Questions on transformations of the complex plane

Question 1

The transformation w = z + 2 − i is applied to the circle |z| = 3. Find the equation of the image in the w-plane.

Accepted Answer

The transformation is a translation by the vector 2 − i, so z = w − 2 + i. Substituting gives |w − 2 + i| = 3, a circle of the same radius 3, now centred at 2 − i. B1 for recognising a translation, M1 for z = w − 2 + i, A1 for the image equation.

Question 2

Describe the transformation w = 2iz geometrically, and find the image of |z − 1| = 1.

Accepted Answer

Since 2i has modulus 2 and argument π/2, the map is an enlargement scale factor 2 about the origin combined with a rotation of 90° anticlockwise. Substituting z = w/(2i) gives |w/(2i) − 1| = 1, and multiplying through by |2i| = 2 gives |w − 2i| = 2. B1 for the enlargement, B1 for the rotation, M1 for substituting for z, A1 for the image circle. The centre 1 has moved to 2i and the radius has doubled.

Question 3

State the condition on the constants a, b, c and d for w = (az + b)/(cz + d) to be a Möbius transformation, and state what the map does when the condition fails.

Accepted Answer

The condition is ad − bc ≠ 0. B1. If instead ad − bc = 0 then the numerator is a fixed multiple of the denominator, so w takes the same value at every z for which it is defined: the map is constant and sends the whole plane to one point. B1 for the constant map. A constant map has no inverse, which is why the condition is imposed.

Question 4

Find the image of the point z = 1 + i under w = z², and describe the image of the whole real axis under the same transformation.

Accepted Answer

(1 + i)² = 1 + 2i + i² = 2i. M1 for squaring, A1 for 2i.
A point of the real axis is z = t with t real, and w = t², which is real and never negative. So the image is the non-negative real axis, and each positive value is reached twice, once from t and once from −t. A1 for the image with that description. Squaring doubles arguments, so the two halves of the real axis, at arguments 0 and π, land on arguments 0 and 2π, which are the same ray.

Question 5

Show that w = 1/z maps the line Re(z) = 1/2 onto a circle, and give its centre and radius.

Accepted Answer

Write w = u + iv, so z = 1/w = (u − iv)/(u² + v²) and Re(z) = u/(u² + v²). Setting that equal to 1/2 gives u² + v² = 2u, that is (u − 1)² + v² = 1: a circle of centre 1 and radius 1. M1 for z = 1/w, A1 for Re(z) = u/(u² + v²), M1 for u² + v² = 2u, A1 for the circle equation, B1 for the centre and radius. The image passes through the origin, which is the image of the point at infinity along the line.

Question 6

Find the image of the circle |z| = 2 under w = 1/z, and explain what happens to the direction of travel.

Accepted Answer

Taking moduli, |w| = 1/|z| = 1/2, so the image is the circle |w| = 1/2. Since arg w = −arg z, a point moving anticlockwise round the original circle traces the image clockwise. M1 for taking moduli, A1 for |w| = 1/2, M1 for arg w = −arg z, A1 for the reversed sense. Inversion reflects in the real axis as well as reciprocating the modulus.

Question 7

Find the image of the line Re(z) = 1 under w = z².

Accepted Answer

Put z = 1 + it, so w = 1 − t² + 2it. Then u = 1 − t² and v = 2t, so t = v/2 and u = 1 − v²/4. Rearranged, v² = 4(1 − u): a parabola with vertex at u = 1 opening in the negative u direction. M1 for z = 1 + it, A1 for u = 1 − t² and v = 2t, M1 for eliminating t, A1 for v² = 4(1 − u).

Question 8

Show that w = (z − i)/(z + i) maps the real axis onto the unit circle |w| = 1 apart from one missing point, and identify that point.

