Problem 1. Show that the set of points $$\left|z-z_{0}\right|=\lambda\left|z-z_{1}\right| \quad (1)$$ where $\lambda>0, \lambda \neq 1$ and $z_{0} \neq z_{1}$ is the circle with center and radius $$\frac{z_{0}-\lambda^{2} z_{1}}{1-\lambda^{2}} \quad \text { and } \quad \frac{\lambda}{\left|1-\lambda^{2}\right|}\left|z_{0}-z_{1}\right|$$ respectively. Show that, conversely, a line or a circle is of the form (1) for some choice of $\lambda$ and of $z_{0}, z_{1}$.
Proof . Equality (1) is equivalent to $$|z|^{2}+\left|z_{0}\right|^{2}-2 \operatorname{Re} z \overline{z_{0}}=\lambda^{2}\left(|z|^{2}+\left|z_{1}\right|^{2}-2 \operatorname{Re} z \overline{z_{1}}\right)$$ that is, since $\lambda>0$ and $\lambda \neq 1$ $$|z|^{2}-2 \operatorname{Re} z \frac{\overline{z_{0}-\lambda^{2} z_{1}}}{1-\lambda^{2}}=\frac{\lambda^{2}\left|z_{1}\right|^{2}-\left|z_{0}\right|^{2}}{1-\lambda^{2}}$$ Completing the square we obtain $$\left|z-\frac{z_{0}-\lambda^{2} z_{1}}{1-\lambda^{2}}\right|^{2}-\left(\frac{\left|z_{0}-\lambda^{2} z_{1}\right|}{1-\lambda^{2}}\right)^{2}=\frac{\lambda^{2}\left|z_{1}\right|^{2}-\left|z_{0}\right|^{2}}{1-\lambda^{2}}$$ and hence we get the circle of center $\frac{z_{0}-\lambda^{2} z_{1}}{1-\lambda^{2}}$ and radius $R$ defined by $$R^{2}=\left(\frac{\left|z_{0}-\lambda^{2} z_{1}\right|}{1-\lambda^{2}}\right)^{2}+\frac{\lambda^{2}\left|z_{1}\right|^{2}-\left|z_{0}\right|^{2}}{1-\lambda^{2}}=\left(\frac{\lambda}{\left|1-\lambda^{2}\right|}\left|z_{0}-z_{1}\right|\right)^{2}$$ We now study the converse, and focus only on the case of a circle. Let $$|z-\Omega|=R$$ be the circle of center $\Omega$ and radius $R>0 .$ We are looking for $\lambda>0$ and $z_{0}, z_{1} \in \mathbb{C}$ (with $z_{0} \neq z_{1}$ ) such that $$\begin{array}{l} \Omega=\frac{z_{0}-\lambda^{2} z_{1}}{1-\lambda^{2}} \\ R=\frac{\lambda}{\left|1-\lambda^{2}\right|}\left|z_{0}-z_{1}\right| \end{array}$$ From the first equation we get $$z_{0}=\left(1-\lambda^{2}\right) \Omega+\lambda^{2} z_{1}$$ Plugging this expression in the second equation we obtain $$R=\lambda\left|\Omega-z_{1}\right|$$ We take $$z_{1}=\Omega+\frac{R}{\lambda}$$ Then $$z_{0}=\left(1-\lambda^{2}\right) \Omega+\lambda^{2} \Omega+\lambda R=\Omega+\lambda R$$ which ends the proof.
