Use \(n\)th roots of unity to show that \(2^{n-1}\sin\frac{\pi}{n}\sin\frac{2\pi}{n}\cdots\sin\frac{(n-1)\pi}{n}=n\).
Let \(f\) be continuous on \[ D=\{z:|z|>R,\ 0\le\arg z\le\theta\},\qquad 0\le\theta\le2\pi, \] and suppose \(zf(z)\to k\) as \(z\to\infty\) in \(D\). If \(L_{R'}\) is the part of \(|z|=R'\) lying in \(D\), show that \[ \lim_{R'\to\infty}\int_{L_{R'}}f(z)\,dz=i\theta k. \]
Let \(f(z)=\sum_{n=0}^\infty c_nz^n\) be analytic and one-to-one on \(\DD\). For \(0<r_0<1\), let \(\overline D_{r_0}=\{|z|\le r_0\}\). Show that the area of \(f(\overline D_{r_0})\) is finite and equals \[ \pi\sum_{n=1}^\infty n|c_n|^2r_0^{2n}. \]
Prove the parallelogram identity \[ |z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2), \] and explain its geometric meaning.
Find the Laurent expansions of \[ f(z)=\frac{z+1}{z(z-1)} \] about \(z=0\) and \(z=1\) in the punctured disks determined by the other singularity.
Prove that on \(|z|=1\):
- \(\sum_{n\ge1}nz^n\) diverges everywhere;
- \(\sum_{n\ge1}z^n/n^2\) converges everywhere;
- \(\sum_{n\ge1}z^n/n\) converges everywhere except at \(z=1\).
Let \(\gamma\) be a positively oriented piecewise smooth simple closed curve with interior \(\Omega_1\) and exterior \(\Omega_2\). Suppose \(f\) is holomorphic on an open set containing \(\gamma\cup\Omega_2\) and \(f(z)\to A\) as \(z\to\infty\). Prove \[ \frac1{2\pi i}\int_\gamma\frac{f(\xi)}{\xi-z}\,d\xi =\begin{cases} A,&z\in\Omega_1,\\ -f(z)+A,&z\in\Omega_2. \end{cases} \]
Let \(D\) contain the closed unit disk in its interior, and let \(f\) be analytic in \(D\) except for finitely many simple poles \(z_1,\dots,z_k\) on \(|z|=1\), with residues \(s_1,\dots,s_k\). If \[ f(z)=\sum_{n=0}^\infty a_nz^n\qquad(|z|<1), \] prove that there is \(M>0\) such that \(|a_n|\le M\) for all \(n\).
Let \(f:[0,\infty)\to\RR\) be continuous, differentiable on \((0,\infty)\), satisfy \(f(0)=0\), and suppose \(f'\) is increasing. For \(x>0\) define \(g(x)=f(x)/x\). Prove that \(g\) is increasing.
Prove or disprove: there is a sequence of analytic polynomials \(p_n\) such that \(p_n(z)\to\bar z\) uniformly for \(z\in\partial\DD\).
For \(R>0\), show that there is \(N_R\) such that for \(n>N_R\), \[ P_n(z)=1+z+\frac{z^2}{2!}+\cdots+\frac{z^n}{n!}\ne0 \qquad(|z|\le R). \]
Let \(f\) be analytic on \(\DD\), with \(f(0)=f'(0)=0\). Show that \[ g(z)=\sum_{n=1}^\infty f(z/n) \] defines an analytic function on \(\DD\). Moreover, prove that \[ g(z)=f(z)\sum_{n=1}^\infty\frac1{n^2} \] for all \(z\in\DD\) if and only if \(f(z)=cz^2\) for some constant \(c\).
Solution
Let \(\zeta=e^{2\pi i/n}\). From \[ z^n-1=(z-1)\prod_{k=1}^{n-1}(z-\zeta^k) \] we obtain, after dividing by \(z-1\) and setting \(z=1\), \[ n=\prod_{k=1}^{n-1}(1-\zeta^k). \] Taking absolute values gives \[ n=\prod_{k=1}^{n-1}|1-e^{2\pi i k/n}|. \] For real \(\theta\), \[ |1-e^{i\theta}|=2\left|\sin\frac\theta2\right|. \] Since \(0<k\pi/n<\pi\) for \(1\le k\le n-1\), all these sines are positive. Therefore \[ n=\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n} =2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}, \] as required.
