Complex mappings

Just for a change, an FP3 topic. I’ve been struggling to tutor complex mappings properly (mainly because I’ve been too lazy to look them up), but have finally seen - I think - how to solve them with minimal headache.

A typical question gives you a mapping from the (complex) $z$-plane to the $w$-plane of $w=\frac{z-i}{z}$, and asks what happens to a given line in one plane or the other.

My recipe for approaching this boils down to three steps:

1. Cross-multiply to get rid of the ugly fraction
2. Multiply out the brackets to get a real equation and a complex equation
3. Use the equation of the line to eliminate something you don’t want
4. Eliminate the other variable you don’t want, leaving you with a relation between two variables

Sounds complicated? Let’s see what happens to $y=x$. First, I multiply up the $z$:

$wz=z+i$, or $(u+vi)(x+yi)=x+(y+1)i$

Then expand:

$(ux-vy)+(vx+uy)i)=x+(y+1)i$, giving two equations:

$ux-vy=x$ (1) and; $vx+uy=y+1$ (2)

Since $y=x$, substituting into (1) and dividing by $x$ gives: $u-v=1$; the substitution gives you a straight line.

Another? How about $x+y+1=0$? I have good news: we can start from (1) and (2) rather than working it all out again. We also know $y=(-1-x)$, so:

$ux+v(1+x)=x$ $vx-u(1+x)=-x$

Group the $x$s together and divide:

$x(u+v-1)=-v$ $x(v-u+1)=u$

$\frac{u+v-1}{v-u+1}=\frac{-v}{u}$

Cross-multiply:

$u(u+v-1)=-v(v-u+1)$ $u^2+uv-u=-v^2+vu-v$ $u^2+v^2-u+v=0$

… so we get a circle.