The method below uses the idea that all transformations like w = (az+b)/(cz+d) transform a “circle-plus” into a “circle-plus”, if by a “circle-plus” we mean either an ordinary circle or a circle with centre at infinity and infinite radius, i.e. a line: click here for proofs.

You identify whether the locus of w is a circle or a line by seeing whether z takes a value which will make w go to infinity (so its locus is a line). Or, equivalently, whether the centre of inversion (z=−d/c) is on the z-locus.

Then you identify the w-circle by finding a diameter, given by the w-values got from z-values at two ends of a diameter in line with the centre of inversion (z=−d/c).

Or, if z moves on a line, a diameter of the w-circle is given by the w-values got from z at infinity and from z at the foot of the perpendicular to the line from the centre of inversion.

You identify the w-line by using any two convenient z-values to calculate two w-values.

Click here for the underlying geometric theorems

Here is a two-and-a-half-minute video giving insight into the type of transformations studied in FP2, transformations like w=(az+b)/(cz+d), called Möbius transformations.

and here is an eight-minute video of the great geometer Donald Coxeter (when he was very old: he died aged 96 in 2003) talking about how the Möbius transformation of inversion transforms a half-plane into the unit disk, creates a new geometry inside the unit disk, and inspired the art work of Maurits Escher.

To my mind this way has two advantages over the algebra in the textbook. It is much quicker (Example 37 has well over a whole page of working in the book, but two lines this way: you could do it in your head). And you can see what you’re doing. You can see the general patterns rather than every bout of algebraic manipulation being a one-off surprise.

Alternative worked answers, using the above method, to all recent FP2 complex-loci exam questions

**Example 35, page 55**

We want to find the locus w = (5iz+i)/(z+1) when locus of z is |z|=1

z=−1 is on locus of z, so w goes to infinity, so locus of w is a line.

When z=1, w=3i. When z=i, w=−4+3i. Therefore the locus of w is the line Im(w)=3.

**Alternatively:** z = (i−w)/(w−5i) as in book

|z|=1 ⇒ |w−i| = |i−w| = |w−5i|

Therefore w moves along the line equidistant from i and 5i, which is the line v=3 if w = u+iv.

**Example 36, page 56**

We want to find locus of w = (3z−2)/(z+1) when locus of z is |z|=2

z=−1 is not on the locus of z, so w does not go to infinity, and so locus of w is a circle.

z=2 and z=−2 are two ends of a diameter of the z-circle which goes through the centre of inversion, z=−1. When z=2, w=4/3, and when z=−2, w=8

Therefore w-circle has as a diameter the line-segment from 4/3 to 8. The centre of that circle is the midpoint, i.e. 14/3, and the radius is half the length, i.e. 10/3.

**Alternatively:** z = (w+2)/(3−w) as in book

|z|=2 ⇒ |w+2| = 2|3−w| = 2|w−3|

Therefore w moves along the circle of Apollonius defined by:

the point M between B=−2 and C=3 where BM=2.MC i.e. 4/3

and the point L beyond C where BL=2.CL, i.e. 8

The circle’s centre is the midpoint of ML, i.e. 14/3, and its radius is ½ML, i.e. 10/3

**Example 37, page 57**

w = (iz−2)/(1−z)

The value z=1 is in the locus of z (the real axis), so w goes to infinity, and the locus of w must be a line.

w=−i when z→∞ and w=−2 when z=0, so the line goes through those points.

**Ex.3I Q.17 p.63**

w = (z+i)/z

(a) z moves along the line y=x, and therefore through z=0, so the locus of w must include movement to infinity. Therefore, w moves on a line.

When z→∞, w→1, and when z = 1+i, w = (1+2i)/(1+i) = 1½+½i

So w moves along the line v=u−1 if w = u+iv.

*Alternatively:* w=1+i/z

As z moves along y=x, 1/z* moves along the same line. 1/z moves along the reflection in the x-axis, the line y=−x. i/z moves along the same line rotated by π/2 anticlockwise, i.e. y=x again. 1+i/z moves along the same line shifted by +1 along the x-axis, i.e. y=x−1.

(b) z does not move through z=0, and so the locus of w does not include movement to infinity. Therefore, the w-locus is a circle.

A diameter is given by the w-values when z→∞ and when z=−½−½i, the foot of the perpendicular to the line x+y+1=0 from the centre of inversion z=0. Therefore [1,−i] is a diameter, and the circle is C.

*Alternatively:* When z→∞, w→1; when z=−1, w=1−i; and when z=−i, w=0. Therefore w moves along the circle through the three points 0, 1−i, 1. By substituting those three values of w into the equation, we see that circle is C.

