Solution. This is an integral with respect to arc length, not \(dz\). On \(\left |z\right | = 1\) put \(z = e^{i\theta }\), so \(\left |dz\right | = d\theta \) and \[\left |z - \tfrac 12\right |^{2} = \left (\cos \theta - \tfrac 12\right )^{2} + \sin ^{2}\theta = \tfrac 54 - \cos \theta .\] Hence \[\int _{\left |z\right |=1}\frac {\left |dz\right |}{\left |z-\frac 12\right |^{2}} = \int ^{2\pi }_{0}\frac {d\theta }{\frac 54 - \cos \theta }.\] Using \(\displaystyle \int ^{2\pi }_{0}\frac {d\theta }{A - \cos \theta } = \frac {2\pi }{\sqrt {A^{2}-1}}\) for \(A > 1\), with \(A = \frac 54\) and \(\sqrt {\frac {25}{16}-1} = \frac 34\), \[\boxed {\int _{\left |z\right |=1}\frac {\left |dz\right |}{\left |z-\frac 12\right |^{2}} = \frac {8\pi }{3} \approx 8.3776 .}\]

There is a slicker route. For \(\left |a\right | < 1\), \[\int _{\left |z\right |=1}\frac {\left |dz\right |}{\left |z-a\right |^{2}} = \frac {2\pi }{1 - \left |a\right |^{2}},\] which is the total mass of the Poisson kernel. With \(a = \frac 12\) this gives \(\frac {2\pi }{3/4} = \frac {8\pi }{3}\) at once. Problem 11 is the same identity on a circle of radius \(\rho \).

Solution. Fix \(z_0 \notin \{\gamma \}\). Since \(\{\gamma \}\) is compact and \(z_0\) is not on it, \[d := \operatorname {dist}\big (z_0, \{\gamma \}\big ) > 0 .\] Let \(\left |z - z_0\right | < d/2\). Then for every \(w\) on \(\gamma \), \[\left |w - z\right | \geq \left |w - z_0\right | - \left |z - z_0\right | \geq d - \tfrac {d}{2} = \tfrac {d}{2}.\] Now \[f(z) - f(z_0) = \int _{\gamma }\varphi (w)\left [\frac {1}{w-z} - \frac {1}{w-z_0}\right ]dw = \left (z - z_0\right )\int _{\gamma }\frac {\varphi (w)}{(w-z)(w-z_0)}\,dw .\] Let \(M = \max _{\gamma }\left |\varphi \right |\), finite because \(\varphi \) is continuous on a compact set, and let \(L\) be the length of \(\gamma \), finite because \(\gamma \) is rectifiable. The estimation inequality gives \[\left |f(z) - f(z_0)\right | \leq \left |z - z_0\right | \cdot \frac {M}{(d/2)(d)} \cdot L = \left |z-z_0\right |\frac {2ML}{d^{2}} .\] The right-hand side tends to \(0\) as \(z \rightarrow z_0\), so \(f\) is continuous at \(z_0\). \(\blacksquare \)

The estimate in fact shows \(f\) is Lipschitz near each point, and the same device — differencing the Cauchy kernel and factoring out \(z - z_0\) — proves that \(f\) is analytic with \(f'(z) = \int _\gamma \varphi (w)(w-z)^{-2}dw\). Note that nothing was assumed about \(\varphi \) beyond continuity: it need not be analytic, or even defined off the curve.

Solution. Both singularities lie inside \(C\): \(z = 0\) and \(z = -2i\), with \(\left |-2i\right | = 2 < 3\). Both are simple. \[\operatorname *{Res}_{z=0}\frac {e^{\pi z}}{z(z+2i)} = \frac {e^{0}}{0 + 2i} = \frac {1}{2i},\] \[\operatorname *{Res}_{z=-2i}\frac {e^{\pi z}}{z(z+2i)} = \frac {e^{-2\pi i}}{-2i} = \frac {1}{-2i} = -\frac {1}{2i},\] using \(e^{-2\pi i} = 1\). The residues cancel: \[\oint _{C}\frac {e^{\pi z}}{z(z+2i)}\,dz = 2\pi i\left (\frac {1}{2i} - \frac {1}{2i}\right ) = \boxed {0 .}\]

The cancellation is not an accident of arithmetic — it is \(e^{-2\pi i} = 1\) doing the work, and the same integral with \(e^{z}\) in place of \(e^{\pi z}\) would not vanish. It is worth checking such a result numerically before believing it, and this one does come out to zero to machine precision.

Solution. Write \(h(s) = 2s^{2} - s - 2\), which is entire.

