= Solution
Write $z=x+iy$. Squaring the nonnegative <complex modulus> on each side gives
$$
(x-4)^2+y^2=4\big[(x-1)^2+y^2\big].
$$
The linear terms cancel, leaving $3(x^2+y^2)=12$. Therefore the <locus> is
$$
\boxed{|z|=2,\quad\text{the circle of radius }2\text{ centred at the origin}.}
$$
Its real and imaginary axis intersections are $\pm2$ and $\pm2i$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-complex-motion.png]
{title=The circular complex locus and an anticlockwise tangential velocity at a general point}
Back to article page