*Alternatively, again:* w = 1+i/z

When z moves on x+y+1=0

1/z* moves on a circle with diameter (0,0) to (−1, −1)

1/z moves on the reflection in the x-axis of that, i.e. a circle with diameter (0,0) to (−1, 1)

i/z moves on that circle rotated anti-clockwise by π/2, i.e. the circle with diameter (0,0) to (−1, −1) again.

1+i/z moves on that circle shifted by 1 unit along the x-axis, so the circle with diameter (1,0) to (0,−1)

which is the required circle C.

**Ex.3I Q.15 p.63**

We want to find locus of w = (3iz+6)/(1−z) when locus of z is |z|=2

z=1 is not on the locus of z, so locus of w is a circle.

z=2 and z=−2 are ends of a diameter in line with the centre of inversion, z=1. When z=2, w=−6−6i, and when z=−2, w=2−2i

So a w-circle diameter is [−6−6i, 2−2i]. The centre of that circle is the midpoint, i.e. −2−4i, and the radius is half the length, i.e. 2√5.

**Alternatively:** z = (w−6)/(w+3i)

|z|=2 ⇒ |w−6| = 2|w+3i|

Therefore w moves along the circle of Apollonius defined by

the point M between C=−3i and B=6 where BM=2.MC i.e. 2−2i

and the point L beyond C where BL=2.LC, i.e. −6−6i

The centre is the midpoint of LM, i.e. −2−4i, and the radius is half |LM|, i.e. 2√5

**Ex.3I Q.14 p.63**

We want to find locus of w = (4−z)/(z+i) when locus of z is |z|=1

z=−i is on locus of z, so w goes to infinity, so locus of w is a line.

When z=1, w=(3/2)−(3/2)i, and when z=−1, w=−(5/2)−(5/2)i

Locus of w is the line through those two points, which is 2u−8v−15=0 if w=u+iv

**Alternatively:** z = (−iw+4)/(w+1)

|z|=1 ⇒ |w+4i| = |w+1|

Therefore w moves along the line ℓ equidistant from −4i and −1

ℓ goes through the midpoint of −4i and −1, which is −½−2i

It is perpendicular to the line joining −4i and −1, so its slope is ¼

So, if w=u+iv, ℓ is (v+2) = ¼(u+½)

or 2u−8v−15=0.

**Review Questions**

**Question 50**

**(a)** z^{2}−2iz−2=0

(z−i)^{2}=1

z=i±1

**(b)** w=(az+b)/(z+d)

when z=0, w=i, so i=b/d and d=−ib

when z=1+i, w=1+i, so 1+i=(a+ia+b)/(1+i−ib)

2i+b(1−i)=a(1+i)+b

a(1+i)+ib−2i=0 …. [1]

when z=−1+i, w=−1+i, so −1+i=(−a+ia+b)/(−1+i−ib)

−2i+b(1+i)=−a+ia+b

a(1−i)−ib+2i=0 …. [2]

Add [1] and [2] ⇒ a=0, ⇒ b=2, d=−2i

**(c)** So w=2/(z−2i)

wz−2iw=2

z=2i(−i+w)/w

|z|=2|w−i|/|w|

**(d): method 1, with geometry**

The two ends of the diameter which is in line with −2i (centre of inversion) are i and −i

w=2/(z−2i), so when z=i, w=2i, and when z=−i, w=^{2}⁄_{3}i.

So [2i,^{2}⁄_{3}i] is the diameter of the w-circle. Therefore centre is ^{4}⁄_{3}i and radius is ^{2}⁄_{3}.

**Alternatively:** When z moves on the circle |z|=1, w moves so that its distance from 0 is twice its distance from i

The two points on the line connecting 0 and i through which w moves are ^{2}⁄_{3}i and 2i (Circle of Apollonius). The circle on which w moves must be symmetrical about that line, so the diameter of that circle is [^{2}⁄_{3}i, 2i].

Therefore centre is ^{4}⁄_{3}i and radius is ^{2}⁄_{3}

**Method 2, straight algebra**

When |z|=1, |w|=2|w−i|

so if w=u+iv

u^{2}+v^{2}=4u^{2}+4(v−1)^{2}

3u^{2}+3v^{2}−8v+4=0

u^{2}+v^{2}−^{8}⁄_{3}v+^{4}⁄_{3}=0

u^{2}+(v−^{4}⁄_{3})^{2}=^{4}⁄_{9}

Centre is ^{4}⁄_{3}i and radius is ^{2}⁄_{3}

**Question 49**

w=(z+1)/(z+i)

**(a): Method 1, with a bit of geometry**

The foot of the perpendicular from the centre of inversion −i to the z-line is z=−1/√2−i/√2.

