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Transformations of the complex plane questions
A transformation such as w = z² or w = 1/z maps each point of one Argand diagram to a point of another, and a line or circle in the z-plane maps to a curve in the w-plane. Substituting and separating real and imaginary parts finds that image curve.
6 original questions · 26 marks · the transformations of the complex plane notes · Further Pure 2
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The transformation w = z + 2 − i is applied to the circle |z| = 3. Find the equation of the image in the w-plane.
Worked 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. Translations move a locus without changing its shape or size.Describe the transformation w = 2iz geometrically, and find the image of |z − 1| = 1.
Worked 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.Show that w = 1/z maps the line Re(z) = 1/2 onto a circle, and give its centre and radius.
Worked 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.Find the image of the circle |z| = 2 under w = 1/z, and explain what happens to the direction of travel.
Worked 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.Find the image of the line Re(z) = 1 under w = z².
Worked 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). Squaring does not keep straight lines straight.Show that w = (z + 1)/(z − 1) maps the unit circle |z| = 1 onto the imaginary axis, apart from one missing point.
Worked 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.
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