Accepted Answer

Take moduli: |w| = |z − i|/|z + i|. A real z lies on the real axis, which is the perpendicular bisector of the segment joining i and −i, so |z − i| = |z + i| and |w| = 1 for every real z. Checking points: z = 0 gives w = −1 and z = 1 gives w = −i, both on the unit circle.
The point w = 1 is never reached, since w = 1 would need z − i = z + i, which no finite z satisfies; w approaches 1 as z grows large in either direction, so the image is the unit circle with the single point 1 removed. M1 for taking moduli, A1 for |z − i| = |z + i| on the real axis, A1 for |w| = 1, B1 for the missing point w = 1.

Question 9

Find the image of the circle |z − 1| = 1 under w = 1/z, and explain why the image is a line even though the original is a circle.

Accepted Answer

Invert first: z = 1/w, so the condition becomes |1/w − 1| = 1. Multiplying through by |w| gives |1 − w| = |w|, which says w is equidistant from 0 and 1: the perpendicular bisector Re w = 1/2. Checking: z = 2 is on the original circle and maps to w = 1/2, and z = 1 + i maps to (1 − i)/2, both with real part 1/2.
The original circle passes through the origin, and the origin has no finite image under inversion, so the image cannot close up into a circle: it is the open end that stretches to infinity. This is the mirror of the standard result, in that w = 1/z swaps lines not through the origin with circles through it. M1 for z = 1/w and substituting, A1 for |1 − w| = |w|, A1 for Re w = 1/2, B1 for the origin-to-infinity explanation.

Question 10

The transformation w = (3z − 2)/(z + 1) is applied to the circle |z| = 2. Verify that the transformation is a Möbius transformation, express z in terms of w, and find the centre and radius of the image.

Accepted Answer

Here a = 3, b = −2, c = 1 and d = 1, so ad − bc = 3 + 2 = 5, which is not zero and the map is a Möbius transformation. B1 for the determinant.
Rearranging, w(z + 1) = 3z − 2, so wz + w = 3z − 2 and z(w − 3) = −2 − w, giving z = (w + 2)/(3 − w). M1 for collecting the z terms, A1 for the inverse.
Substituting into |z| = 2 gives |w + 2|/|3 − w| = 2, that is |w + 2| = 2|w − 3|. This is a ratio of distances from −2 and 3 with the ratio not equal to 1, so the image is a circle. Writing w = u + iv and squaring: (u + 2)² + v² = 4[(u − 3)² + v²], so 3u² + 3v² − 28u + 32 = 0, and completing the square gives (u − 14/3)² + v² = 100/9: a circle of centre (14/3, 0) and radius 10/3. M1 for squaring into cartesian form, A1 for the centre and radius.
Checking two points: z = 2 maps to 4/3 and z = −2 maps to 8, and those are the two ends of the horizontal diameter, since 8 − 4/3 = 20/3. Substituting for z rather than for w is the step to get right: the locus is a condition on z, so z is the letter that has to be replaced.

Question 11

Find the image of the circle |z| = 3 under w = z², and the image of the half-line arg z = π/6.

Accepted Answer

Taking moduli, |w| = |z|² = 9, so the image of the circle is the circle |w| = 9. M1 for taking moduli, A1 for |w| = 9. Since arg w = 2 arg z, a point going once round the original circle takes its image twice round the image circle. B1 for that.
On the half-line, arg z = π/6 and the modulus is free, so arg w = π/3 with the modulus still free: the image is the half-line arg w = π/3 from the origin. A1.

Question 12

Show that w = (z + 1)/(z − 1) maps the unit circle |z| = 1, apart from one point, onto the imaginary axis.

Accepted Answer

Put z = cos θ + i sin θ. Then w = (cos θ + 1 + i sin θ)/(cos θ − 1 + i sin θ); multiplying top and bottom by the conjugate of the denominator makes the denominator (cos θ − 1)² + sin²θ = 2 − 2cos θ, and the real part of the numerator is (cos²θ − 1) + sin²θ = 0. So w is purely imaginary for every θ. The point z = 1 is excluded, since the denominator vanishes there, and no finite w corresponds to it. M1 for z = cos θ + i sin θ, M1 for multiplying by the conjugate, A1 for the denominator, M1 for the real part of the numerator, A1 for showing it vanishes, B1 for excluding z = 1.