So a diameter of the circle on which w moves is given by w when z=−1/√2−i/√2 and w when z→∞, i.e. [-1,1]. The circle is |w|=1.

**Method 2, straight algebra**

If z=x+iy and arg z=π/4, then x=y and z=x+ix

Let w=u+iv, then u+iv=(x+ix+1)/(x+ix+i)

ux+iux+iu+ivx−vx−v=x+ix+1

equating real parts: ux−vx−v=x+1

equating imaginary parts: ux+u+vx=x

x=(−1−v)/(1−u+v)=u/(1−u−v)

−1+v^{2}+u−uv=u−u^{2}+uv

u^{2}+v^{2}=1, i.e. the circle |w|=1

**(b)** If the z-locus is |z|=1, then z=−i is on the locus, and so the locus of w goes to infinity and is a line.

When z=i, w=½−½i, and when z=1, w=1−i

Therefore the locus of w is the line Im(w)=−Re(w).

**Question 48**

Let r=(z−2i)/(z+2)

**(a): Method 1, using a bit of geometry**

The centre of inversion of the transformation from r to z is when z→∞, i.e. r=1.

Therefore the foot of the perpendicular from the centre of inversion to the r-line is r=0

A diameter of the z-circle is given by the z-values when r=0 and r→∞, i.e. [2i,−2]

Which part of that circle corresponds to the *half-*line arg(r)=π/2? The point z=0 is *not* in that part of the circle, since r=−i when z=0. Therefore it must be other (upper) part of the circle.

*Alternatively:* Since the angle in a semicircle is a right angle

arg(z−2i)/(z+2)=π/2 ⇒ r is on a semicircle with diameter [−2,2i].

So: centre −1+i and radius √2

Also, the line from 2i to z is π/2 *anticlockwise* from the line from −2, so it is the upper semicircle.

**Method 2: straight algebra**

arg(w)=π/2 ⇒ Re(r)=0 and w=iq for some real q≥0

If z=x+iy, iq=(x+iy−2i)/(x+iy+2)

iqx−qy+2iq=x+iy−2i

equating real parts −qy=x

equating imaginary parts, qx+2q=y−2

so −x^{2}/y−2x/y=y−2

x^{2}−2x+1−y^{2}+2y+1=4

so z is on a circle with centre (1,−i) and radius 2√2. Which part of the circle? q≥0, so x and y are of opposite signs on the locus of P, and so it’s the upper half of the circle.

**(b)** |z+1−i|=√2

**(c): Method 1, using a bit of geometry**

w=2(1+i)/(z+2)

Since the locus of z includes z=−2, the locus of w goes to infinity. Therefore the locus of w is part of a straight line rather than part of a circle.

When z=2i, w=1

When z=−2+2i, w=1−i

and, as we’ve already seen, when z=−2, w→∞

Therefore the locus of w is a half-line going from 1 through 1−i to infinity, in other words the half-line parallel to the imaginary axis downwards from w=1

**Method 2: just algebra**

w=2(1+i)/(z+2)

so z+2=2(1+i)/w

z+1−i = 2(1+i)/w − 1−i, so | 2(1+i)/w − 1−i|=1

Let w=u+iv, then √2=|[(2−u+v)+i(2−u−v)]/(u+iv)|

2u^{2}+2v^{2} = (2−u+v)^{2}+(2−u−v)^{2}

= 8−4u+4v−4u−4v+2u^{2}+2v^{2}

so 0=8−8u, u=1 defines the line.

Which segment of that line?

arg(z+2) = arg(1+i) − arg(w)

arg(z+2)>π/4 from part (a) and arg(1+i)=π/4, so arg(w)≤0

Therefore it is the half-line below the real axis.

**Question 47**

**(a)** If |z|=1, locus of z is a circle of centre 0 and radius 1.

**(b)** w = 1/(z−1) = 1/(exp[iθ]−1)

= 1/(cos θ + i sin θ − 1)

= (cos θ − i sin θ − 1) / (cos^{2} θ − 2 cos θ + 1 + sin^{2} θ)

= (cos θ − i sin θ − 1) / (2 − 2 cos θ)

= (−2 sin^{2}(θ/2) − 2i sin(θ/2) cos(θ/2)) / 4 sin^{2}(θ/2)

= −½ − ½i cot (θ/2)

Since cot (θ/2) can take any value from −∞ to +∞ as θ moves from 0 to 2π, the locus of w is a line parallel to the imaginary axis through −½