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Hint: Consider the cases where \(\infty\) is one of the fixed points and \(\infty\) is not a fixed point for \(T\).
8. Find the Mobius transformation that maps \(0,1,\infty\) to the following respective points:
1. \(0,i,\infty\)
2. \(0,1,2\)
3. \(-i,\infty,1\)
4. \(-1,\infty,1\)
9. Discuss the image of the circle \(|z-2|=1\) and its interior under the following transformations:
1. \(w=z-2i\)
2. \(w=3iz\)
3. \(w=\dfrac{z-2}{z-1}\)
4. \(w=\dfrac{z-4}{z-3}\)
5. \(w=\dfrac{1}{z}\)10. Prove that any linear fractional transformation \(T\) with \(c\neq 0\) can be written as:
\[T=T_{4}\circ T_{3}\circ T_{2}\circ T_{1},\]
where
\[T_{1}(z)=z+\frac{d}{c},\quad T_{2}(z)=\frac{1}{z},\quad T_{3}(z)=\left[\frac{( bc-ad)}{c^{2}}\right]z\quad\text{ and }\quad T_{4}(z)=z+\frac{a}{c}.\]
Interpret \(T\) geometrically.
11. Show that the map \(w=\overline{z}\) is not a conformal mapping at the point \(z_{0}=1+i\).
12. Use the scalar product given in (3.1) to prove Theorem 126.
13. Show that if \(F\) and \(G\) are linear fractional maps, then so is \(F\circ G\).
### 3.6 Integration in the Complex Plane
#### Integration Along Paths
It is well known that continuous real-valued functions on compact intervals in \(\mathbb{R}\) are integrable. However, continuous functions are not so adequate for the integration of complex-valued functions. We say that a real or complex-valued function \(h\) is _piecewise continuous_ on a compact interval \([\alpha,\beta]\subset\mathbb{R}\) if there exists points
\[\alpha=t_{0} and continuous functions \(h_{k}\) on \([t_{k},t_{k+1}]\) such that \(h_{k}(t)=h(t)\) for \(t\in(t_{k},t_{k+1})\) for \(k=0,\dots,n-1\). Basically this means that \(h\) is continuous except possibly for a finite number of jump discontinuities. A real-valued piecewise continuous function \(h\) is integrable with \[\int_{\alpha}^{\beta}h(t)\,dt=\sum_{k=0}^{n-1}\int_{t_{k}}^{t_{k+1}}h_{k}(t)\,dt.\] On the other hand, if \(g\) is a complex-valued function defined on \([\alpha,\beta]\subset\mathbb{R}\) where \(g=\operatorname{Re}g+i\mathrm{Im}\,g\), we say \(g\) is integrable if and only if \(\operatorname{Re}g\) and \(\mathrm{Im}\,g\) are both integrable and we define \[\int_{\alpha}^{\beta}g(t)dt:=\int_{\alpha}^{\beta}\operatorname{Re}g(t)dt+i \int_{\alpha}^{\beta}\mathrm{Im}\,g(t)dt.\] For example, \[\int_{0}^{2\pi}e^{it}dt=\int_{0}^{2\pi}\cos t\,dt+i\int_{0}^{2\pi}\sin t\,dt= \sin t|_{0}^{2\pi}+i[-\cos t]|_{0}^{2\pi}=0.\] **Definition 110**.: Let \([\alpha,\beta]\,(-\infty<\alpha\leq\beta<\infty)\) be a closed and bounded interval in \(\mathbb{R}\). A _curve_\(\gamma\) with parameter interval \([\alpha,\beta]\) is a continuous function \[\gamma:[\alpha,\beta]\to\mathbb{C}\] with initial point \(\gamma(\alpha)\) and terminal point \(\gamma(\beta)\). The curve \(\gamma\) is _closed_ if \(\gamma(\alpha)=\gamma(\beta)\). It is _simple_ if it does not intersect itself ( if \(\alpha\leq t_{1} The curve \(\gamma\) has a certain orientation determined as \(t\) increases from \(\alpha\) to \(\beta\). Given \(\gamma\), there exists a curve \(-\gamma\) with the same image set but the opposite orientation: \[-\gamma(t):=\gamma(\alpha+\beta-t)\quad\text{where}\quad t\in[\alpha,\beta].\] We are now ready to define the integral of a function along a path: let \(\gamma\) be a path with parameter interval \([\alpha,\beta]\). Suppose \(\alpha=t_{0} \[\int_{\gamma}f(z)\,dz:=\int_{\alpha}^{\beta}f(\gamma(t))\gamma^{\prime}(t)\,dt.\] Note that since \[(f\circ\gamma)\gamma^{\prime}:t\mapsto f(\gamma(t))\gamma^{\prime}(t)\] is piecewise continuous and hence integrable, and the above integral on the right makes sense. Given \(f(z)=u(z)+iv(z)\), if we want to know the relationship between the line integrals of \(u(x(t),y(t))\) and \(v(x(t),y(t))\) along \(\gamma\) and the complex integral \(\int_{\gamma}f(z)dz\) we simply write: \[\int_{\gamma}f(z)dz =\int_{\alpha}^{\beta}(u+iv)(x^{\prime}+iy^{\prime})dt\] \[=\int_{\alpha}^{\beta}(ux^{\prime}-vy^{\prime})dt+i\int_{\alpha} ^{\beta}(uy^{\prime}+vx^{\prime})dt\] \[=\int_{\gamma}udx-\int_{\gamma}vdy+i\int_{\gamma}vdx+i\int_{ \gamma}udy.