Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 11

Write $z_1=a+ib$ and $z_2=c+id$ with real $a,b,c,d$. By the definition of complex conjugation,

With $z_1=a+ib$ and $z_2=c+id$, direct multiplication gives $z_1{z}_2=(ac-bd)+i(ad+bc)$. Taking the complex conjugate yields

The given complex numbers have positive real parts, so $\arg z_1=\arctan A$ and $\arg z_2=\arctan B$. Their product isBy the assumed complex argument ranges, this product also has positive real part, hence $1-AB>0$. Its complex argument is consequently $\arctan [(A+B)/(1-AB)]$, with no extra multiple of $\pi$. Since multiplication adds complex arguments,This proves the arctangent addition formula under the stated branch restriction. For arbitrary real $A,B$, the principal inverse tangent formula can require adding or subtracting $\pi$ when $AB>1$.

A counterclockwise rotation through $\alpha$ about the origin in the complex plane sends $z_j$ to $e^{i\alpha }{z}_j$. By part (a), and because the complex conjugate of $e^{i\alpha }$ is $e^{-i\alpha }$,Thus $Q$ is invariant under the common rotation.

Write $z_j=x_j+i{y}_j$. Multiplication givesTherefore the real part is the dot product of the corresponding vectors:For nonzero vectors, this is $|z_1||z_2|\cos (\arg z_1-\arg z_2)$, measuring their alignment.

From the same multiplication, the imaginary part isHere the planar vectors are embedded in three dimensions with zero third coordinate. Its magnitude is the area of their parallelogram; its sign is positive when the oriented angle from ${\mathbf{v}}_2$ to ${\mathbf{v}}_1$ has positive sine. The reversed order matters: $\Im (z_1{z}_2^{\ast })$ is the negative of the usual signed area for the order $({\mathbf{v}}_1,{\mathbf{v}}_2)$.

Since $x=\cos y$ with real $y$, necessarily $-1\le x\le 1$. Use Euler's formula and set $w=e^{iy}\ne 0$. Thenso the quadratic equation gives $w=x\pm i\sqrt{1-x^2}$. Let $q=x+i\sqrt{1-x^2}$. It lies on the upper half of the unit circle, so $q=e^{i\theta }$ with $0\le \theta \le \pi$. Choose the complex logarithm value $\ln q=i\theta$, taking $\ln (-1)=i\pi$ at $x=-1$. The two roots are $e^{\pm i\theta }$, so $e^{iy}=e^{\pm i\theta }$ gives $y=\pm \theta +2n\pi$.

Because $i\ln q=-\theta$, this set of real solutions is exactlyThe formula includes $x=\pm 1$: the two sign families then coincide after adjusting $n$. No real $y$ exists if $|x|>1$.

For $\theta \notin 2\pi \mathbb{Z}$, use the finite geometric series with ratio $e^{i\theta }$:The second equality follows by factoring $1-e^{it}=-2i{e}^{it/2}\sin (t/2)$ in numerator and denominator. Taking real parts proves the finite trigonometric sum identityWhen $\theta =2p\pi$, every cosine is $1$, so the sum is $\boxed{N}$. This also gives the continuous limiting value at the excluded arguments.