I am currently studying about the inversion function in complex numbers and i am reading about a proof that says straight lines are mapped to straight lines through the inversion function which is $f(z)=\frac{1}{z}$. The book says that the equation of an arbiary straight line l which does not pass through $(0,0)$ is this: $$Re(\bar\zeta w)=k,$$ where $k\neq 0$. Why is this an equation of a straight line? Sorry if this question is too easy to be answered but i cant figure out why this represents a line in $\mathbb{C}$.

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    $\begingroup$ No need to apologize - a) MSE is a Q&A site and there's plenty of worse questions, b) I didn't find this obvious either. $\endgroup$ Commented Jan 5, 2022 at 0:08
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    $\begingroup$ Any straight line not passing $(0,0)$ can be rotated (i.e. multiplied by a norm-$1$ complex number) to a vertical line $x=k$ (i.e. $\Re (z)=k$), where $k>0$ is the distance between $(0,0)$ and the original line. $\endgroup$
    – Zerox
    Commented Jan 5, 2022 at 0:11

1 Answer 1


Suppose that $w$ is a fixed complex number. $\mathbb{C}$ has an analogue of a dot product, thinking of it as a vector space over itself. This Hermitian inner product is written as $\langle z, w \rangle = \bar{z}w $. This is clearly $0$ if and only if one of the arguments is $0$. However, writing $z = a + bi, w = c + di$, we see $\Re(\bar{z}w) = ac + bd = 0$ when $ac = -bd$, which is the same as $a/b = -d/c$, i.e. these complex numbers are perpendicular as vectors in the real plane. In fact, you can check that $\Re{(\bar{z}}w)$ is nothing more than the familiar inner product on the plane. So for a fixed $z$ or $w$, this is saying to take all vectors that have a given inner product against that fixed vector. The locus of such vectors is a line, being defined by linear equations.

  • $\begingroup$ Very helpful and in that way, way easier to understand. Thank you very much! $\endgroup$ Commented Jan 5, 2022 at 0:33

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