Problem 2. Let $\varphi$ be a possibly degenerate Moebius map, and let $$M_{1}=\left(\begin{array}{ll} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}\right) \quad \text { and } \quad M_{2}=\left(\begin{array}{ll} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array}\right)$$ be such that $$\varphi(z)=T_{M_{1}}(z)=T_{M_{2}}(z)$$ for every $z$ in their common domain of definition. Show that there is a complex number $\lambda \neq 0$ such that $$M_{2}=\lambda M_{1}$$
Proof . Solution of Exercise $2.3 .2 .$ We have for all $z \in \mathbb{C}$ where both functions are defined: $$\varphi(z)=\frac{a_{1} z+b_{1}}{c_{1} z+d_{1}}=\frac{a_{2} z+b_{2}}{c_{2} z+d_{2}}$$ Hence, $$z^{2}\left(a_{1} c_{2}-a_{2} c_{1}\right)+z\left(a_{1} d_{2}-a_{2} d_{1}+c_{2} b_{1}-b_{2} c_{1}\right)+b_{1} d_{2}-b_{2} d_{1} \equiv 0$$ One can proceed by remarking that the coefficients of the above polynomial are all equal to $0,$ and then by distinguishing various cases. We will chose another avenue to solve the problem (we will make use of limit at infinity; technically speaking, we will see this notion really only in the next chapter). With $$X(z)=\frac{c_{1} z+d_{1}}{c_{2} z+d_{2}}$$ and taking into account the previous formula we have \begin{aligned} \left(\begin{array}{ll} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}\right)\left(\begin{array}{l} z \\ 1 \end{array}\right) &=\left(c_{1} z+d_{1}\right)\left(\begin{array}{c} \varphi(z) \\ 1 \end{array}\right) \\ &=X(z)\left(c_{2} z+d_{2}\right)\left(\begin{array}{c} \varphi(z) \\ 1 \end{array}\right) \\ &=X(z)\left(\begin{array}{ll} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array}\right)\left(\begin{array}{l} z \\ 1 \end{array}\right) \end{aligned} Take now two points $u$ and $v,$ with $u \neq v,$ at which all the expressions make sense. We have $$\left(\begin{array}{ll} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}\right)\left(\begin{array}{ll} u & v \\ 1 & 1 \end{array}\right)=\left(\begin{array}{ll} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array}\right)\left(\begin{array}{cc} u X(u) & v X(v) \\ X(u) & X(v) \end{array}\right)$$ Thus \begin{aligned} \left(\begin{array}{ll} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}\right) &=\left(\begin{array}{ll} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array}\right)\left(\begin{array}{cc} u X(u) & v X(v) \\ X(u) & X(v) \end{array}\right) \frac{1}{u-v}\left(\begin{array}{cc} 1 & -v \\ -1 & u \end{array}\right) \\ &=\left(\begin{array}{cc} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array}\right)\left(\begin{array}{cc} \frac{u X(u)-v X(v)}{u-v} & \frac{u v(X(v)-X(u))}{u-v} \\ \frac{X(u)-X(v)}{u-v} & \frac{u X(v)-v X(u)}{u-v} \end{array}\right) \end{aligned} It will appear from the proof that $X(z)$ is a constant, but we do not know this at this stage. Without loss of generality, we may assume that $$\lim _{z \rightarrow \infty} X(z) \neq \infty$$ Indeed, if $$\lim _{z \rightarrow \infty} X(z)=\infty$$ then $$\lim _{z \rightarrow \infty} \frac{1}{X(z)}=0$$ and replacing $X$ by $1 / X$ amounts to interchange the indices $_{1}$ and $2 .$ Letting $u \rightarrow \infty,$ and then $v \rightarrow \infty$ in $(2.4 .11)$ we obtain $$\left(\begin{array}{ll} a_{1} & b_{1} \\ c_{1} & d_{1} \end{array}\right)=X(\infty)\left(\begin{array}{ll} a_{2} & b_{2} \\ c_{2} & d_{2} \end{array}\right)$$ Hence, $X(\infty) \neq 0$ (otherwise, $\varphi$ will not be defined) and we obtain the result by setting $\lambda=\frac{1}{X(\infty)}$
Problem 3. Let $u$ and $v$ be in the open unit disk $\mathbb{D} .$ Show that $$w=\frac{u+v}{1+u \bar{v}} \in \mathbb{D}$$ and that $$b_{u}\left(b_{v}(z)\right)=\frac{1+u \bar{v}}{1+v \bar{u}} b_{w}(z)$$
Proof . Solution of Exercise 2.3.3. It suffices to note that $$\left(\begin{array}{cc} 1 & -u \\ -\bar{u} & 1 \end{array}\right)\left(\begin{array}{cc} 1 & -v \\ -\bar{v} & 1 \end{array}\right)=\left(\begin{array}{cc} 1+u \bar{v} & -(u+v) \\ -(\bar{u}+\bar{v}) & 1+\bar{u} v \end{array}\right)$$ Then, the associated transformation is $$\frac{(1+u \bar{v}) z-(u+v)}{-z(\bar{u}+\bar{v})+1+\bar{u} v}=\frac{1+u \bar{v}}{1+v \bar{u}} b_{w}(z)$$
Problem 4. Show that four points are on the same complex circle (or on the same complex line) if and only if the number $$\frac{\left(\frac{\left(z_{1}-z_{2}\right)}{\left(z_{1}-z_{3}\right)}\right)}{\left(\frac{\left(z_{2}-z_{4}\right)}{\left(z_{3}-z_{4}\right)}\right)}$$ is real.
