Here is a persuasive advertisement for ghostwriting services specifically for the topic of applying the maximum modulus theorem to find the maximum value of a polynomial function on a bounded region:
**Unlock the Secrets of Polynomial Functions with Our Expert Ghostwriting Services!**
Are you struggling to grasp the maximum modulus theorem and its applications in finding the maximum value of a polynomial function on a bounded region? Do you need help with a complex math assignment or project?
Our team of expert ghostwriters is here to help! With years of experience in mathematics and a deep understanding of complex polynomial functions, we can provide you with high-quality, custom-written content that meets your specific needs.
**Our Services:**
* Custom-written solutions to complex polynomial function problems
* Step-by-step explanations of the maximum modulus theorem and its applications
* Expert guidance on finding the maximum value of a polynomial function on a bounded region
* Original, plagiarism-free content tailored to your specific needs
* Fast turnaround times and flexible deadlines
**Why Choose Us?**
* Our ghostwriters hold advanced degrees in mathematics and have a proven track record of producing high-quality content
* We understand the nuances of complex polynomial functions and can provide clear, concise explanations
* We offer competitive pricing and flexible payment options
* Our customer service team is available 24/7 to ensure your complete satisfaction
Don’t let complex polynomial functions hold you back any longer! Trust our expert ghostwriters to help you succeed. Contact us today to learn more about our services and how we can help you achieve academic success!
The function \(f(z)\) is a polynomial and thus holomorphic on the region defined by \(|z|\leq 2\). By the maximum modulus theorem its maximum occurs on the boundary \(|z|=2\). Then from \(|2z+5i|^{2}=4|z|^{2}+20\text{Im}(z)+25\), and \(|z|=2\) we get
\[|2z+5i|=\sqrt{16+20\text{Im}(z)+25}.\]
As \(z=x+iy\), and \(|\text{Im}(\,z)|\leq|z|=2\), \(\text{Im}(z)\) attains its maximum at the point \(z=2i\). Thus,
\[\max_{|z|\leq 2}|2z+5i|=\sqrt{81}=9.\]
### 38 Schwarz’ Lemma
The maximum modulus principle is a powerful tool for obtaining bounds for the size of holomorphic, non-constant functions. There is a simple but important consequence of the maximum modulus theorem called Schwarz’ lemma.
**Lemma 17**.: _Let \(f\) be holomorphic in the unit disc \(\mathbb{D}(0,1)\), with \(f(0)=0\) and \(|f(z)|<1\) in \(\mathbb{D}(0,1)\), Then, \(|f^{\prime}(0)|\leq 1\) and \(|f(z)|\leq|z|\) in \(\mathbb{D}(0,1)\). Strict inequality holds in both estimates unless \(f\) is a rotation of the disc: \(f(z)=e^{i\theta}z\)._
For the proof of the Schwarz Lemma, we refer the reader to [5], pp. 135-136. Ahlfors gave a generalization of the proof of Schwarz’s lemma in terms of curvature of the metric [4]. This extended proof can be thought of as a starting point of an area in complex analysis called Geometric Function Theory (GFT). To learn more about GFT consult [34].
## Exercises
1. Show that
\[\int_{\gamma}\frac{\cos z}{z(z^{2}+1)}\ dz=\left\{\begin{array}{ll}2\pi i(1- \cos i)&\gamma:|z|=3,\\ \\ 2\pi i&\gamma:|z|=\frac{1}{3},\\ \\ 0&\gamma:|z-1|=\frac{1}{3}.\end{array}\right.\]
2. Let \(M\geq 0\) and let \(f\) be an entire function such that \(|f(z)|\leq M\) for all \(z\in\mathbb{C}\). Show that \(f\) is constant.
3. Show the following identities:
\[\mbox{i.} \int_{|z|=2}\frac{9z^{2}-iz+4}{z(z^{2}+1)}\ dz=18\pi i,\] \[\mbox{ii.} \int_{\gamma}\frac{e^{3z}+3\cosh z}{\left(z-\frac{i\pi}{2} \right)^{4}}\ dz=8\pi.\]
where \(\gamma\) is a simple closed contour containing \(\frac{i\pi}{2}\) in its interior.
4. Prove that every polynomial equation \(p(z)=a_{0}+a_{1}z+\cdots+a_{n}z^{n}=0\) with degree \(n\geq 1\) and \(a_{n}\neq 0\) has exactly \(n\) roots.
5. Prove that if \(f(z)\) is holomorphic inside and on a circle \(\gamma\) with center \(a\) and radius \(r\), then \(f(a)\) is the mean of the values of \(f(z)\) on \(\mathbb{C}\),
\[\mbox{i.e.,}\ \ \ f(a)=\frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{it})\,dt.\]
(This is called _Gauss’ Mean Value Theorem_.)
6. Suppose \(f\) is holomorphic inside and on a positively oriented contour \(\gamma\). Let \(a\) lie inside \(\gamma\). Show that \(f^{\prime}(a)\) exists and \(f^{\prime}(a)=\frac{1}{2\pi i}\int\frac{f(w)}{(w-a)^{2}}\,dw\).
(Hint: Use Cauchy’s integral formula and show that
\[\left|\frac{f(a+h)-f(a)}{h}-\frac{1}{2\pi i}\int\frac{f(w)}{(w-a)^{2}}\,dw \right|\longrightarrow 0\]
as \(h\longrightarrow 0\).)Let \(f(z)\) be entire and let \(|f(z)|\geq 1\) on the whole complex plane. Prove that \(f\) is constant.
