Do any example from here
Prove that no sequence of polynomials converges uniformly to \(1/z\) on \(S^1\).
Anything from the homeworks
Show that \(f'=0 \implies f\) is constant using integrals and primitives (i.e. antiderivatives).
See S&S Corollary 3.4.
Suppose \(f\) is continuously complex differentiable on \(\Omega\), and \(T\) is a triangle whose interior is contained in \(\Omega\). Use Green’s theorem to prove \(\int_{\partial T}f(z)\,dz=0\).
Show that for \(0<a<1\), \[ \int_{-\infty}^{\infty}{e^{ax}\over1+e^x}\,dx ={\pi\over\sin\pi a}. \]
Solution
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.