For \(\left |z\right | < 3\). Cauchy’s integral formula applied to \(h\) on the circle \(C\) gives \[\int _{C}\frac {h(s)}{s-z}\,ds = 2\pi i\,h(z).\] At \(z = 2\), which lies inside \(C\), \[g(2) = 2\pi i\,h(2) = 2\pi i\big (2(4) - 2 - 2\big ) = 2\pi i (4) = \boxed {8\pi i .}\]

For \(\left |z\right | > 3\). Now the point \(s = z\) lies outside \(C\), so the integrand \(\frac {h(s)}{s-z}\) is analytic as a function of \(s\) everywhere inside and on \(C\). By Cauchy’s theorem \[\boxed {g(z) = 0 \qquad \text {for } \left |z\right | > 3 .}\]

So \(g\) is \(2\pi i\,h\) inside the circle and identically zero outside — one formula, two entirely different functions, separated by the contour. \(g\) is of course not continuous across \(\left |z\right | = 3\); the jump is exactly the statement of the Cauchy integral formula.

Solution. The contour is the circle of centre \(4\) and radius \(2\), so it encloses the real interval \((2,6)\).

Where are the singularities? \(z^{2} - 25 = 0\) at \(z = \pm 5\). Only \(z = 5\) lies inside, since \(\left |5 - 4\right | = 1 < 2\) while \(\left |-5-4\right | = 9 > 2\).

Is \(\log z\) analytic here? The principal branch is analytic on \(\mathbb {C}\setminus (-\infty ,0]\). The disc \(B(4,2)\) lies in \(\operatorname {Re}z > 2\), well clear of the cut, so \(\log z\) contributes no singularity of its own — which is the point of the question.

The pole at \(5\) is simple, and \[\operatorname *{Res}_{z=5}\frac {\log z}{z^{2}-25} = \frac {\log 5}{2z}\bigg |_{z=5} = \frac {\log 5}{10}.\] Hence \[\int _{\gamma }\frac {\log z}{z^{2}-25}\,dz = 2\pi i\cdot \frac {\log 5}{10} = \boxed {\frac {\pi i \log 5}{5} \approx 1.0112\,i .}\]

Had the contour been large enough to reach the negative axis, the branch cut would have had to be accounted for and the residue theorem could not have been applied in this form.

Solution. (a). Parametrise \(\gamma \) by \(\gamma (t)\), \(t \in [a,b]\); then \(\overline {\gamma }\) is parametrised by \(\overline {\gamma (t)}\) with derivative \(\overline {\gamma '(t)}\).

Continuity of \(z \mapsto \overline {f(\overline {z})}\) is immediate: it is the composition of the continuous maps \(z \mapsto \overline {z}\), then \(f\), then conjugation again.

For the identity, compute both sides: \[\int _{\overline {\gamma }} f(z)\,dz = \int ^{b}_{a} f\big (\overline {\gamma (t)}\big )\,\overline {\gamma '(t)}\,dt,\] \[\overline {\int _{\gamma }\overline {f\big (\overline {z}\big )}\,dz} = \overline {\int ^{b}_{a}\overline {f\big (\overline {\gamma (t)}\big )}\,\gamma '(t)\,dt} = \int ^{b}_{a} f\big (\overline {\gamma (t)}\big )\,\overline {\gamma '(t)}\,dt,\] using that conjugation commutes with the real integral and reverses products. The two agree. \(\blacksquare \)

(b). On the unit circle \(\overline {z} = 1/z\), so differentiating, \[\overline {dz} = d\left (\frac {1}{z}\right ) = -\frac {dz}{z^{2}} .\] Therefore, conjugating an integral over \(\gamma \) directly, \[\overline {\int _{\left |z\right |=1} f(z)\,dz} = \int _{\left |z\right |=1}\overline {f(z)}\ \overline {dz} = -\int _{\left |z\right |=1}\overline {f(z)}\,\frac {dz}{z^{2}} .\] \(\blacksquare \)

A caution about this part. The relation \(\overline {z} = 1/z\) holds only on the unit circle and is false off it, so this identity does not extend to other contours; the general statement is part (a). Checked against \(f(z) = 1/z\), where both sides equal \(-2\pi i\), and against \(f(z) = e^{z}\), where both vanish.

Solution. Statement. Let \(f\) be continuous on an open set \(\Omega \). If \[\int _{\partial \Delta } f(z)\,dz = 0\] for every closed triangle \(\Delta \subseteq \Omega \), then \(f\) is analytic on \(\Omega \).

