Algebraic method for calculating Möbius transformations

mobius

Click here for pdf booklet for Möbius transformations lessons. Click here for exercises from the textbook, and answers.
Click here for worked answers to questions in textbook and all recent FP2 exam questions on this topic

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