Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 1B (Vectors and Matrices)

The triangle inequality givesHence the left-hand side of the proposed equation is always at least $|a-b|$. There are no solutions when $c<|a-b|$.

Equality in the triangle inequality holds exactly when the two displacement vectors $a-z$ and $z-b$ point in the same direction. Thus $z$ lies on the closed line segment joining $a$ and $b$. Conversely, every point of that segment satisfiesThe solution set is therefore precisely that segment in the complex plane.

By the focal definition of an ellipse, the solution set is the ellipse with foci $a$ and $b$. Its centre is $(a+b)/2$, its major axis lies along the line through the foci, and its semiaxes areThe sketch is consequently a nondegenerate ellipse symmetric about both the focal line and its perpendicular bisector.

The required root of unity issoTherefore

The logarithms of $1+\omega =e^{i\pi /3}$ areThe definition of complex exponentiation therefore gives all values asSince $1+\omega =\frac{1}{2}+i\frac{\sqrt{3}}{2}$, these can be written