1. Sector arc. Parametrize the arc by \(z=R'e^{it}\), \(0\le t\le\theta\). Then \[ \int_{L_{R'}}f(z)\,dz =i\int_0^\theta R'e^{it}f(R'e^{it})\,dt. \] The hypothesis \(zf(z)\to k\) as \(z\to\infty\) in the region means that the integrand converges uniformly in \(t\in[0,\theta]\) to \(k\). Therefore \[ \boxed{\lim_{R'\to\infty}\int_{L_{R'}}f(z)\,dz=i\theta k}. \]
2. Area of a univalent image. Since \(f\) is holomorphic, its real Jacobian is \(|f'(z)|^2\). Injectivity means the change-of-variables formula has multiplicity one, so \[ \operatorname{Area}(f(D_{r_0})) =\int_{|z|<r_0}|f'(z)|^2\,dA(z). \] The derivative is bounded on the compact disk \(|z|\le r_0\), so this area is finite. Since \[ f'(z)=\sum_{n=1}^\infty nc_nz^{n-1}, \] orthogonality of \(e^{in\theta}\) on circles gives \[ \int_0^{2\pi}|f'(re^{i\theta})|^2\,d\theta =2\pi\sum_{n=1}^\infty n^2|c_n|^2r^{2n-2}. \] Integrating \(r\,dr\) from \(0\) to \(r_0\) yields \[ \boxed{\operatorname{Area}(f(D_{r_0})) =\pi\sum_{n=1}^\infty n|c_n|^2r_0^{2n}}. \]
3. Parallelogram law. Expanding with complex conjugates, \[ |z_1+z_2|^2+|z_1-z_2|^2 =2|z_1|^2+2|z_2|^2. \] Geometrically, if \(z_1,z_2\) are adjacent side vectors of a parallelogram, \(z_1\pm z_2\) are its diagonal vectors; the identity says that the sum of the squares of the diagonal lengths equals the sum of the squares of the four side lengths.
4. Laurent expansions at \(0\) and \(1\). Partial fractions give \[ \frac{z+1}{z(z-1)}=-\frac1z+\frac2{z-1}. \] Thus for \(0<|z|<1\), \[ \boxed{f(z)=-\frac1z-2\sum_{n=0}^\infty z^n}. \] Writing \(w=z-1\), we have \[ f(z)=\frac2w-\frac1{1+w}, \] so for \(0<|z-1|<1\), \[ \boxed{f(z)=\frac2{z-1}-\sum_{n=0}^\infty(-1)^n(z-1)^n}. \]
5. Boundary power series. For \(|z|=1\), the terms of \(\sum nz^n\) have modulus \(n\), so they do not tend to zero; hence the series diverges everywhere. The series \(\sum z^n/n^2\) converges absolutely because \(\sum1/n^2<\infty\). Finally, at \(z=1\), \(\sum z^n/n\) is the divergent harmonic series. If \(|z|=1\) and \(z\ne1\), the partial sums of \(\sum_{n=1}^N z^n\) are \[ \frac{z(1-z^N)}{1-z}, \] and are uniformly bounded in \(N\); since \(1/n\searrow0\), Dirichlet’s test gives convergence.
6. Exterior Cauchy formula. Put \(h=f-A\), so \(h(z)\to0\) at infinity. Choose a large circle \(C_R\) containing \(\gamma\) and the point \(z\) when \(z\in\Omega_2\). Since \(h(\xi)/(\xi-z)=o(1/R)\) uniformly on \(C_R\), \[ \int_{C_R}\frac{h(\xi)}{\xi-z}\,d\xi\to0. \] If \(z\in\Omega_1\), the integrand is holomorphic in the annulus between \(\gamma\) and \(C_R\), hence \[ \int_{C_R}\frac{h(\xi)}{\xi-z}\,d\xi -\int_\gamma\frac{h(\xi)}{\xi-z}\,d\xi=0, \] and the second integral is therefore zero. If \(z\in\Omega_2\), the same annulus contains the simple pole at \(\xi=z\), so \[ \int_{C_R}\frac{h(\xi)}{\xi-z}\,d\xi -\int_\gamma\frac{h(\xi)}{\xi-z}\,d\xi =2\pi i h(z). \] Letting \(R\to\infty\) gives \((2\pi i)^{-1}\int_\gamma h/(\xi-z)=0\) in \(\Omega_1\) and \(-h(z)\) in \(\Omega_2\). Adding the constant term \(A\), whose Cauchy integral is \(A\) in \(\Omega_1\) and \(0\) in \(\Omega_2\), gives the claimed formula.
