Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 1C (Vectors and Matrices)

Write $z=x+iy$ and $\alpha =p+iq$. The complex conjugates in the PDF giveSince $(p,q)\ne (0,0)$, the locus is the straight line through the origin perpendicular to the position vector of $\alpha$ in the complex plane. Equivalently,

Write $z=x+iy$. Dividing by two givesThe left side is nonnegative, so squaring is reversible only when $x+a\ge 0$. Squaring yields $y^2=4ax$.

If $a>0$, this equation implies $x\ge 0$, so the sign condition holds automatically. The locus is the parabolawith vertex at the origin, focus of a conic $(a,0)$ and directrix of a conic $x=-a$. Indeed the original equation equates distance from the focus of a conic with distance from the directrix of a conic on its right side.

If $a<0$, the squared equation forces $x\le 0$, making $x+a<0$. This violates the sign condition. Thus the locus is empty for $a<0$; squaring without checking that condition would introduce a spurious parabola.

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.