8. Let \(f\) be continuous on a region \(A\) and holomorphic on \(A\setminus\{z_{0}\}\) for a point \(z_{0}\in A\). Show that \(f\) is holomorphic on \(A\).
9. Show that if \(F\) is holomorphic on \(A\), then so is \(f\) where \(f(z)=\dfrac{F(z)-F(z_{0})}{z-z_{0}}\), if \(z\neq z_{0}\) and \(f(z_{0})=F^{\prime}(z_{0})\) where \(z_{0}\) is some point in \(A\).
10. Let \(f\) be holomorphic on a region \(A\) and let \(\gamma\) be a circle with radius \(R\) and center \(z_{0}\) that lies in \(A\). Assume that the disc \(\{z:|z-z_{0}| \[|f^{(k)}(z_{0})|\leq\dfrac{k!}{R^{k}}\,M.\] 11. Find the maximum of \(|\mbox{sin}z|\) on \([0,2\pi]\times[0,2\pi]\). 12. Let \(f(z)=z^{2}+4z-1\) be defined on \(|z|\leq 1\). Show that the maximum modulus of \(f(z)\) over the region is \(6\). 13. If \(f\) is holomorphic in a closed region G bounded by a simple closed curve \(\gamma\) and \(f(z)\neq 0\) for all \(z\in G\), prove that the modulus \(|f(z)|\) attains its minimum on \(\gamma\). 14. Let the function \(f\) be holomorphic in the entire complex plane and satisfy \[\int_{0}^{2\pi}|f(re^{i\theta}|\,d\theta\leq r^{\frac{17}{3}}\] for all \(r>0\). Prove that \(f\) is a zero function. ### 3.9 Laurent Expansion and Singularities #### Uniformly Convergent Series and Weierstrass’ M-test Let \(S\) be a nonempty subset of \(\mathbb{C}\) and let \(\{f_{n}\}\) be a sequence of complex-valued functions defined on \(S\). We say \(\{f_{n}\}\) converges uniformly to \(f\) on \(S\) if \[d_{n}:=\sup\Big{\{}|f_{n}(z)-f(z)|:z\in S\Big{\}}\to 0\quad\mbox{ as }n\to\infty.\] As usual, we say \(\sum_{k=0}^{\infty}u_{k}(z)\) converges uniformly on \(S\) if the sequence of partial sums \(f_{n}=u_{1}+u_{2}+\dots+u_{n}\) converges uniformly on \(S\). Fortunately, there is one easy sufficient condition for uniform convergence of series, called the Weierstrass M-test (see Section 2.3). **Theorem 143** (Weierstrass M-test).: _The series \(\sum u_{k}\) converges uniformly on \(S\) if there exist real numbers \(M_{k}\) such that for all \(k\in\mathbb{N}\)_ \[|u_{k}(z)|\leq M_{k}\quad\text{for all $z\in S$ \ and \ $\sum_{k=1}^{\infty}M_{k}$ converges.}\] The proof of this theorem is very similar to the one we gave in Chapter 2. We invoke the fact that \(\mathbb{C}\) is complete, thus a sequence \(\{z_{n}\}\) converges if and only if it is a Cauchy sequence. The next theorem is about the relationship between the radius of convergence of a power series and uniform convergence. **Theorem 144**.: _Let \(\sum c_{k}z^{k}\) be a complex power series with a radius of convergence \(R>0\). Then \(\sum c_{k}z^{k}\) converges uniformly on any closed disc \(\overline{D}(0;R-\epsilon)\) for any \(\epsilon>0\) (Figure 3.40)._ : Recall that if \(\sum c_{k}z^{k}\) has a radius of convergence \(R\), then \(\sum c_{k}z^{k}\) converges absolutely for all \(|z| \[|c_{k}z^{k}|\leq M_{k}:=|c_{k}|(R-\epsilon)^{k}\quad\text{since $|z|\leq R- \epsilon Therefore \(\sum M_{k}\) converges by the above fact and thus \(\sum c_{k}z^{k}\) converges uniformly on any curve \(\gamma=\gamma(0;r)\) where \(0\leq r **Remark 77**.: Note that in general \(\sum c_{k}z^{k}\) does not converge uniformly on \(D(0;R)\). Consider the geometric series \(\sum_{k=0}^{\infty}z^{k}\), where \(R=1\). For \(|z|<1\), \[f_{n}(z)=1+z+z^{2}+\cdots+z^{n}=\frac{1-z^{n+1}}{1-z}\] Figure 3.40: Pointwise convergence versus uniform convergencewhich converges pointwise to \(f(z)=\dfrac{1}{1-z}\). However, the convergence is not uniform. To see this, consider \[d_{n}:=\sup\left\{|f_{n}(z)-f(z)|:|z|<1\right\}=\sup\left\{\left| \dfrac{1-z^{n+1}}{1-z}-\dfrac{1}{1-z}\right|:|z|<1\right\}\\ =\sup\left\{\left|\dfrac{z^{n+1}}{1-z}\right|:|z|<1\right\}\geq \sup\left\{\dfrac{x^{n+1}}{1-x}:0\leq x<1\right\}.\]
Proof
发表回复