Proof. Analyticity is local, so it is enough to prove it on a disc \(D \subseteq \Omega \); fix \(a \in D\) and define \[F(z) = \int _{[a,z]} f(w)\,dw \qquad (z \in D),\] the integral along the straight segment, which lies in \(D\) because a disc is convex.

For \(z, z+h \in D\) the three points \(a\), \(z\), \(z+h\) span a triangle contained in \(D\), and the hypothesis says the integral of \(f\) round it vanishes. Reading that statement as “\([a,z]\) then \([z,z+h]\) equals \([a,z+h]\)”, \[F(z+h) - F(z) = \int _{[z,\,z+h]} f(w)\,dw .\] Hence \[\frac {F(z+h)-F(z)}{h} - f(z) = \frac {1}{h}\int _{[z,\,z+h]}\big (f(w) - f(z)\big )\,dw,\] and estimating, \[\left |\frac {F(z+h)-F(z)}{h} - f(z)\right | \leq \sup _{w \in [z,z+h]}\left |f(w)-f(z)\right | \xrightarrow [h \to 0]{} 0\] by continuity of \(f\) at \(z\). So \(F'(z) = f(z)\): \(F\) is analytic on \(D\).

An analytic function is infinitely differentiable, so \(F'' \) exists — that is, \(f' \) exists — and \(f\) is analytic on \(D\). \(\blacksquare \)

Morera is the converse of Cauchy’s theorem and is the standard route to analyticity when no derivative is in sight: it is what proves a locally uniform limit of analytic functions analytic, and what carries the Schwarz reflection principle across the real axis.

Solution. On the circle \(\left |z\right | = R\) we have \(z\overline {z} = R^{2}\), so \[\overline {z} = \frac {R^{2}}{z}, \qquad \text {and differentiating along the curve,}\qquad d\overline {z} = -\frac {R^{2}}{z^{2}}\,dz .\] Therefore \[\int _{C} p(z)\,d\overline {z} = -R^{2}\int _{C}\frac {p(z)}{z^{2}}\,dz .\] Write \(p(z) = c_0 + c_1 z + c_2 z^{2} + \cdots \). Then \[\frac {p(z)}{z^{2}} = \frac {c_0}{z^{2}} + \frac {c_1}{z} + c_2 + \cdots ,\] whose residue at \(0\) is the coefficient of \(1/z\), namely \(c_1 = p'(0)\). By the residue theorem, \[\int _{C}\frac {p(z)}{z^{2}}\,dz = 2\pi i\, p'(0),\] and so \[\boxed {\int _{C} p(z)\,d\overline {z} = -2\pi i R^{2} p'(0).}\]

Two remarks. The integral is not of an analytic differential — \(d\overline {z}\) is not \(dz\) — and it is emphatically not zero, even though \(p\) is entire; Cauchy’s theorem says nothing here. And the answer depends on \(R\), growing like \(R^{2}\), which no genuine contour integral of an entire function against \(dz\) would do.

Solution. (a). Cauchy’s Theorem. If \(G\) is a simply connected region and \(f\) is analytic on \(G\), then \[\int _{\gamma } f(z)\,dz = 0\] for every closed rectifiable curve \(\gamma \) in \(G\).

(b). Fix \(a \in G\) and define \[F(z) = \int _{\gamma _z} f(w)\,dw,\] where \(\gamma _z\) is any rectifiable path in \(G\) from \(a\) to \(z\).

\(F\) is well defined. If \(\sigma \) is another such path, then \(\gamma _z\) followed by \(\sigma \) reversed is a closed curve in \(G\), so by part (a) its integral vanishes and the two paths give the same value. This is exactly where simple connectedness is used, and the conclusion is false without it: on \(G = \mathbb {C}\setminus \{0\}\) with \(f(z) = 1/z\), two paths from \(1\) to \(-1\) on opposite sides of the origin differ by \(2\pi i\), which is why no single-valued logarithm exists there.

