Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 11

Using the complex conjugate, $b\bar{z}+\bar{b}z=2\mathrm{Re}(\bar{b}z)$. The locus is therefore the straight lineIn the complex plane, the quantity $\mathrm{Re}(\bar{b}z)/|b|$ is the signed scalar projection of $z$ in the direction $b$. Thus the straight line is perpendicular to $b$ and passes through $-cb/(2|b{|}^2)$. This description uses no separation of $z$ into real coordinates.

Multiplying the complex numbers gives $b\bar{z}+\bar{b}z=2(b_{1}x+b_{2}y)$. Hence the real linear equation is

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

Since $z\bar{z}=x^2+y^2$, the real quadratic equation isCompleting the square makes its geometry explicit:

A squared complex modulus is nonnegative. Conversely, every nonnegative right-hand side of $|z+b{|}^2=|b{|}^2-c$ can be attained. The necessary and sufficient condition is therefore

There is a single point exactly when the radius is zero:Then $|z+b{|}^2=0$ forces this unique value of the complex number $z$.

Under the given rotation, $\bar{z}=\bar{\alpha }\bar{w}$ and $z\bar{z}=|\alpha {|}^{2}w\bar{w}=w\bar{w}$. Collecting terms givesSince $\overline{b\bar{\alpha }}=\bar{b}\alpha$, the transformed coefficients are

For the real positive dilation factor, $\bar{z}=\alpha \bar{w}$ and $z\bar{z}={\alpha }^{2}w\bar{w}$. ThusandThe original PDF specifies a dilation; the differently named transformation in the OCR is an error.

For the translation, expand the products of complex numbers and collect the terms in $w\bar{w}$, $\bar{w}$, $w$ and the constant:It follows thatThe last coefficient is real because $b\bar{\alpha }+\bar{b}\alpha =2\mathrm{Re}(\bar{b}\alpha )$.

For the reciprocal Möbius transformation, substitute $z=1/w$, $\bar{z}=1/\bar{w}$ and multiply by $w\bar{w}$, which is nonzero on the stipulated domain. This yieldsConsequentlyFor a nondegenerate original circle, $c=0$ says that it passes through the origin, and its image is a straight line. The restriction $w\ne 0$ must be retained: multiplying through does not authorize adding that excluded point. Degenerate choices can give a constant or an identically zero equation instead of a straight line.