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:

- Cross-multiply to get rid of the ugly fraction
- Multiply out the brackets to get a real equation and a complex equation
- Use the equation of the line to eliminate something you don't want
- 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.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## manmathmo

Hi Colin,

A bit heavy for A-Level of course, but just for interest this reminded me that Charles Walkden has good notes on Mobius maps (complex transformations). The “circles-and-lines to circles-and-lines” thing happens in lecture 3.

http://www.maths.manchester.ac.uk/~cwalkden/hyperbolic-geometry/hyperbolic-geometry.html

He does it using the complex equation for a circle or line, so avoiding considering the real and complex components.

## Colin

Oo! I’ll check that out đź™‚