Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1 / 2C (Vectors and Matrices)

The polar form of a complex number and the roots of unity givesince $512=2^9$. Using the complex conjugate to compute the squared complex modulus,The residue $3j-k$ runs through every class modulo $9$, already with $j=0$. The cosine is even, so the nine angles give five distinct values, represented by $r=0,1,2,3,4$. They are distinct because cosine is strictly decreasing on $[0,\pi ]$ and these representative angles lie in that interval. HenceEvery listed value is attained by choosing $z_1=1$ and $z_2=2e^{2\pi ir/9}$.

For any complex number $w$, $\mathrm{Re}w\le |w|$, while the complex modulus is multiplicative and unchanged by complex conjugation. ConsequentlyBoth sides are nonnegative, so taking square roots proves the triangle inequality:The proposed further inequality fails for perpendicular vectors in the complex plane. Taking $z_1=1$, $z_2=i$ gives