 down vote accepted

The characterization of the imaginary axis is correct. In fact, the following are equivalent for a complex number $z$:

  1. $z$ is purely imaginary;
  2. there is a non-zero real number $x$ such that $|z+x|=|z-x|$;
  3. $|z+x|=|z-x|$ for all $x\in\Bbb R$.

Your proof shows only that $z$ is purely imaginary iff $|z+x|=|z-x|$ where $x$ is specifically the real part of $z$. More generally, let $z=u+iv$ and $x\in\Bbb R\setminus\{0\}$ be such that $|z+x|=|z-x|$. Then $$|(u+x)+iv|=|(u-x)+iv|\;,$$ so $$(u+x)^2+v^2=(u-x)^2+v^2\;,$$ and (after a little algebra) $ux=-ux$. Since $x\ne 0$, this immediately implies that $u=0$ and $z=vi$.

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