Past exam of the mathematical methods I course of the University of Oxford / 2019 / 9

Put $w=e^{i\theta }$. By Euler's formula, $w-w^{-1}=2i\sin \theta$. The binomial theorem givesGroup opposite powers and use $w^j-w^{-j}=2i\sin (j\theta )$:Since $(2i{)}^7=-128i$, division by $-2i$ yieldsThis derivation is an application of the same exponential representation underlying De Moivre's theorem.

To invert the complex cosine, let $z=\cos w$ and put $u=e^{iw}$. Since the exponential function is nonzero, $u\ne 0$, and Euler's formula gives $2z=u+u^{-1}$. Multiplying by $u$ yields the quadratic equation $u^2-2zu+1=0$. ThusChoosing either square root of $1-z^2$ merely exchanges the two signs. The two values have product one, so neither vanishes; taking a complex logarithm is legitimate.

For each choice of sign, all solutions of $e^{iw}=u$ satisfy $iw=\mathrm{Log}u+2\pi in$. Therefore the full inverse relation isConversely, exponentiating any displayed value recovers one root of the quadratic equation, and hence $(e^{iw}+e^{-iw})/2=z$, proving completeness. With a multivalued complex logarithm, the $2\pi n$ term is already included and this is the source's compact formula for inverse cosine.

The principal inverse cosine requires compatible choices of logarithm branch, square root and sign. Both signs cannot represent that same principal function indiscriminately: at $z=0$, the two principal-log values are $\pi /2$ and $-\pi /2$, whereas the principal $\arccos 0$ is $\pi /2$. Thus the printed $\pm$ is correct for the inverse relation, not a pair of interchangeable principal-value formulas.

By Vieta's formulas, the sum of all five polynomial roots, counted with multiplicity of a root, is minus the coefficient of $z^4$ divided by the leading coefficient. Hence

For a monic polynomial of degree five, Vieta's formulas give the product of its polynomial roots as $(-1{)}^5$ times its constant coefficient. Therefore

The coefficients are real, so the supplied polynomial root $2+2i$ has its complex conjugate $2-2i$ as another polynomial root. Their monic polynomial factor of degree two isPolynomial division then givesThus all five polynomial roots, with multiplicity of a root, areTheir sum is $5$ and their product is $8\cdot 1\cdot (-1)=-8$, agreeing with solution and solution.