Maths › Further Pure 2 › Transformations of the complex plane
Transformations of the complex plane
Find an image locus by solving the transformation for z and substituting into the original condition. For inversion, distinguish the ordinary complex plane from the extended plane: zero and infinity are interchanged. Record any point left out of the image, either because the transformation is undefined there or because no finite z maps to it.
Builds on Further loci and regions and Complex arithmetic.
IN THIS TOPIC
- Find the image of a line or circle under w = z² by eliminating between real and imaginary parts.
- Invert a Möbius transformation to express z in terms of w before substituting, and state the condition ad − bc ≠ 0.
- Recognise that on the extended plane w = 1/z sends a line not through the origin to a circle through the origin, and say which point an ordinary Argand diagram omits.
COMMON MISCONCEPTION
A transformation of the complex plane sends straight lines to straight lines.
The affine maps w = az + b with a non-zero send every line to a line, being an enlargement, a rotation and a translation together. Squaring and inversion do not: under w = z² the line x = 1 becomes the parabola v² = 4(1 − u), and under w = 1/z the line Re(z) = 1/2 becomes the circle (u − 1)² + v² = 1.
Squaring bends the plane
Under w = z², write z = x + iy and separate into u = x² − y² and v = 2xy. Fix one of x or y, eliminate the other, and the image curve follows. A straight line does not in general map to a straight line. The maps met here that do send every line to a line are the affine ones, w = az + b with a ≠ 0, which combine an enlargement, a rotation and a translation; squaring is not one of them.
WORKED EXAMPLE
A line becomes a parabola
Find the image of the line x = 1 under w = z².
On the line, z = 1 + iy, so w = 1 − y² + 2iy, giving u = 1 − y² and v = 2y.
Eliminating y: y = v/2, so u = 1 − v²/4.
The image is v² = 4(1 − u), a parabola with vertex at (1, 0) opening towards negative u. Check with y = 2: w = −3 + 4i, and 16 = 4(1 + 3) as required.
Circles centred at the origin behave more simply. Squaring squares the modulus and doubles the argument, so |z| = r maps to |w| = r². The image is a circle again; only its radius and the number of times it is traced change.
Möbius transformations
A Möbius transformation has the form w = (az + b)/(cz + d) with ad − bc ≠ 0. Without that condition the numerator is a multiple of the denominator and the map is constant. On the extended complex plane it maps generalised circles to generalised circles, a family that counts a line as a circle through the point at infinity, although individual members change type.
To find an image locus, rearrange for z in terms of w before substituting into the given condition. Under w = 1/z, a line not through z = 0 maps to a circle through w = 0 on the extended plane; for finite z, omit w = 0 from an ordinary Argand diagram, because no finite point of the line has image 0.
WORKED EXAMPLE
A line becomes a circle
Find the image of the line Re(z) = 1/2 under w = 1/z.
Inverting gives z = 1/w. Writing z* for the conjugate, Re(z) = 1/2 says z + z* = 1, so 1/w + 1/w* = 1.
Combining over |w|²: (w + w*)/|w|² = 1, that is 2u = u² + v².
Completing the square gives (u − 1)² + v² = 1, a circle of centre (1, 0) and radius 1. On the extended plane it passes through the origin, which is the image of the point at infinity along the line; on an ordinary Argand diagram omit w = 0, since no finite z on the line maps to it. Check with z = 1/2, where w = 2 and (2 − 1)² = 1.
GUIDED PRACTICE
A translation and a rotation
Describe the transformations w = z + 3 − 2i and w = 2iz geometrically.
Show the working
Adding a constant is a translation, so every point moves 3 right and 2 down.
Multiplying by 2i multiplies the modulus by 2 and adds π/2 to the argument, giving an enlargement of scale factor 2 about the origin combined with a quarter turn anticlockwise.
Both have the affine form w = az + b with a ≠ 0, so both send lines to lines and circles to circles; squaring and inversion do not in general.
ASSESSMENT FOCUS
- Rearrange for z in terms of w before substituting, since substituting w into the original condition describes the wrong set of points.
- Separate into real and imaginary parts early, then eliminate the parameter between them.
- Name the image curve and give its equation, since 'a circle' alone is incomplete without a centre and radius.
- Quote ad − bc ≠ 0 with a Möbius transformation; without it the map is constant.
- Say whether an answer is on the extended plane or on an ordinary Argand diagram, and name any point the finite image omits.
CHECK YOURSELF
Under w = z², find the image of the circle |z| = 3.
Show a hint
Squaring squares the modulus.
Show the answer
|w| = |z|² = 9, so the image is the circle of radius 9 centred at the origin. The argument doubles, so the image is traced twice as z goes once round.
Under w = z², separate u = x² − y² and v = 2xy, then eliminate to find the image curve.
A Möbius map w = (az + b)/(cz + d) needs ad − bc ≠ 0, and on the extended plane it sends generalised circles to generalised circles.
Invert to z in terms of w before substituting; under w = 1/z a line missing the origin maps to a circle through the origin on the extended plane, with w = 0 omitted from a finite Argand diagram.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the transformations of the complex plane questions page.
CHECK YOUR PROGRESS
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- Find the image of a line or circle under w = z² by eliminating between real and imaginary parts.
- Invert a Möbius transformation to express z in terms of w before substituting, and state the condition ad − bc ≠ 0.
- Recognise that on the extended plane w = 1/z sends a line not through the origin to a circle through the origin, and say which point an ordinary Argand diagram omits.
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