Transformations w = (az + b)/(cz + d)

Make z the subject

w(cz + d) = az + b, so z(cw - a) = b - dw, and z = (b - dw)/(cw - a).

Do this carefully: a slip here spoils every later step.

Image of a circle |z - p| = r

Substitute z = g(w) and write the left side as a single fraction. Then use |A/B| = |A|/|B| to get |A| = r|B|.

• If this has the form |w - α| = |w - β|, the image is a line (perpendicular bisector).
• Otherwise it has the form |w - α| = k|w - β| with k ≠ 1, which is a circle: put w = u + iv, square and complete the square.

Image of a line

For a line such as Re z = 1 or y = x, put z = g(w), write it as X + iY in terms of u and v (multiply top and bottom by the conjugate of the denominator), and apply the condition to X or Y.

Line or circle?

The point z = -d/c is sent to infinity. A circle or line through z = -d/c maps to a line; any other circle or line maps to a circle. Use this as a check on your answer. The word "hence" in a question tells you to use the previous part.

Worked example

Find the image of |z| = 1 under w = (z + i)/(z - 2).

1. w(z - 2) = z + i, so z(w - 1) = 2w + i and z = (2w + i)/(w - 1).
2. |z| = 1 gives |2w + i| = |w - 1|.
3. With w = u + iv: 4u² + (2v + 1)² = (u - 1)² + v², so 3u² + 3v² + 2u + 4v = 0.
4. u² + v² + (2/3)u + (4/3)v = 0, so (u + 1/3)² + (v + 2/3)² = 5/9.

Answer: A circle, centre -1/3 - (2/3)i, radius √5/3

Check yourself

1. For w = (z + 3)/(z - 1), which expression gives z in terms of w?
   • z = (w - 3)/(w + 1)
   • z = (w + 3)/(w - 1)
   • z = (w - 1)/(w + 3)
   • z = (3 - w)/(1 + w)
   Show answer

   z = (w + 3)/(w - 1). wz - w = z + 3, so wz - z = w + 3, z(w - 1) = w + 3.

2. The circle |z| = 1 is mapped by w = 3z/(z + 1) onto a straight line. The line is u = k. Find k.
   Show answer

   1.5. w(z + 1) = 3z gives z = w/(3 - w). |z| = 1 gives |w| = |w - 3|, the perpendicular bisector of 0 and 3: u = 3/2.

3. Find the radius of the image of |z| = 2 under w = 1/(z - 1). Give your answer as a fraction in its simplest form.
   Show answer

   2/3. z = 1 + 1/w = (w + 1)/w. |w + 1| = 2|w|: (u + 1)² + v² = 4u² + 4v², so 3u² + 3v² - 2u - 1 = 0, (u - 1/3)² + v² = 4/9. Radius 2/3.

4. For the same image, find the real part of its centre. Give your answer as a fraction in its simplest form.
   Show answer

   1/3. From (u - 1/3)² + v² = 4/9 the centre is 1/3 + 0i.

5. Under w = (az + b)/(cz + d), when is the image of a circle a straight line?
   • When the circle passes through z = -d/c
   • When the circle is centred at the origin
   • When a = d
   • When the circle has radius 1
   Show answer

   When the circle passes through z = -d/c. z = -d/c is the point sent to infinity; a circle through it has an unbounded image, which is a line.

6. What is the image of the real axis under w = (z + i)/(z - i)?
   • The real axis
   • The imaginary axis
   • The circle |w| = 1
   • The line u = 1
   Show answer

   The circle |w| = 1. For real z, z is equidistant from i and -i, so |z + i| = |z - i| and |w| = 1.

7. Which point of the z-plane has no image under w = (2z + 1)/(z - 3)?
   • z = 3
   • z = -1/2
   • z = 2
   • z = 0
   Show answer

   z = 3. At z = 3 the denominator is zero.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit F2
• Pearson Edexcel IAL Further Pure Mathematics 2 scheme of work

Last updated 8 November 2026. Spotted a mistake? Tell us.