Proof . We follow the hints given after the exercice, and focus on the case of a circle. We first assume that the four points are on a common circle. For the unit circle, we can always assume that one of the points is $z_{1}=1$, and so we have to check that for any $t_{2}, t_{3}$ and $t_{4}$ in $(0,2 \pi)$, $$\frac{\left(1-e^{i t_{2}}\right)\left(e^{i t_{3}}-e^{i t_{4}}\right)}{\left(1-e^{i t_{3}}\right)\left(e^{i t_{3}}-e^{i t_{4}}\right)}$$ is real, that is, to check that $$\frac{\left(1-e^{i t_{2}}\right)\left(e^{i t_{3}}-e^{i t_{4}}\right)}{\left(1-e^{i t_{3}}\right)\left(e^{i t_{3}}-e^{i t_{4}}\right)}=\frac{\left(1-e^{-i t_{2}}\right)\left(e^{-i t_{3}}-e^{-i t_{4}}\right)}{\left(1-e^{-i t_{3}}\right)\left(e^{-i t_{3}}-e^{-i t_{4}}\right)}$$ This is checked by multiplying both the numerator and denominator of the right side by $e^{i t_{2}} e^{i t_{3}} e^{i t_{4}}$ Let now $\left|z-z_{0}\right|=R$ be the equation of another circle. The Moebius map $\varphi(z)=\frac{z-z_{0}}{R}$ maps this circle onto the unit circle. We now check the invariance of the cross ratio using the formula. Indeed using this equation we have $$\begin{array}{l} \varphi\left(z_{1}\right)-\varphi\left(z_{2}\right)=\frac{(a d-b c)\left(z_{1}-z_{2}\right)}{\left(c z_{1}+d\right)\left(c z_{2}+d\right)} \\ \varphi\left(z_{1}\right)-\varphi\left(z_{3}\right)=\frac{(a d-b c)\left(z_{1}-z_{3}\right)}{\left(c z_{1}+d\right)\left(c z_{3}+d\right)} \\ \varphi\left(z_{2}\right)-\varphi\left(z_{4}\right)=\frac{(a d-b c)\left(z_{2}-z_{4}\right)}{\left(c z_{2}+d\right)\left(c z_{4}+d\right)} \\ \varphi\left(z_{3}\right)-\varphi\left(z_{4}\right)=\frac{(a d-b c)\left(z_{3}-z_{4}\right)}{\left(c z_{4}+d\right)\left(c z_{3}+d\right)} \end{array}$$ and hence the result by a direct computation. Still for the case of a circle, we consider the converse statement: Let there be therefore four pairwise points for which the previous formula is real. The first three points determine uniquely a circle, which we move, via a Moebius map, to be the unit circle. The corresponding quotient does not change. So we are left with the following question: Given four pairwise different points for which the quotient is real, three of them being on the unit circle, show that the fourth is also on the unit circle. We set $z_{1}=e^{i \theta_{1}}, z_{2}=e^{i \theta_{2}}, z_{3}=e^{i \theta_{3}},$ where $\theta_{1}, \theta_{2}, \theta_{3} \in[0,2 \pi)$ are pairwise different. We have thus: $$\frac{\left(e^{-i \theta_{1}}-e^{-i \theta_{2}}\right)\left(e^{-i \theta_{1}}-e^{-i \theta_{3}}\right)}{\left(e^{-i \theta_{2}}-e^{-i \theta_{4}}\right)\left(e^{-i \theta_{3}}-\overline{z_{4}}\right)}=\frac{\left(e^{i \theta_{1}}-e^{i \theta_{2}}\right)\left(e^{i \theta_{1}}-e^{i \theta_{3}}\right)}{\left(e^{i \theta_{2}}-e^{i \theta_{4}}\right)\left(e^{i \theta_{3}}-z_{4}\right)}$$ Applying the previous exercise with the Moebius map $\varphi(z)=1 / z$ we see that $$\varphi\left(e^{i \theta_{j}}\right)=e^{-i \theta_{j}}, \quad j=1,2,3$$ and so $\overline{z_{4}}=\varphi\left(z_{4}\right),$ i.e., $\left|z_{4}\right|=1 .$ This concludes the proof.

E-mail: [email protected]  微信:shuxuejun

uprivate™是一个服务全球中国留学生的专业代写公司 专注提供稳定可靠的北美、澳洲、英国代写服务 专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务