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

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.