7. Bounded Taylor coefficients. Define \[ H(z)=f(z)-\sum_{j=1}^k\frac{s_j}{z-z_j}. \] The simple poles are removed, so \(H\) is holomorphic on a neighborhood of the closed unit disk. Hence there is \(\rho>1\) such that \(H(z)=\sum b_nz^n\) on \(|z|<\rho\), and Cauchy’s estimates give \(|b_n|\le C\rho^{-n}\) for some \(C\). For \(|z|<1\) and \(|z_j|=1\), \[ \frac{s_j}{z-z_j} =-\sum_{n=0}^\infty\frac{s_j}{z_j^{n+1}}z^n. \] Thus \[ a_n=b_n-\sum_{j=1}^k\frac{s_j}{z_j^{n+1}}, \] and therefore \[ |a_n|\le C+\sum_{j=1}^k|s_j|. \] This constant is independent of \(n\).
8. Monotonicity of \(f(x)/x\). Let \(0<x<y\). By the mean value theorem there are \(\xi\in(0,x)\) and \(\eta\in(x,y)\) such that \[ \frac{f(x)-f(0)}x=f'(\xi), \qquad \frac{f(y)-f(x)}{y-x}=f'(\eta). \] Since \(f'\) is increasing, the second slope is at least the first. Using \(f(0)=0\), \[ f(y)\ge f(x)+(y-x)\frac{f(x)}x =y\frac{f(x)}x, \] so \(f(y)/y\ge f(x)/x\).
9. Polynomial approximation to \(\bar z\) on the circle. Such a sequence does not exist. On \(|z|=1\), \(\bar z=1/z\). If \(p_n\to\bar z\) uniformly there, then \[ zp_n(z)\to1 \] uniformly on \(|z|=1\). By the maximum modulus principle applied to \(zp_n(z)-1\), the convergence would be uniform on the whole closed disk. But at \(z=0\) every term equals \(-1\), a contradiction.
10. Exponential partial sums. The polynomials \(P_n\) converge uniformly to \(e^z\) on every closed disk. On \(|z|\le R\), \[ |e^z|=e^{\Re z}\ge e^{-R}. \] Choose \(N_R\) so that \[ \sup_{|z|\le R}|P_n(z)-e^z|<e^{-R}/2 \qquad(n>N_R). \] Then \(|P_n(z)|\ge e^{-R}/2>0\) on the disk, as required.
11. The series \(\sum f(z/n)\). Since \(f(0)=f'(0)=0\), the function \(F(z)=f(z)/z^2\) extends holomorphically across \(0\). On every compact disk \(|z|\le r<1\), let \(M_r=\max_{|w|\le r}|F(w)|\). Then \[ |f(z/n)|\le M_r\frac{|z|^2}{n^2}, \] so the series defining \(g\) converges uniformly on compact subsets by the \(M\)-test and hence defines a holomorphic function.
Write \[ f(z)=\sum_{m=2}^\infty a_mz^m. \] Normal convergence permits summing coefficientwise: \[ g(z)=\sum_{m=2}^\infty a_m\zeta(m)z^m, \qquad \zeta(m)=\sum_{n=1}^\infty n^{-m}. \] Thus \(g=\zeta(2)f\) exactly when \[ a_m(\zeta(m)-\zeta(2))=0 \qquad(m\ge2). \] For every \(m>2\), \(\zeta(m)<\zeta(2)\), so \(a_m=0\). Hence \(f(z)=a_2z^2\). Conversely, for \(f(z)=cz^2\) one has \(g(z)=cz^2\sum n^{-2}=\zeta(2)f(z)\).