Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 1B (Vectors and Matrices)

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its centre is $2\sqrt{\alpha \beta }$ on the real axis and its radius is $\alpha +\beta$ in the complex plane.

It remains to exclude extraneous points introduced by squaring. On this circle, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus taking the nonnegative square root recovers the original equality at every point of the circle. This is a circle from an affine complex-modulus equation, with the sign check essential to the geometric conclusion.

Put $q=e^{i\theta }$. Since $q\ne 1$, the finite geometric series giveswhere the second equality multiplies numerator and denominator by $1-\overline{q}$. Taking imaginary parts gives the finite trigonometric sumTaking real parts of the same finite geometric series givesThe denominator is nonzero by the restriction on $\theta$. Equivalently, these finite trigonometric sums are $\sin (N\theta /2)\sin ((N+1)\theta /2)/\sin (\theta /2)$ and $\sin (N\theta /2)\cos ((N+1)\theta /2)/\sin (\theta /2)$, respectively.