Compute \(\int_\gg \frac{ \diff{z} }{ z }\) where \(\gg\) is the unit circle, oriented counterclockwise. More generally, show that for any \(w \in \C\) and \(r > 0\text{,}\)
Evaluate the integrals \(\int_\gamma x\,\diff{z}\text{,}\)\(\int_\gamma y\,\diff{z}\text{,}\)\(\int_\gamma z\,\diff{z}\) and \(\int_\gamma \conj z\,\diff{z}\) along each of the following paths.
\(f(z) = z + \frac 1 z\) and \(\gamma\) is parametrized by \(\gamma(t)\text{,}\)\(0\leq t\leq 1\text{,}\) and satisfies \(\Im \gamma(t) > 0\text{,}\)\(\gamma(0) = -4+i\text{,}\) and \(\gamma(1) = 6+2i\text{.}\)
Assume \(\gg\text{,}\)\(\sigma\text{,}\) and \(\tau\) are smooth. Start with the definition of \(\int_\sigma f\text{,}\) apply the chain rule to \(\sigma=\gg\circ\tau\text{,}\) and then use the change of variables formula, Theorem A.0.6.
Prove the following integration by parts statement: Let \(f\) and \(g\) be holomorphic in \(G\text{,}\) and suppose \(\gg \subset G\) is a piecewise smooth path from \(\gg(a)\) to \(\gg(b)\text{.}\) Then
\begin{equation*}
\int_\gg f g' \ = \ f(\gg(b)) g(\gg(b)) - f(\gg(a)) g(\gg(a)) -
\int_\gg f' g \, \text{.}
\end{equation*}
Show that \(\int_\gg z^n \, \diff{z}=0\) for any closed piecewise smooth \(\gg\) and any integer \(n\ne-1\text{.}\) (If \(n\) is negative, assume that \(\gg\) does not pass through the origin, since otherwise the integral is not defined.)
Exercise 4.5.13 excluded \(n=-1\) for a good reason: Exercise 4.5.4 gives a counterexample. Generalizing these, if \(m\) is any integer, find a closed path \(\gg\) so that \(\int_\gg
z^{-1}\,\diff{z}=2m\pi i\text{.}\)
Taking the previous two exercises one step further, fix \(z_0 \in \C\) and let \(\gg\) be a simple, closed, positively oriented, piecewise smooth path such that \(z_0\) is inside \(\gg\text{.}\) Show that, for any integer \(n\text{,}\)
\begin{equation*}
\int_\gg (z-z_0)^n \, \diff{z} \ = \ \begin{cases}2 \pi i \amp
\text{ if } n = -1 \, , \\ 0 \amp \text{
otherwise. } \end{cases}
\end{equation*}
Show that \(F(z)=\frac i2\Log(z+i) -\frac i2\Log(z-i)\) is an antiderivative of \(\frac1{1+z^2}\) for \(\Re(z)>0\text{.}\) Is \(F(z)\) equal to \(\arctan z\text{?}\)
Suppose \(a \in \C\) and \(\gamma_0\) and \(\gamma_1\) are two counterclockwise circles so that \(a\) is inside both of them. Give a homotopy that proves \(\gg_0 \sim_{ \C \setminus \{ a
\} } \gg_1\text{.}\)
Suppose that \(\gg\) is a closed path in a region \(G\text{,}\) parametrized by \(\gg(t),\,t\in[0,1]\text{,}\) and \(\tau\) is a continuous increasing function from \([0,1]\) onto \([0,1]\text{.}\) Show that \(\gg\) is \(G\)-homotopic to the reparametrized path \(\gg\circ\tau\text{.}\)
This exercise gives an alternative proof of Corollary 4.3.8 via Green’s Theorem A.0.10. Suppose \(G \subseteq \C\) is a region, \(f\) is holomorphic in \(G\text{,}\)\(f'\) is continuous, \(\gg\) is a simple piecewise smooth closed curve, and \(\gg\sim_G0\text{.}\) Explain that we may write
\begin{equation*}
\int_\gg f(z) \, \diff{z} \ = \ \int_\gg (u + i \, v)
(\diff{x} + i \, \diff{y}) \ = \ \int_\gg u \, \diff{x} - v
\, \diff{y} \ + \ i \int_\gg v \, \diff{x} + u \, \diff{y}
\end{equation*}
and show that these integrals vanish, by using Green’s Theorem A.0.10 together with Proposition 4.4.6, and then the Cauchy–Riemann equations (2.2).
by arguing that this integral does not change if we replace \(C[0,2]\) by \(C[0,r]\) for any \(r > 1\text{,}\) then use Proposition 4.1.8Item d to obtain an upper bound for \(|\int_{ C[0,r] } \frac{ \diff{z} }{ z^3 + 1 }|\) that goes to 0 as \(r \to \infty\text{.}\)
by writing the sine function in terms of the exponential function and making the substitution \(z = e^{i \phi}\) to turn the real integral into a complex integral.
Suppose \(f\) and \(g\) are holomorphic in the region \(G\) and \(\gg\) is a simple piecewise smooth \(G\)-contractible path. Prove that if \(f(z) = g(z)\) for all \(z \in \gg\text{,}\) then \(f(z) = g(z)\) for all \(z\) inside \(\gg\text{.}\)
Let \(f(z) = \frac{ 1 }{ z^2 - 1 }\) and define the two paths \(\gg = C[1,1]\) oriented counter-clockwise and \(\sigma =
C[-1,1]\) oriented clockwise. Show that \(\int_\gg f \ = \ \int_\sigma f\) even though \(\gg \not\sim_G \sigma\) where \(G = \C \setminus \{ \pm
1 \}\text{,}\) the region of holomorphicity of \(f\text{.}\)
This exercise gives an alternative proof of Cauchy’s Integral Formula (Theorem 4.4.5) that does not depend on Cauchy’s Theorem (Theorem 4.3.4). Suppose the region \(G\) is convex; this means that, whenever \(z\) and \(w\) are in \(G\text{,}\) the line segment between them is also in \(G\text{.}\) Suppose \(f\) is holomorphic in \(G\text{,}\)\(f'\) is continuous, and \(\gg\) is a positively oriented, simple, closed, piecewise smooth path, such that \(w\) is inside \(\gg\) and \(\gg \sim_G 0\text{.}\)