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.

Year FMEDEXCEL 9FM0 FP2

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.

WHAT YOU PROBABLY THINK

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: u = x² − y² and v = 2xy. Fixing one of x or y and eliminating the other gives the image curve. Straight lines rarely survive as straight lines, which is where the opening claim fails: 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: that is 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. Checking y = 2: w = −3 + 4i, and 16 = 4(1 + 3) as required.

Circles centred at the origin behave more simply: |z| = r maps to |w| = r², since squaring squares the modulus and doubles the argument. The circle survives; 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: 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: 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: (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. Checking z = 1/2: w = 2, and (2 − 1)² = 1.

YOUR TURN

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: every point moves 3 right and 2 down.

Multiplying by 2i multiplies the modulus by 2 and adds π/2 to the argument: an enlargement by scale factor 2 about the origin, combined with a quarter turn anticlockwise.

Both preserve lines and circles as they are, which is exactly why the interesting cases are squaring and inversion.

THE EXAM BIT

  • 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: 'a circle' alone will not do without centre and radius.
  • Check one convenient point through the mapping; it catches sign and factor slips immediately.

CHECK YOURSELF

Under w = z², find the image of the circle |z| = 3.

Show a hint

Squaring squares the modulus.

Show the answer

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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.

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.

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