\(F' = f\). Given \(z \in G\), take a disc about \(z\) inside \(G\). For small \(h\) we may reach \(z+h\) by first following \(\gamma _z\) and then the segment \([z, z+h]\), so \[F(z+h) - F(z) = \int _{[z,\,z+h]} f(w)\,dw,\] and the estimate of Morera’s proof gives \[\left |\frac {F(z+h)-F(z)}{h} - f(z)\right | \leq \sup _{w\in [z,z+h]}\left |f(w)-f(z)\right | \longrightarrow 0 .\] Hence \(F\) is analytic on \(G\) with \(\frac {dF}{dz} = f\). \(\blacksquare \)

Solution. Since \(f\) is holomorphic on \(\left |z\right | < R\) with \(R > 1\), its Taylor series \[f(z) = \sum ^{\infty }_{n=0} a_n z^{n},\qquad a_n = \frac {f^{(n)}(0)}{n!},\] converges uniformly on the closed unit circle, so it may be integrated term by term. The only integrals needed are \[\int ^{2\pi }_{0} e^{ikt}\,dt = \begin {cases} 2\pi & k = 0,\\ 0 & k \neq 0.\end {cases}\]

Write the weight in exponentials: \[\cos ^{2}\frac {t}{2} = \frac {1+\cos t}{2} = \frac {1}{2} + \frac {e^{it} + e^{-it}}{4}.\] Then, with \(f(e^{it}) = \sum a_n e^{int}\), \[\int ^{2\pi }_{0} f(e^{it})\cdot \frac {1}{2}\,dt = \frac {1}{2}\big (2\pi a_0\big ) = \pi a_0,\] \[\int ^{2\pi }_{0} f(e^{it})\cdot \frac {e^{it}}{4}\,dt = \frac {1}{4}\sum _n a_n\!\int ^{2\pi }_{0}\! e^{i(n+1)t}dt = 0,\] because \(n + 1 \geq 1\) for every \(n \geq 0\) and no term is constant; \[\int ^{2\pi }_{0} f(e^{it})\cdot \frac {e^{-it}}{4}\,dt = \frac {1}{4}\sum _n a_n\!\int ^{2\pi }_{0}\! e^{i(n-1)t}dt = \frac {1}{4}\big (2\pi a_1\big ) = \frac {\pi }{2}a_1 .\] Adding, and using \(a_0 = f(0)\), \(a_1 = f'(0)\), \[\boxed {\int ^{2\pi }_{0} f\!\left (e^{it}\right )\cos ^{2}\frac {t}{2}\,dt = \pi f(0) + \frac {\pi }{2}f'(0).}\]

For the second, \(\sin ^{2}\frac {t}{2} = \frac {1-\cos t}{2} = \frac 12 - \frac {e^{it}+e^{-it}}{4}\), so the \(a_1\) term changes sign: \[\boxed {\int ^{2\pi }_{0} f\!\left (e^{it}\right )\sin ^{2}\frac {t}{2}\,dt = \pi f(0) - \frac {\pi }{2}f'(0).}\]

The mechanism is worth naming: integrating against \(e^{-ikt}\) over a full period selects the \(k\)-th Taylor coefficient. Adding the two results recovers \(\int _0^{2\pi } f(e^{it})dt = 2\pi f(0)\), the mean value property.

Solution. Recognise the denominator. With \(z = x+iy\), \[x^{2}+y^{2}-2ax+a^{2} = \left |z\right |^{2} - 2a\operatorname {Re}z + a^{2} = \left |z - a\right |^{2},\] so the integral is \(\displaystyle \int _{\left |z\right |=\rho } \frac {\left |dz\right |}{\left |z-a\right |^{2}}\), the same shape as Problem 1.

On \(\left |z\right | = \rho \) put \(z = \rho e^{i\theta }\), so \(ds = \rho \,d\theta \) and \[\left |z-a\right |^{2} = \rho ^{2} + a^{2} - 2a\rho \cos \theta .\] Hence \[\int _{\gamma }\frac {ds}{\left |z-a\right |^{2}} = \rho \int ^{2\pi }_{0}\frac {d\theta }{\big (\rho ^{2}+a^{2}\big ) - 2a\rho \cos \theta } = \rho \cdot \frac {2\pi }{\sqrt {\big (\rho ^{2}+a^{2}\big )^{2} - (2a\rho )^{2}}},\] using \(\int ^{2\pi }_{0}\frac {d\theta }{A - B\cos \theta } = \frac {2\pi }{\sqrt {A^2-B^2}}\) for \(A > \left |B\right |\). The discriminant is a perfect square, \[\big (\rho ^{2}+a^{2}\big )^{2} - 4a^{2}\rho ^{2} = \big (\rho ^{2}-a^{2}\big )^{2},\] and \(\rho > a > 0\) makes \(\rho ^{2} - a^{2}\) positive, so \[\boxed {\int _{\gamma }\frac {ds}{x^{2}+y^{2}-2ax+a^{2}} = \frac {2\pi \rho }{\rho ^{2}-a^{2}} .}\]

Setting \(\rho = 1\) and \(a = \frac 12\) returns \(\frac {2\pi }{3/4} = \frac {8\pi }{3}\), which is Problem 1. The blow-up as \(\rho \downarrow a\) is the point \(z = a\) approaching the contour.

