Complex analysis

The Cartesian Cauchy–Riemann equations for \(f=u+iv\) are \[ u_x=v_y,\qquad u_y=-v_x. \] A function on an open set is holomorphic if it is complex differentiable at every point; meromorphic if it is holomorphic except for isolated poles; analytic if it is locally represented by a convergent power series (equivalently, in one complex variable, holomorphic); harmonic if it is \(C^2\) and satisfies \(u_{xx}+u_{yy}=0\); entire if it is holomorphic on all of \(\CC\). Uniform continuity means that for every \(\epsilon>0\) there is a \(\delta>0\) working simultaneously for all pairs of points in the domain; uniform boundedness means that one constant \(M\) bounds the modulus everywhere. A sequence \(f_n\to f\) converges uniformly if \(\sup_x|f_n(x)-f(x)|\to0\); a series converges uniformly when its partial sums do. The Laplace equation is \(u_{xx}+u_{yy}=0\). The Dirichlet problem asks for a harmonic function on a domain with prescribed boundary values.

For partial fractions, factor the denominator over \(\CC\) and write one term \(A_{j,k}/(z-a_j)^k\) for every multiplicity \(k\) of every pole \(a_j\); multiplying through by the denominator and comparing coefficients (or evaluating derivatives at the poles) determines the constants.

A local branch of \(\sqrt z\) about \(z_0=4+3i\) has radius of convergence \[ \boxed{5}. \] Indeed, the open disk of radius \(5\) about \(z_0\) avoids \(0\) and is simply connected, so it admits a holomorphic square root. A larger Taylor disk would contain \(0\); if a holomorphic \(g\) there satisfied \(g^2=z\), then \(g(0)=0\) and differentiating would give \(2g(0)g'(0)=1\), impossible.

The logarithmic derivative of a nonzero holomorphic function is \(f'/f\). An example with \(f^2\) analytic but \(f\) not analytic is \[ f(z)=\begin{cases}1,&\Im z\ge0,\\-1,&\Im z<0, \end{cases} \qquad f^2\equiv1. \]

For \(f(z)=z^2\) and \(|z|,|w|<R\), \[ |z^2-w^2|\le2R|z-w|, \] so \(z^2\) is uniformly continuous on that disk. On \(\CC\), take \(z_n=n\) and \(w_n=n+1/n\); then \(|z_n-w_n|\to0\) but \(|z_n^2-w_n^2|\to2\), so uniform continuity fails.

If a real-differentiable map \(f=u+iv\) is given at \((a,b)\), then it is complex differentiable there exactly when its real derivative is \(\CC\)-linear, equivalently exactly when the Cauchy–Riemann equations hold at \((a,b)\).

In polar coordinates, the chain rule gives \[ u_r=u_x\cos\theta+u_y\sin\theta, \qquad v_\theta=r(-v_x\sin\theta+v_y\cos\theta). \] Using \(u_x=v_y\) and \(u_y=-v_x\) yields \[ u_r={1\over r}v_\theta, \qquad v_r=-{1\over r}u_\theta. \]

For the radius-one power-series questions:

1. \(\sum_{n\ge1}z^n/n^2\) has radius \(1\) and converges absolutely at every point of \(S^1\).
2. \(f(z)=1/(1+z)=\sum_{n\ge0}(-1)^nz^n\) has radius \(1\) and is analytic at \(1\), but \(\sum_{n\ge0}(-1)^n\) diverges.
3. If \(f\) were analytic at every point of \(S^1\), compactness of \(S^1\) would give finitely many neighborhoods on which \(f\) is analytic, together containing an annulus around \(S^1\). Joined with the unit disk, these give analyticity on a disk \(|z|<1+\epsilon\) for some \(\epsilon>0\), contradicting the assumed Taylor radius \(1\).

