Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 12

The complex number $-1+i$ has complex modulus $\sqrt{2}$ and argument $3\pi /4$. A cube root has modulus $(\sqrt{2}{)}^{1/3}=2^{1/6}$, and its argument is one third of $3\pi /4+2k\pi$. Hence the three distinct roots areTheir displayed arguments are $\pi /4$, $11\pi /12$, and $19\pi /12$; the last is equivalently $-5\pi /12$ in the principal range.

Use the exponential definition of the hyperbolic tangent:The complex exponential has modulus one here, so $\mathrm{Re}z=0$. Its arguments satisfy $2\mathrm{Im}z=-\pi /2+2k\pi$. Thus all solutions areThe denominator is then $1-i\ne 0$, so no pole was admitted by cross-multiplication. Substitution also follows from $\tanh (i\theta )=i\tan \theta$ and the period $i\pi$.

Divide the complex polynomial by its known factor $z-(2+i)$:The quadratic factors further as $(z-1)(z-1-i)$, because its two proposed roots sum to $2+i$ and multiply to $1+i$. Therefore the remaining solutions areAll three roots are distinct. Complex coefficients do not require nonreal roots to occur in conjugate pairs.

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^4=\cos 4\theta +i\sin 4\theta$. Take real parts of the binomial expansion:Eliminate ${\sin }^2\theta$ using $1-{\cos }^2\theta$ and collect powers: