Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 11

Use the full set of complex logarithm values of $i$:By the definition of complex exponentiation, the inner power can therefore take any value $i^i=\exp [-(\pi /2+2\pi k)]$, a positive real number. Its complex logarithm values are $-(\pi /2+2\pi k)+2\pi im$, where $m\in \mathbb{Z}$. ThusAll possible real parts and imaginary parts are consequentlyChoosing the principal complex logarithm at both stages gives the particular value $-i$. The independent logarithm choices are needed because the question asks for all possible values.

The inner power takes the values $r_k=e^{-(\pi /2+2\pi k)}$. For each such real exponent, all values of the outer complex exponentiation are obtained from all complex logarithm values of its base $i$:Writing ${\Theta }_{k,m}=(\pi /2+2\pi m)e^{-(\pi /2+2\pi k)}$, the answer isAll values lie on the unit circle. The principal choice is $k=m=0$, giving $\cos [(\pi /2)e^{-\pi /2}]+i\sin [(\pi /2)e^{-\pi /2}]$. In particular the two nestings in parts (i) and (ii) are different: complex exponentiation is not associative.

The complex modulus $|z-(-1-i)|$ is the Euclidean distance from $z$ to $-1-i$. Thus the locus isa circle with centre $(-1,-1)$ and radius $8$. Its extreme horizontal coordinates are $-9$ and $7$, and its extreme vertical coordinates are also $-9$ and $7$.

Write $z=x+iy$ and use the complex conjugate $z^{\ast }=x-iy$. ThenThus the real part is stretched by a factor of two while the imaginary part is unchanged. If $u=X+iY$, substitution into the circle equation givesThis is an ellipse with centre $(-2,-1)$, horizontal semiaxis $16$ and vertical semiaxis $8$. It is the second panel in the figure in part (b).

Adding $4+3i$ is a translation by $(4,3)$ in the complex plane. The ellipse of part (i) therefore keeps its semiaxes and axis directions, with new centre $(2,2)$. Writing $v=X+iY$, its equation isThe third panel in the figure in part (b) shows the translated ellipse.

Multiplication by $i$ is a counterclockwise rotation through $\pi /2$: it maps coordinates $(X,Y)$ to $(-Y,X)$. The centre $(2,2)$ therefore becomes $(-2,2)$, and the long axis becomes vertical. For $w=X+iY$, the resulting ellipse isIts horizontal semiaxis is $8$ and its vertical semiaxis is $16$, as shown in the final panel of the figure in part (b).

By De Moivre's theorem, $\sin 5\theta$ is the imaginary part of $(\cos \theta +i\sin \theta {)}^5$. The binomial theorem gives, with $s=\sin \theta$,Use ${\cos }^2\theta =1-s^2$ and simplify:The requested equation cancels the $16s^5$ term, leaving $5s(1-4s^2)=0$. Thus $s=0$ or $s=\pm 1/2$. The complete list in $0\le \theta <2\pi$ is