$z_1 = 1 + i$ and $z_2 = -1 + i$
I am told:
$w = \dfrac{az + b}{z + d}$ where $z \not= -d$
Where a, b and d are complex numbers, maps the complex number $z$ onto the complex number $w$. Given that $z_1$ and $z_2$ are invariant under this transformation and that $z = 0$ maps to $w = i$ find the values of a, b and d.
I first tried substituting $z_1$ and $z_2$ into the equation:
$1 + i = \dfrac{a(1+i)+b}{(1+i)+d}$
$\therefore (1+i)(1+i+d) = a+ai+b$
$\therefore (d - a-b) +i(2+d-a) = 0$
For real:
$d-a-b = 0 \Rightarrow a = d-b$
For imaginary:
$2+d-a = 0 \Rightarrow a=2+d$
If I equate these:
$2+d = d-b \Rightarrow b=-2$
Subbing this into $a=d-b$ gives $a=d+2$ so I don't think there's anymore I can do here.
Doing the same thing for $z_2$ gives:
$(a-d-b)+i(d-a-2) = 0$
For real:
$a = d+b$
For imaginary:
$a = d - 2$
So for $z_1$ I have that $a=d+2$ and for $z_2$ I have that $a=d-2$. So I'm guessing I've gone wrong somewhere or misunderstood how this works? Can anyone point me in the right direction?
Thank you :)
You forgot that $a,b,d$ are complex numbers. So the splitting into imaginary and real components is wrong, plus, you don't have enough unknowns.
What you need are three complex equations for three unknowns - excatly what you have. Equations (linearized in $a,b,d$) are of the form
$$-az-b+dw=-zw$$
Plug in all three conditions:
$$-(1+i)a-b+(1+i)d=-(1+i)^2$$ $$-(-1+i)a-b+(-1+i)d=-(-1+i)^2$$ $$-b+id=0$$ You can use $b=id$ immediately, put it into the first two and simplify: $$-(1+i)a+d=-2i$$ $$-(-1+i)a-d=2i$$ Add equations: $$-2ia=0\rightarrow a=0$$ This gives $$d=-2i$$ and $$b=2$$
Your conformal mapping is
$$w=\frac{2}{z-2i}$$
Verify: $$w(1+i)=\frac{2}{1+i-2i}=\frac{2}{1-i}=\frac{2(1+i)}{2}=1+i$$ $$w(-1+i)=\frac{2}{-1+i-2i}=\frac{2}{-1-i}=\frac{2(1-i)}{-2}=-1+i$$ $$w(0)=\frac{2}{-2i}=i$$