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1988 Paper 2 Q4
D: 1600.0 B: 1516.0
complex numbers locus Argand diagram curve sketching real and imaginary parts quadratic mapping parabola

The complex number \(w\) is such that \(w^{2}-2w\) is real.

  1. Sketch the locus of \(w\) in the Argand diagram.
  2. If \(w^{2}=x+\mathrm{i}y,\) describe fully and sketch the locus of points \((x,y)\) in the \(x\)-\(y\) plane.
The complex number \(t\) is such that \(t^{2}-2t\) is imaginary. If \(t^{2}=p+\mathrm{i}q\), sketch the locus of points \((p,q)\) in the \(p\)-\(q\) plane.

Solution:

  1. Suppose \(w = u+ vi\) then \(w^2 - 2w = u^2-v^2-2u+(2uv-2v)i\) so to be purely real we must have \(2uv-2v = 2v(u-1) = 0\) ie either \(v = 0\) or \(u = 1\). Therefore the locus is the real axis and the line \(1 + ti\):
    TikZ diagram
  2. If \(w^2 = x+yi\) then we must have \(x = u^2-v^2\) and \(y = 2uv\), so either \(v = 0, y = 0, x = u^2-2u \geq -1\) or \(u = 1, x = 1-v^2, y = 2v\) which is a parabola:
    TikZ diagram
If \(t = u+iv\) then \(t^2-2t = u^2-v^2-2u + (2uv-2v)i\). For this to be purely imaginary, we need \(u^2-v^2 - 2u = 0 \Rightarrow (u-1)^2-v^2 = 1\), ie points on a hyperbola. Then \(p = u^2-v^2\) and \(q = 2uv\). We can parameterise our hyperbola as \(u = 1 \pm \cosh s, v = \sinh s\) and so \(p = 1 + 2 \cosh s\) and \(q = \sinh 2s\) or \(q = \pm (p-1) \sqrt{(\frac{p-1}{2})^2-1}\) where \(p \geq 3\)
TikZ diagram

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