Contents page for FP2 transforming loci in the complex plane

mobius

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

And 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.


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.

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

ex36

Example 37, page 57

37

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.

l

Ex.3I Q.17 p.63

17

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

q15

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.

q14


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


Review Questions

q50

Question 50

(a) z2−2iz−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=23i.

So [2i,23i] is the diameter of the w-circle. Therefore centre is 43i and radius is 23.

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 23i and 2i (Circle of Apollonius). The circle on which w moves must be symmetrical about that line, so the diameter of that circle is [23i, 2i].

Therefore centre is 43i and radius is 23

Method 2, straight algebra

When |z|=1, |w|=2|w−i|

so if w=u+iv

u2+v2=4u2+4(v−1)2

3u2+3v2−8v+4=0

u2+v283v+43=0

u2+(v−43)2=49

Centre is 43i and radius is 23


q49

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+v2+u−uv=u−u2+uv

u2+v2=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).


q48

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.

semicircle

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 −x2/y−2x/y=y−2

x2−2x+1−y2+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

halfline

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)|

2u2+2v2 = (2−u+v)2+(2−u−v)2

= 8−4u+4v−4u−4v+2u2+2v2

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.


q47

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) / (cos2 θ − 2 cos θ + 1 + sin2 θ)

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

= (−2 sin2(θ/2) − 2i sin(θ/2) cos(θ/2)) / 4 sin2(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