Finally, \[ {z+1\over z(z-1)^2}={1\over z}-{1\over z-1}+{2\over(z-1)^2}. \] About \(0\) this gives \[ {1\over z}+\sum_{n\ge0}(2n+3)z^n, \qquad 0<|z|<1, \] and \[ \sum_{n\ge2}(2n-3)z^{-n}, \qquad |z|>1. \] Writing \(w=z-1\), the expansions about \(1\) are \[ {2\over w^2}-{1\over w}+\sum_{n\ge0}(-1)^nw^n, \qquad 0<|w|<1, \] and \[ {1\over w^2}+\sum_{n\ge3}(-1)^{n-1}w^{-n}, \qquad |w|>1. \]

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 \(g(z)=cz^2\sum n^{-2}=\zeta(2)f(z)\).

Partial fractions give \[ \frac1{(z-1)(z-2)} =-\frac1{z-1}+\frac1{z-2}. \] On the annulus \(1<|z|<2\), expand the first term in powers of \(z^{-1}\) and the second in powers of \(z\): \[ -\frac1{z-1} =-\frac1z\frac1{1-z^{-1}} =-\sum_{n=1}^{\infty}z^{-n}, \] because \(|z^{-1}|<1\), while \[ \frac1{z-2} =-\frac12\frac1{1-z/2} =-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}}, \] because \(|z/2|<1\). Hence the Laurent series valid precisely on \(1<|z|<2\) is \[ \boxed{ \frac1{(z-1)(z-2)} =-\sum_{n=1}^{\infty}z^{-n} -\sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}}.} \]

1. Equilateral-triangle criterion. Let \(\omega=e^{2\pi i/3}\). Since \(1+\omega+\omega^2=0\), \[ z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1 =(z_1+\omega z_2+\omega^2z_3) (z_1+\omega^2z_2+\omega z_3). \] Thus the displayed quadratic identity holds exactly when one factor vanishes. For example, \[ z_1+\omega z_2+\omega^2z_3=0 \iff z_1-z_3=-\omega(z_2-z_3). \] The multiplier \(-\omega=e^{-i\pi/3}\) has modulus \(1\) and rotates by \(-\pi/3\). Hence the two sides issuing from \(z_3\) have equal length and meet at angle \(\pi/3\), so the triangle is equilateral. The other factor gives the opposite orientation. Conversely, an equilateral triangle has one of these two oriented side relations, so one factor vanishes and the identity follows.

2. Apollonius circle. Squaring both sides of (a), and using \(|w_1|^2=c^2|w_2|^2\), gives \[ |w_1-c^2w_2|^2 =c^2|w_1-w_2|^2, \] which proves (a). If \(|z-z_1|=c|z-z_2|\), apply (a) to \(w_1=z-z_1\), \(w_2=z-z_2\): \[ |(1-c^2)z-(z_1-c^2z_2)|=c|z_1-z_2|. \] Since \(c\ne1\), this is the circle with center and radius \[ \boxed{\frac{z_1-c^2z_2}{1-c^2}}, \qquad \boxed{\frac{c|z_1-z_2|}{|1-c^2|}}. \]

3. Three Laurent regions. For \(|z|<1\), \[ \frac1{1-z^2}+\frac1{z-3} =\sum_{k=0}^\infty z^{2k} -\sum_{k=0}^\infty\frac{z^k}{3^{k+1}}. \] For \(1<|z|<3\), \[ \frac1{1-z^2}+\frac1{z-3} =-\sum_{k=1}^\infty z^{-2k} -\sum_{k=0}^\infty\frac{z^k}{3^{k+1}}. \] For \(|z|>3\), \[ \frac1{1-z^2}+\frac1{z-3} =-\sum_{k=1}^\infty z^{-2k} +\sum_{k=0}^\infty 3^kz^{-k-1}. \] Each follows from the geometric series in the indicated region.

