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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|u^{2}+(v−1)|^{2}

⇒ (1+v)^{2}+u^{2}=4[(u^{2}+(v−1)^{2}]

⇒ 1+2v+v^{2}+u^{2}=4u^{2}+4v^{2}−8v+4

⇒ 3u^{2}+3v^{2}−10v+3=0

⇒ u^{2}+v^{2}−(10/3)v+1=0

⇒ u^{2}+v^{2}−(10/3)v+(25/9)=(25/9)−(9/9) (completing the square)

⇒ u^{2}+(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)

⇒ −u^{2}−(v^{2}−1)=0 (the denominator of the fraction is now u^{2}+(v−1)^{2}, which is real)

⇒ u^{2}+v^{2}=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)= −u^{2}−(v^{2}−1)

⇒ 2u=−u^{2}−(v^{2}−1) and u≥0

⇒ u^{2}+2u+v^{2}=1 and u≥0

⇒ (u+1)^{2}+v^{2}=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*