\] ### Basic Properties of Integrals Along Paths * Join of paths: Suppose \(\gamma\) is a path with parameter interval \([\alpha,\beta]\) and is the join of paths \(\gamma_{1},\gamma_{2},\ldots,\gamma_{n}\), and let \(f:\gamma^{*}\to\mathbb{C}\) be continuous. Then \[\int_{\gamma}f(z)\,dz=\sum_{k=1}^{n}\int_{\gamma_{k}}f(z)\,dz.\]* Reversal: Suppose \(\gamma\) is a path with parameter interval \([\alpha,\beta]\) and let \(f:\gamma^{*}\to\mathbb{C}\) be continuous. Then \[\int_{-\gamma}f(z)dz=-\int_{\gamma}f(z)\,dz.\] Proofs of the above properties can be found in any standard book on complex analysis, e.g., [37], pp. 94-111. What we will be interested is the path integrals of holomorphic functions around simple closed curves oriented counterclockwise. The following easy examples motivate our main theorem. **Example 105**.: We will notice soon the reason \(\int_{\gamma}\frac{1}{z}=2\pi i\) for the unit circle is that the function \(f(z)=\frac{1}{z}\) is not holomorphic in the interior of the unit circle, namely at \(z=0\), otherwise the integral would be zero. The integral in the next example is often called the _fundamental integral_ and it combines and generalizes the two integrals given in the above example. **Example 106**.: Let \(a\in\mathbb{C}\) and \(r>0\), then we claim: \[\int_{\gamma}(z-a)^{n}dz=\begin{cases}0&\text{if }n\neq-1\\ 2\pi i&\text{if }n=-1\end{cases}.\]Here \(\gamma=\gamma(a,r)\) denotes the circle centered at \(a\) with radius \(r\). Since for \(t\in[0,2\pi]\), \(\gamma(a,r)=a+re^{it}\), using the very definition of the integral along a path we get \[\int_{\gamma}(z-a)^{n}dz =\int_{0}^{2\pi}(re^{it})^{n}rie^{it}dt\] \[=ir^{n+1}\int_{0}^{2\pi}e^{i(n+1)t}dt\] \[=ir^{n+1}\int_{0}^{2\pi}[\cos(n+1)t+i\sin(n+1)t]dt.\] For the case \(n=-1\) the above integral \(\int_{\gamma}(z-a)^{n}dz=i\int_{0}^{2\pi}\cos 0tdt=i2\pi\) and for the case \(n\neq-1\) \[\int_{\gamma}(z-a)^{n}dz=ir^{n+1}\left(\left[\frac{\sin(n+1)t}{n+1}\right]_{0 }^{2\pi}-i\left[\frac{\cos(n+1)t}{n+1}\right]_{0}^{2\pi}\right)=0.\] ### The Complex Fundamental Theorem of Calculus Recall that by the Fundamental Theorem of Calculus for real integrals, if \(F(x)\) is an antiderivative of a function \(f\), that is \(F\) is a continuous function for which \(F^{\prime}(x)=f(x)\), then the integral of \(f\) on the interval \([a,b]\) is the number \[\int_{a}^{b}f(x)\,dx=F(x)|_{a}^{b}=F(b)-F(a).\] Suppose \(\gamma\) is a path with parameter interval \([\alpha,\beta]\) and that \(G(z)\) is a function defined on an open set which also contains \(\gamma^{*}=\gamma([\alpha,\beta])\). Suppose \(G^{\prime}(z)\) exists and is continuous at each point of \(\gamma^{*}\). Then \[\int_{\gamma}G^{\prime}(z)\,dz=\begin{cases}G(\gamma(\beta))-G(\gamma(\alpha) )&\text{in general}\\ 0&\text{if $\gamma$ is closed}.\end{cases}\] Clearly, this is an analog of the Fundamental Theorem of Calculus for real integrals. However, it is not as “fundamental” as in the real case. It is rather a stepping stone to Cauchy’s Theorem and its consequences. For the proof of the Complex Fundamental Theorem of Calculus we refer the reader to [41], p. 124. **Example.** Let \(\gamma(t)=e^{it}\) for \(0\leq t\leq 2\pi\) and \(f(z)=\text{cosec}^{2}z\); we know \(f(z)=\frac{d}{dz}(-\cot z)\) in an open set containing \(\gamma(t)\). Thus, \(\int\limits_{\gamma}\text{cosec}^{2}z\,dz=0\) by the Complex Fundamental Theorem of Calculus.
* Reparametrization: Let \(\tilde{\gamma}\) be another path with parameter interval \([\tilde{\alpha},\tilde{\beta}]\) and \(g\) be a function which maps \([\tilde{\alpha},\tilde{\beta}]\) onto \([\alpha,\beta]\); moreover assume \(g\) has a positive continuous derivative. Then \[\int_{\gamma}f(z)\,dz=\int_{\tilde{\gamma}}f(z)\,dz.\]
1. Suppose we have \(f(z)=z\) and our simple closed curve \(\gamma\) is the unit circle \(\gamma(0,1)=e^{it}=|z|=1\) oriented counterclockwise. Then, \[\int_{\gamma}f(z)\,dz=\int_{\gamma}z\,dz=\int_{0}^{2\pi}e^{it}\,ie^{it}dt=0.\]
2. Suppose we have \(f(z)=\frac{1}{z}\) and our simple closed curve \(\gamma\) is the unit circle \(\gamma(0,1)=e^{it}=|z|=1\) oriented counterclockwise. Since \(|z|^{2}=z\,\overline{z}=1\), our function \(f(z)=\frac{1}{z}=\overline{z}\); hence \[\int_{\gamma}f(z)\,dz=\int_{\gamma}\overline{z}\,dz=\int_{0}^{2\pi}(\cos t-i \sin t)ie^{it}dt=2\pi i.\]
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