Solution. Let \(z_k\) be a zero of \(f\) of multiplicity \(m_k\), so near \(z_k\) \[f(z) = (z-z_k)^{m_k} g(z),\qquad g(z_k) \neq 0,\ g \text { analytic}.\] Differentiating and dividing, \[\frac {f'(z)}{f(z)} = \frac {m_k}{z - z_k} + \frac {g'(z)}{g(z)},\] the second term being analytic at \(z_k\). Multiplying by \(z\), \[\frac {z f'(z)}{f(z)} = \frac {z\,m_k}{z-z_k} + z\frac {g'(z)}{g(z)},\] which has a simple pole at \(z_k\) with \[\operatorname *{Res}_{z=z_k}\frac {zf'(z)}{f(z)} = m_k z_k .\] These are the only singularities inside \(\gamma \): \(f\) is analytic there, and \(f \neq 0\) on \(\gamma \), so no pole sits on the contour. Summing over the zeros and applying the residue theorem, \[\boxed {\int _{\gamma }\frac {z f'(z)}{f(z)}\,dz = 2\pi i\sum ^{n}_{k=1} m_k z_k .}\]

With the \(z\) removed this is the argument principle, counting zeros with multiplicity; the extra factor weights each zero by its location, so the integral returns the sum of the roots with multiplicity. Dividing by the argument principle’s value \(2\pi i \sum m_k\) gives their centroid.

Solution. The contour is the circle of centre \(1\) and radius \(\frac 12\), which lies in \(\operatorname {Re}z > 0\). The principal branch of \(z^{1/m}\) is therefore analytic on and inside \(\gamma \), and the only singularity of the integrand is the pole of order \(m\) at \(z = 1\).

For a pole of order \(m\), \[\operatorname *{Res}_{z=1}\frac {z^{1/m}}{(z-1)^{m}} = \frac {1}{(m-1)!}\lim _{z\to 1}\frac {d^{m-1}}{dz^{m-1}}\left [z^{1/m}\right ].\] Differentiating the power repeatedly, \[\frac {d^{m-1}}{dz^{m-1}}z^{1/m} = \left [\prod ^{m-2}_{j=0}\left (\frac {1}{m}-j\right )\right ] z^{\frac 1m-(m-1)},\] and at \(z = 1\) the power is \(1\). Hence \[\boxed {\int _{\gamma }\frac {z^{1/m}}{(z-1)^{m}}\,dz = \frac {2\pi i}{(m-1)!}\prod ^{m-2}_{j=0}\left (\frac {1}{m}-j\right ).}\]

The first few values, with the empty product equal to \(1\): \[m=1:\ 2\pi i,\qquad m=2:\ \pi i,\qquad m=3:\ -\frac {2\pi i}{9},\qquad m=4:\ \frac {7\pi i}{64}\ \ (\approx 0.3436\,i).\] The sign alternates from \(m = 3\) onwards, because every factor \(\frac 1m - j\) with \(j \geq 1\) is negative.

Solution. An antiderivative is available: \(\frac {d}{dz}\left (2z^{1/2}\right ) = z^{-1/2}\), so the integral is a difference of values of \(2z^{1/2}\) — provided the branch is followed continuously along the path.

The path runs from \(1\) to \(-i\) and then from \(-i\) to \(-1\), so it lies in the closed lower half plane. Its argument decreases continuously, \[\arg z:\quad 0 \ \longrightarrow \ -\frac {\pi }{2} \ \longrightarrow \ -\pi ,\] and stays within \((-\pi , 0]\), where the principal branch is continuous. So we may use the principal branch throughout, taking the limiting value \(\arg = -\pi \) at the endpoint.

With \(F(z) = 2z^{1/2} = 2e^{\frac 12 \operatorname {Log} z}\), \[F(1) = 2e^{0} = 2,\qquad F(-1) = 2e^{\frac 12(-i\pi )} = 2e^{-i\pi /2} = -2i,\] and therefore \[\boxed {\int _{\gamma } z^{-1/2}\,dz = F(-1) - F(1) = -2 - 2i .}\]

The trap. Evaluating \(F(-1)\) by the principal convention \(\arg (-1) = +\pi \) would give \(+2i\) and the answer \(-2 + 2i\) — the complex conjugate, and wrong. The path arrives at \(-1\) from below, so the argument that must be used is \(-\pi \), not \(+\pi \). On a multivalued integrand it is the branch carried along the path that counts, never the value a convention assigns at the endpoint in isolation.