# Straight Lines in Complex Plane/ Inversion Function

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 $$\mathbb{C}$$.

• 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. Commented Jan 5, 2022 at 0:08
• 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. Commented Jan 5, 2022 at 0:11

## 1 Answer

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.

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