MathsFurther Pure 2 › Transformations of the complex plane

Transformations of the complex plane

Feed a line into w = z² and a parabola comes out. Feed it into w = 1/z and a circle appears. Mapping one plane to another turns geometry into algebra you can actually do.

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.
  • Recognise that w = 1/z sends lines not through the origin to circles through it.

COMMON MISCONCEPTION

A transformation of the complex plane sends straight lines to straight lines.

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 drops out. Straight lines rarely survive as straight lines. Only very special maps preserve straightness, and squaring is not one of them.

The line x = 1 under w = z²: every point lands on the parabola v² = 4(1 − u), opening leftwards from (1, 0)x = 1z-planew = z²v² = 4(1 − u)w-plane
FIG. 1The line x = 1 under w = z²: each point maps to u = 1 − y² and v = 2y, tracing the parabola v² = 4(1 − u) opening leftwards.

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 far more simply. Squaring squares the modulus and doubles the argument, so |z| = r maps to |w| = r². The circle survives as a circle, and only its size and the speed of tracing change.

Möbius transformations

For w = (az + b)/(cz + d), rearrange to get z in terms of w before substituting into the given locus. These maps have a remarkable property, in that they send the family of lines and circles to itself, although individual members swap type. In particular w = 1/z sends a line not through the origin to a circle through the origin.

The line Re(z) = 1/2 under w = 1/z: the image is the circle of radius 1 centred at (1, 0), through the originRe(z) = ½z-planew = 1/z2(u − 1)² + v² = 1through 0w-plane
FIG. 2The line Re(z) = 1/2 under w = 1/z: it maps to the circle of radius 1 centred at (1, 0), which passes through the origin.

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. It passes through the origin, which is the image of the point at infinity along the line. 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 send lines to lines and circles to circles, which is what makes squaring and inversion the interesting cases.

ASSESSMENT FOCUS

  • Rearrange for z in terms of w before substituting; substituting the wrong way round is the standard error.
  • 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.

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.

For a Möbius map, invert to z in terms of w first; w = 1/z turns lines missing the origin into circles through it.

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.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

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.
  • Recognise that w = 1/z sends lines not through the origin to circles through it.

Open the full revision checklist to see every objective in the course in one place.