Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 11

In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, with $r=\sqrt{x^2+y^2}$. Thus the polar form of a complex number isHere $r=|z|\ge 0$ is its complex modulus; for $z\ne 0$, the complex argument $\theta$ is determined modulo $2\pi$. At $z=0$ the angle is undefined, although every value of $\theta$ gives the same point when $r=0$.

For $z\ne 0$, the complex logarithm is $\ln z=\ln r+i(\theta +2\pi k)$, where $k\in \mathbb{Z}$. ConsequentlyA specified branch chooses one continuous determination of the complex argument on an appropriate cut domain. For the usual principal-value convention use the principal argument $\mathrm{Arg}(x+iy)\in (-\pi ,\pi ]$. A formula using only $\arctan (y/x)$ would lose the quadrant information; it is a valid local angle only with the appropriate additive multiple of $\pi$.

Multiply numerator and denominator by the complex conjugate:Therefore, away from the origin,

Use $|z{|}^2=z\bar{z}$ and $z-\bar{z}=2iy$. ThenThe real part and imaginary part are therefore

Expand the defining complex exponential functions:Thus the real part and imaginary part of the hyperbolic cosine are

A constant real part $\mathrm{Re}\ln z=L$ means $\ln |z|=L$, so its level sets are the concentric circlesTheir radii are strictly positive. The real part is independent of the logarithm's branch, so there is no branch ambiguity in these circles.

For $\mathrm{Re}(1/z)=K$, the equation is $x=K(x^2+y^2)$. If $K\ne 0$, complete the square to obtainThese circles have centres on the real axis and are tangent to the imaginary axis at the excluded origin. Positive $K$ puts the circle to the right, negative $K$ to the left. The separate case $K=0$ gives the punctured imaginary axis $\boxed{x=0,y\ne 0}$. The sketch uses an open marker at the origin to emphasize that $1/z$ is never defined there.

Set $w=e^z\ne 0$. The hyperbolic cosine equation becomes $w+w^{-1}=-4$, hence $w^2+4w+1=0$. Its two polynomial roots are $w=-2\pm \sqrt{3}$, both negative. Taking every value of the complex logarithm yieldsSince $(2-\sqrt{3})(2+\sqrt{3})=1$, their real logarithms have opposite signs. All solutions are thereforeThe positive real constant is also $\mathrm{arcosh}2$, the inverse hyperbolic cosine of $2$. Substitution gives $\cosh z=-\cosh (\ln (2+\sqrt{3}))=-2$ for both signs and every integer $k$.