Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 3

Since $(1-i{)}^2=-2i$, division in the complex numbers gives

Multiplying by the complex conjugate of the denominator gives $(1+i)/(i-1)=-i$. Cubing yields

Write $-8i=8e^{-i\pi /2}$. The three distinct complex roots are $z=2e^{i(-\pi /6+2k\pi /3)}$, $k=0,1,2$. In Cartesian form,

The right-hand side has modulus $4$ and argument $\pi /3$. Its square roots have modulus $2$ and arguments $\pi /6$ and $7\pi /6$, giving

Complex exponentiation uses a branch of the complex logarithm. All logarithms of $i$ are $i(\pi /2+2\pi k)$ with $k\in \mathbb{Z}$, soSince the trigonometric factor is always $i$, all values in Cartesian form areThe principal logarithm selects $k=0$ and gives the principal value $i{e}^{\pi /2}$.

Use the complex exponential forms of the sine and cosine, and put $w=e^{2iz}$. Multiplying the equation by $2e^{iz}$ gives $(2-i)(w-1)/i+(w+1)=0$, which simplifies to $w=1-i$. Taking all branches of the complex logarithm,Thus every solution isThe exponential factor used to multiply the equation is nonzero, so no solutions were introduced or lost.