Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 8A (Algebra and Geometry)

Write a straight line in Cartesian form $Ax+By+C=0$, where $A,B,C$ are real and $(A,B)\ne (0,0)$. With $z=x+iy$, set $c=(A+iB)/2$ and $r=C$. Using the complex conjugate,Thus the real linear equation of a line in the complex plane isConversely, writing $c=u+iv$ gives $2ux+2vy+r=0$, an ordinary straight line with normal vector $(u,v)$. The nonzero condition on $c$ matters: $c=0$ would give either the whole complex plane or the empty set.

The quotient is defined only when $z+\overline{u}\ne 0$. Its unit-modulus condition then gives $|z+u{|}^2=|z+\overline{u}{|}^2$. On expanding with complex conjugates,Hence at least one imaginary part is zero, and either $z$ or $u$ is real. Conversely, either reality condition makes the two squared moduli equal; provided the denominator is nonzero, the quotient has modulus one. No conclusion is drawn from an undefined quotient.

For real $x$ at which the denominator is nonzero, mapping to the unit circle gives $|ax+b{|}^2=|cx+d{|}^2$. There are infinitely many such real $x$, so the resulting quadratic polynomial identity has all coefficients zero:If $c=0$, the first equality gives $a=0$, contradicting the nonzero determinant $ad-bc$. Thus $a,c\ne 0$. Put $\lambda =a/c$, $k=b/a$ and $\ell =d/c$. The equalities becomeTwo complex numbers with equal real parts and equal moduli have imaginary parts equal in magnitude. Consequently $\ell =k$ or $\ell =\overline{k}$. The first option would give $ad-bc=ac(\ell -k)=0$, so it is excluded. The second givesMoreover $k$ must be nonreal, since otherwise the two options coincide and the map would be constant. This proves the Möbius transformations from the real axis to the unit circle representation directly from coefficients.

Conversely, for real $x$, the numbers $x+k$ and $x+\overline{k}$ are complex conjugates, and their moduli agree. Nonreal $k$ ensures the denominator is nonzero on the real line; multiplication by $\lambda$ preserves modulus. At the point at infinity on the Riemann sphere, the map takes the value $\lambda$. Thus the representation also respects the extended real axis.