$z = 1/w$ transformation for parallel lines $y = x + b$

Question

I am supposed to find the image of the family of parallel lines $ y = x + b $ under the transformation $w = \frac 1 z $.

Attempt:

Replace $x$ and $y$ with $\Re(z)$ and $\Im(z)$, respectively. Then, by the fact that $z = \frac 1 w$, we can say that

$$\Im(\frac 1 w) = \Re(\frac 1 w) + b$$

Since $$\frac 1 w = \frac u {u^2+v^2} -\frac v {u^2+v^2}i$$

we can conclude that the the equation above can be written as

$$-\frac v {u^2+v^2} = \frac u {u^2+v^2} + b$$

However, I'm not sure where to continue from here. I thought that perhaps, we could simplify this further to

$$-\frac v {u^2+v^2} = \frac u {u^2+v^2} + \frac{bu^2+bv^2}{u^2+v^2}$$

and then, write the right hand side as one fraction. However, I'm not sure where else I could go from here.

Answer 1

Note that the line $y=x+b$ can be rewritten as $\Re((1+i)z)=-b$, or:

$$(1+i)z^*+(1-i)z=-2b$$

with $*$ denoting complex conjugation. Now, substitute $z=\frac{1}{w}$:

$$\frac{1+i}{w^*}+\frac{1-i}{w}=-2b\implies -2bww^*=(1+i)w+(1-i)w^*$$

This is close to the equation of a circle in $\mathbb{C}$, so we rearrange it to express it as such by effectively completing the square:

$$ww^*+\frac{1+i}{2b}w+\frac{1-i}{2b}w^*+\frac{2}{4b^2}=\frac{2}{4b^2}$$

$$\left(w+\frac{1-i}{2b}\right)\left(w^*+\frac{1+i}{2b}\right)=\frac{1}{2b^2}$$

$$\left(w+\frac{1-i}{2b}\right)\left(w+\frac{1+i}{2b}\right)^*=\frac{1}{2b^2}$$

$$\left\vert\left(w+\frac{1-i}{2b}\right)\right\vert^2=\frac{1}{2b^2}$$

This expresses the image of the given line as a circle of radius $\frac{1}{b\sqrt2}$, centred at $\frac{-1+i}{2b}$.