Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 1

Write $z=x+iy$, with real $x,y$. Its complex conjugate is $z^{\ast }=x-iy$, so equating real and imaginary parts gives $4x=1$, $2y=1$. Thus the complex plane locus is the single point

The modulus obeys $|z{|}^4=4$, giving $|z|=\sqrt{2}$. The complex argument obeys $4\arg z=\pi /2$ modulo $2\pi$. The four distinct complex roots are thereforeThey form a square centred at the origin, on the circle of radius $\sqrt{2}$, with the first vertex at argument $\pi /8$.

For $z=x+iy$, the equation becomes $\sqrt{(x-2{)}^2+y^2}=x$. Its right side must be nonnegative. Squaring then gives $y^2=4(x-1)$, whose points already have $x\ge 1$, so no extraneous branch has been introduced:This is a parabola opening along the positive real axis, with vertex $(1,0)$, focus $(2,0)$ and directrix $x=0$. The original modulus equation equates distance from the focus with distance from that directrix.