# Algebraic method for calculating Möbius transformations

Change the subject of the w=(az+b)/(cz+d) equation so you get z=(Aw+B)/(Cw+D) where A, B, C, and D are different numbers calculated from a, b, c, d

Then use the equation for the z-locus to give you an equation in w

Put w=u+iv and transform the w-equation into a u-and-v equation

Take w=(iz+1)/(z+i) for an example
Change the subject
wz+wi=iz+1
wz−iz=1−wi
z=(1−iw)/(w−i)

Case 1: |z|=2, i.e. z-locus is a circle of centre O and radius 1
This is the “typical” case: the z-locus is an actual circle – not a line, a “circle at infinity” – and the pole or centre of inversion of the equation, the z-value when the formula “blows up”, is not on the z-locus
|z|=2 ⇒ |(1−iw)/(w−i)|=2
⇒ |1−iw|=2|w−i|
Squaring up, this ⇒ |1−iw|2=4|w−i|2
⇒ |1−i(u+iv)|2=4|u+iv−i|2
⇒ |(1+v)−iu|2=4|u2+(v−1)|2
⇒ (1+v)2+u2=4[(u2+(v−1)2]
⇒ 1+2v+v2+u2=4u2+4v2−8v+4
⇒ 3u2+3v2−10v+3=0
⇒ u2+v2−(10/3)v+1=0
⇒ u2+v2−(10/3)v+(25/9)=(25/9)−(9/9) (completing the square)
⇒ u2+(v−(5/3))2=(16/9)=(4/3)2
⇒ w-locus is a circle with centre (0,(5/3)), and radius 4/3

Case 2: |z|=1, i.e. z-locus is a circle of centre O and radius 1
|z|=1 ⇒ |(1−iw)/(w−i)|=1
⇒ |1−iw|=|w−i|
⇒ |w+i|=|w−i| because dividing through by −i doesn’t change the |…| of a complex number
⇒ w is an equal distance from −i and i
⇒ w is on the real axis, otherwise known as v=0

Case 3: Im(z)=0, i.e. z is on the real axis, or y=0
Im(z)=0 ⇒ Im[(1−iw) / (w−i)]=0
⇒ Im[(1−i(u+iv)) / (u+iv−i)]=0
⇒ Im[((1+v)−iu) / (u+(v−1)i)]=0
⇒ Im[((1+v)−iu).(u−(v−1)i) / (u+(v−1)i) (u−(v−1)i)]=0 (doing complex-number division)
⇒ −u2−(v2−1)=0 (the denominator of the fraction is now u2+(v−1)2, which is real)
⇒ u2+v2=1
⇒ w-locus is a circle with centre 0 and radius 1
Notice that this is the reverse of Case 2

Case 4: Re(z)=0, i.e. z is on the real axis, or x=0
Do similar working to above, except
Re[(1−iw) / (w−i)]=0
so u(1+v)−u(v−1)=0
u+u=0
u=0
w-locus is the imaginary axis.

Case 5: arg(z)=π/4, i.e. z is on a diagonal half-line from the origin at angle π/4 anticlockwise from the positive real axis, and Im(z)=Re(z)
Do similar working to above, except
Re[(1−iw) / (w−i)]=Im[(1−iw) / (w−i)] and both ≥0
u(1+v)−u(v−1)= −u2−(v2−1)
⇒ 2u=−u2−(v2−1) and u≥0
⇒ u2+2u+v2=1 and u≥0
⇒ (u+1)2+v2=2 and u≥0
w-locus is the arc to the right of the y-axis of the circle with centre (−1,0) and radius √2.
Notice it’s less than a semicircle