这是一份bath巴斯大学PH30025作业代写的成功案

Suppose the complex numbers $z_{1}$ and $z_{2}$ are represented inby the points $P_{1}$ and $P_{2}$ respectively. Also, let $A$ be the point $z=1$. Then, with $O P_{2}$ as base corresponding to $\mathrm{OA}$, the triangle $\mathrm{OP}{2} \mathrm{P}$ is constructed to be similar to triangle $\mathrm{OAP}{1}$. Now,
$$
\begin{gathered}
\mathrm{OP} / r_{2}=\mathrm{OP} / \mathrm{OP}{2}=\mathrm{OP}{1} / \mathrm{OA}=r_{1} / 1 \
\mathrm{OP}=r_{1} r_{2} .
\end{gathered}
$$
Also
$$
\angle \mathrm{POP}{2}=\angle \mathrm{P}{1} \mathrm{OA}=\theta_{1}
$$
and so
$$
\angle \mathrm{POA}=\theta_{1}+\theta_{2} \text {. }
$$

PH30025 COURSE NOTES ：

$\operatorname{Cosh}^{-1} x$ may be expressed in terms of logarithms as follows:
If
$$
y=\cosh ^{-1} x
$$
then
$x=\cosh y=1 / 2\left(\mathrm{e}^{y}+\mathrm{e}^{-y}\right)$
and so
$$
e^{2 y}-2 x e^{y}+1=0 \text {. }
$$
Regarding this as a quadratic equation in $\mathrm{e}^{y}$,
$$
e^{y}=x \pm\left(x^{2}-1\right)^{1 / 2}
$$