4. Pseudohyperbolic identity. Direct expansion gives \[ |1-\bar z_1z_2|^2-|z_1-z_2|^2 =1-|z_1|^2-|z_2|^2+|z_1|^2|z_2|^2 =(1-|z_1|^2)(1-|z_2|^2). \] If \(|z_1|,|z_2|<1\), the right side is positive, proving (b). If \(z_1\ne z_2\), then whenever the quotient is defined its modulus equals \(1\) exactly when the right side is zero, i.e. exactly when \(|z_1|=1\) or \(|z_2|=1\). Under either of those alternatives the denominator cannot vanish unless \(z_1=z_2\), which is excluded, so this also proves the converse in (c).

5. Compact convergence versus circle \(L^1\) convergence. If \(f_n\to f\) uniformly on compact sets, then on \(|z|=r\), \[ \int_{|z|=r}|f_n-f|\,|dz| \le2\pi r\sup_{|z|=r}|f_n-f|\to0. \] Conversely, fix \(0<\rho<1\) and choose \(r\) with \(\rho<r<1\). Cauchy’s formula applied to \(f_n-f\) gives, for \(|z|\le\rho\), \[ |f_n(z)-f(z)| \le\frac1{2\pi(r-\rho)} \int_{|\zeta|=r}|f_n(\zeta)-f(\zeta)|\,|d\zeta|. \] The right side tends to zero, uniformly for \(|z|\le\rho\). Every compact subset of \(\DD\) lies in such a disk, proving (a).

6. Product of sines. With \(\zeta=e^{2\pi i/n}\), \[ n=\prod_{k=1}^{n-1}(1-\zeta^k). \] Taking moduli and using \(|1-e^{2\pi ik/n}|=2\sin(k\pi/n)\) gives \[ \boxed{2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=n}. \]

7. Harmonic conjugate. The one-form \[ \eta=-u_y\,dx+u_x\,dy \] is closed because \[ \frac{\partial(-u_y)}{\partial y} =-u_{yy}=u_{xx} =\frac{\partial u_x}{\partial x}, \] using \(u_{xx}+u_{yy}=0\). A disk is simply connected, so the integral of \(\eta\) is path-independent. Hence \[ v_x=-u_y,\qquad v_y=u_x, \] which are the Cauchy–Riemann equations for \(u+iv\); therefore \(u+iv\) is analytic. Finally, \[ v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0, \] so \(v\) is harmonic.

8. Complex line integrals and the \(ML\) estimate. By definition, \[ \int_\alpha f(z)\,dz =\int_a^b f(\alpha(t))\alpha'(t)\,dt. \] Therefore \[ \left|\int_\alpha f\right| \le\int_a^b|f(\alpha(t))|\,|\alpha'(t)|\,dt \le M\,\operatorname{length}(\alpha). \] For \(C_R\), use \(z=Re^{it}\), \(0\le t\le2\pi\), and the continuous lift \(L(t)=\log R+it\) along this parametrization. Then \[ |L(t)|\le\log R+2\pi, \qquad |z|=R, \] so the \(ML\) estimate yields \[ \boxed{ \left|\int_{C_R}\frac{L(z)}{z^2}\,dz\right| \le \frac{2\pi}{R}(\log R+2\pi).} \] There is no single-valued holomorphic logarithm on a neighborhood of the full circle, so a determination along the parametrized contour is part of the data needed to interpret the last integral.

The bullets “Do any example from here” and “Anything from the homeworks” are open-ended study prompts rather than propositions with unique answers. The remaining finite tasks are as follows.

If \(f'=0\) on a region \(\Omega\), then for any \(a,b\in\Omega\) choose a piecewise \(C^1\) path \(\gamma\) from \(a\) to \(b\). Since \(f\) is a primitive of the zero function, \[ f(b)-f(a)=\int_\gamma f'(z)\,dz=0. \] Thus \(f\) is constant.

Suppose polynomials \(p_n\) converged uniformly to \(1/z\) on \(S^1\). Then \(q_n(z)=zp_n(z)-1\) converges uniformly to \(0\) on \(S^1\). By the maximum modulus principle, \[ \sup_{|z|\le1}|q_n(z)|\le\sup_{|z|=1}|q_n(z)|\longrightarrow0. \] But \(q_n(0)=-1\) for every \(n\), a contradiction.