Question 13

The transformation w = (z + 2)/(z − 2) is applied to the circle |z| = 1. Show that the image is a circle, and find its centre and radius.

Accepted Answer

Rearrange for z: w(z − 2) = z + 2, so z(w − 1) = 2w + 2 and z = 2(w + 1)/(w − 1). Then |z| = 1 becomes 2|w + 1| = |w − 1|, a fixed ratio of distances from −1 and 1, which is an Apollonius condition, so a circle is expected before any algebra.
With w = u + iv, squaring gives 4[(u + 1)² + v²] = (u − 1)² + v², so 3u² + 3v² + 10u + 3 = 0, that is u² + v² + (10/3)u + 1 = 0. Completing the square gives (u + 5/3)² + v² = 16/9: a circle of centre (−5/3, 0) and radius 4/3.
Checking: z = 1 maps to −3 and z = −1 maps to −1/3, the two ends of a diameter on the real axis, and z = i maps to (−3 − 4i)/5, which also satisfies the equation. M1 for rearranging for z, A1 for z = 2(w + 1)/(w − 1), M1 for the modulus condition 2|w + 1| = |w − 1|, M1 for squaring into cartesian form, A1 for the circle equation, A1 for the centre and radius.

Question 14

The line L has equation y = x + 1 and does not pass through the origin. Show that the image of L under w = 1/z is a circle through the origin, giving its centre and radius, and state which point of that circle is not the image of any finite point of L.

Accepted Answer

Write w = u + iv. Then z = 1/w = (u − iv)/(u² + v²), so x = u/(u² + v²) and y = −v/(u² + v²). M1 for z = 1/w with real and imaginary parts separated, A1 for x and y.
Substituting into y = x + 1 gives −v/(u² + v²) = u/(u² + v²) + 1. Multiplying through by u² + v², which is not zero for any image point, gives −v = u + u² + v², that is u² + v² + u + v = 0. M1 for clearing the denominator, A1 for the collected equation.
Completing the square, (u + ½)² + (v + ½)² = ½, a circle of centre (−½, −½) and radius 1/√2. A1. Putting u = v = 0 satisfies the equation, so the circle passes through the origin.
The origin itself is the point no finite z reaches, since |w| = 1/|z| is never zero. As the point runs off along L in either direction its image creeps towards the origin without arriving, so on an ordinary Argand diagram that single point has to be left out; on the extended plane it is the image of the point at infinity. B1 for naming the origin with that reason. This is why the rule is stated as lines not through the origin going to circles through it: had L passed through the origin, the origin would have had no image and the picture would be a line again.

Question 15

The region R consists of the points with 1 ≤ |z| ≤ 2 and 0 ≤ arg z ≤ π/4. Describe the image of R under w = z², find its exact area, and compare that area with the area of R.

Accepted Answer

In modulus-argument form, w = z² has |w| = |z|² and arg w = 2 arg z. M1 for both statements. So 1 ≤ |z| ≤ 2 becomes 1 ≤ |w| ≤ 4, and 0 ≤ arg z ≤ π/4 becomes 0 ≤ arg w ≤ π/2. A1 for the image region, which is the part of an annulus of inner radius 1 and outer radius 4 lying in the first quadrant.
The area of a sector of angle θ between radii r and R is ½(R² − r²)θ, so the image has area ½(16 − 1)(π/2) = 15π/4, about 11.78 square units. M1 for the sector area formula, A1 for 15π/4.
R itself has area ½(4 − 1)(π/4) = 3π/8. A1. The image is therefore ten times the area of R. B1 for the comparison.
That factor is an average rather than a local one: squaring scales areas by a factor |2z|² = 4|z|² at the point z, which runs from 4 to 16 across R, so the enlargement near the inner arc is much gentler than near the outer one. The map is one-to-one on R because the argument range doubles to π/2 and no two points of R end up with the same image.