Find the Laurent series of \(f(z)=\frac{1}{(z-1)(z-2)}\) valid in the annulus \(1<|z|<2\).
Prove that distinct complex numbers \(z_1,z_2,z_3\) are the vertices of an equilateral triangle if and only if \[ z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1. \]
- Prove that if \(|w_1|=c|w_2|\), where \(c>0\), then \[ |w_1-c^2w_2|=c|w_1-w_2|. \]
- If \(c>0\), \(c\ne1\), and \(z_1\ne z_2\), prove that \[ \left|\frac{z-z_1}{z-z_2}\right|=c \] is a circle, and find its center and radius.
Expand \[ \frac1{1-z^2}+\frac1{z-3} \] as a Laurent series about \(0\) on each of the regions \(|z|<1\), \(1<|z|<3\), and \(|z|>3\).
For \(z_1,z_2\in\CC\):
- prove \[ |1-\bar z_1z_2|^2-|z_1-z_2|^2 =(1-|z_1|^2)(1-|z_2|^2); \]
- deduce that if \(|z_1|,|z_2|<1\), then \[ \left|\frac{z_1-z_2}{1-\bar z_1z_2}\right|<1; \]
- if \(z_1\ne z_2\), prove that the last modulus equals \(1\) if and only if \(|z_1|=1\) or \(|z_2|=1\).
Let \((f_n)\) and \(f\) be holomorphic on \(\DD\). Prove that the following are equivalent:
- \(f_n\to f\) uniformly on compact subsets of \(\DD\);
- for every \(0<r<1\), \[ \int_{|z|=r}|f_n(z)-f(z)|\,|dz|\longrightarrow0. \]
Use \(n\)-th roots of unity to prove \[ 2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=n. \]
Let \(u\) be harmonic on an open disk of radius \(R>0\), with continuous second partial derivatives. Fix \((a,b)\) in the disk and define \[ v(x,y)=\int_{(a,b)}^{(x,y)}(-u_y\,dx+u_x\,dy). \]
- Show that the integral is path-independent. (b) Show that \(u+iv\) is analytic. (c) Show that \(v\) is harmonic.
Let \(f:D\to\CC\) be continuous and let \(\alpha:[a,b]\to D\) be a smooth curve. (a) Define \(\int_\alpha f\). (b) If \(|f|\le M\) on the image of \(\alpha\), prove \[ \left|\int_\alpha f\right|\le M\,\operatorname{length}(\alpha). \]
(c) For \(C_R:|z|=R\), \(R>1\), give an upper bound for \[ \left|\int_{C_R}\frac{\log z}{z^2}\,dz\right|, \] after fixing a continuous determination of the logarithm along the parametrized circle.
Solution
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.