For the Green-theorem exercise, write \(f=u+iv\). Then \[ \int_{\partial T}f(z)\,dz =\int_{\partial T}(u\,dx-v\,dy) +i\int_{\partial T}(v\,dx+u\,dy). \] Green’s theorem turns the real and imaginary parts into \[ \iint_T(-v_x-u_y)\,dxdy, \qquad \iint_T(u_x-v_y)\,dxdy, \] respectively. Both vanish by the Cauchy–Riemann equations, proving the contour integral is zero.

Finally, with \(t=e^x\), \[ \int_{-\infty}^{\infty}{e^{ax}\over1+e^x}\,dx =\int_0^\infty {t^{a-1}\over1+t}\,dt. \] For \(0<a<1\), Euler’s beta integral gives \[ \int_0^\infty {t^{a-1}\over1+t}\,dt =B(a,1-a)=\Gamma(a)\Gamma(1-a) ={\pi\over\sin\pi a}, \] by Euler’s reflection formula.

Part 1.

<1>1. No, not every continuous function on \(\bar\DD\) is uniformly approximable by polynomials in \(z\). <2>1. A uniform limit of holomorphic polynomials on \(\bar\DD\) is holomorphic on the interior \(\DD\).

<2>2. But there are continuous functions on \(\bar\DD\) that are not holomorphic on \(\DD\).

<2>3. Hence \(\bar z\) (or any such function) cannot be uniformly approximated by polynomials in \(z\).

<1>2. Q.E.D. (part 1).

Part 2.

<1>1. For each \(n \ge 0\), let \(E_n = \{z \in \CC : f^{(n)}(z) = 0\}\).

<1>2. Each \(E_n\) is closed, and \(\CC = \bigcup_{n=0}^{\infty} E_n\).

<1>3. By the Baire category theorem, some \(E_n\) has nonempty interior.

<1>4. Hence \(f^{(n)} \equiv 0\) on \(\CC\).

<1>5. Therefore \(f\) is a polynomial of degree at most \(n-1\).

<1>6. Q.E.D. (part 2).

Part 3. Since \(|\alpha|<r<|\beta|\), the integrand has exactly one pole inside \(\gamma\), namely the simple pole at \(\alpha\). Its residue is \[ \operatorname{Res}_{z=\alpha} \frac1{(z-\alpha)(z-\beta)} =\frac1{\alpha-\beta}. \] Therefore the residue theorem gives \[ \boxed{\int_\gamma\frac{dz}{(z-\alpha)(z-\beta)} =\frac{2\pi i}{\alpha-\beta}}. \]

Part 4. Parametrize \(\gamma_x\) by \(z=x+it\), \(0\le t\le b\). Then \[ \int_{\gamma_x}f(z)\,dz=i\int_0^b f(x+it)\,dt. \] Hence \[ \left|\int_{\gamma_x}f(z)\,dz-iAb\right| \le b\sup_{0\le t\le b}|f(x+it)-A|. \] The right-hand side tends to \(0\) by the assumed uniform convergence, so \[ \boxed{\lim_{x\to\infty}\int_{\gamma_x}f(z)\,dz=iAb}. \]

Part 5. Let \(z\in\bar D\). For small \(h\ne0\) with \(z+h\in\bar D\), \[ \frac{g(z+h)-g(z)}h =\frac{\overline{f(\bar z+\bar h)-f(\bar z)}}h =\overline{ \frac{f(\bar z+\bar h)-f(\bar z)}{\bar h} }. \] As \(h\to0\), also \(\bar h\to0\), so the last expression tends to \(\overline{f'(\bar z)}\). Thus \(g\) is complex differentiable at every point of \(\bar D\), with \[ \boxed{g'(z)=\overline{f'(\bar z)}}. \]