Solution. Note first that \(f\) has no singularity at the origin — the poles are at \(z = \pm 1\) — so on the punctured disc the Laurent series is just the Taylor series.

(a) \(0 < \left |z\right | < 1\). Substituting \(z^{2}\) into the geometric series, \[\frac {1}{1-z^{2}} = \sum ^{\infty }_{n=0} z^{2n} = 1 + z^{2} + z^{4} + \cdots ,\] valid for \(\left |z\right | < 1\). There is no principal part.

(b) \(1 < \left |z\right | < 2\). Now \(\left |z\right | > 1\), so expand in powers of \(1/z\): \[\frac {1}{1-z^{2}} = -\frac {1}{z^{2}-1} = -\frac {1}{z^{2}}\cdot \frac {1}{1 - z^{-2}} = -\frac {1}{z^{2}}\sum ^{\infty }_{n=0} z^{-2n} = -\sum ^{\infty }_{n=1} z^{-2n},\] that is \(-z^{-2} - z^{-4} - z^{-6} - \cdots \). This is all principal part.

The two series represent the same function on different annuli and neither is the other rearranged; a Laurent expansion belongs to an annulus, not to a function. The outer one is in fact valid on the whole of \(\left |z\right | > 1\), the bound \(2\) in the question being irrelevant.

Solution. Split into partial fractions: \[\frac {1}{z(z-1)(z-2)} = \frac {1/2}{z} - \frac {1}{z-1} + \frac {1/2}{z-2}.\] On \(1 < \left |z\right | < 2\) each term must be expanded in the direction that converges there — in \(1/z\) for the poles inside the annulus, in \(z\) for the pole outside.

\[-\frac {1}{z-1} = -\frac {1}{z}\cdot \frac {1}{1-\frac 1z} = -\sum ^{\infty }_{n=1} z^{-n} \qquad \left (\left |z\right |>1\right ),\] \[\frac {1/2}{z-2} = -\frac {1}{4}\cdot \frac {1}{1-\frac z2} = -\sum ^{\infty }_{n=0}\frac {z^{n}}{2^{\,n+2}} \qquad \left (\left |z\right |<2\right ).\] Adding, and combining the two contributions to \(z^{-1}\), namely \(\frac 12 - 1 = -\frac 12\), \[\boxed {\frac {1}{z(z-1)(z-2)} = -\frac {1}{2z} \;-\; \sum ^{\infty }_{n=2} z^{-n} \;-\; \sum ^{\infty }_{n=0}\frac {z^{n}}{2^{\,n+2}} .}\]

Spot-checking the coefficients numerically on \(\left |z\right | = 1.5\) gives \(c_{-1} = -0.5\), \(c_{-2} = c_{-3} = -1\), \(c_0 = -0.25\), \(c_1 = -0.125\), exactly as above.

Solution. Let \(D\) be a closed disc about \(a\) contained in \(G\), and let \(\Delta \) be any triangle in \(D\). We show \(\int _{\partial \Delta } f = 0\) and appeal to Morera.

If \(a \notin \Delta \) the integral vanishes because \(f\) is analytic on a neighbourhood of \(\Delta \). If \(a \in \Delta \), subdivide: place \(a\) as a common vertex and split \(\Delta \) into small triangles, isolating \(a\) inside a triangle \(\Delta _\varepsilon \) of diameter \(\varepsilon \). All the others contribute nothing, so \[\left |\int _{\partial \Delta } f\right | = \left |\int _{\partial \Delta _\varepsilon } f\right | \leq M \cdot \operatorname {length}\left (\partial \Delta _\varepsilon \right ) \leq 3M\varepsilon ,\] where \(M = \max _{D}\left |f\right |\), finite because \(f\) is continuous on the compact set \(D\) — this is the only place the continuity hypothesis is used, and it is indispensable. Letting \(\varepsilon \rightarrow 0\) gives \(\int _{\partial \Delta } f = 0\).

By Morera’s theorem \(f\) is analytic on the interior of \(D\), in particular at \(a\). \(\blacksquare \)

The moral: a singularity that is merely a point of continuity is no singularity at all. Compare \(1/z\), which is analytic on \(\mathbb {C}\setminus \{0\}\) but cannot be made continuous at \(0\), and so escapes the theorem.

Solution. The only singularities are \(z = 0\) (a double pole) and \(z = 1\) (simple), so the two maximal annuli centred at the origin are \[0 < \left |z\right | < 1 \qquad \text {and}\qquad \left |z\right | > 1 .\]

On \(0 < \left |z\right | < 1\). Expand \(\frac {1}{1-z}\) in powers of \(z\): \[\frac {1}{z^{2}(1-z)} = \frac {1}{z^{2}}\sum ^{\infty }_{n=0} z^{n} = \sum ^{\infty }_{n=0} z^{\,n-2} = \frac {1}{z^{2}} + \frac {1}{z} + 1 + z + z^{2} + \cdots \]

On \(\left |z\right | > 1\). Expand in powers of \(1/z\): \[\frac {1}{1-z} = -\frac {1}{z}\cdot \frac {1}{1-\frac 1z} = -\sum ^{\infty }_{n=0} z^{-n-1},\] so \[\frac {1}{z^{2}(1-z)} = -\sum ^{\infty }_{n=0} z^{-n-3} = -\frac {1}{z^{3}} - \frac {1}{z^{4}} - \cdots \]

Both are maximal: the first cannot extend past \(\left |z\right | = 1\) because of the pole there, and the second cannot extend inward for the same reason. Numerically, the inner expansion has \(c_{-2} = c_{-1} = c_0 = c_1 = 1\) and the outer has \(c_{-3} = c_{-4} = -1\) with \(c_{-2} = c_{-1} = 0\), confirming both.

Solution. Existence. Define \[f(z) = \begin {cases} \frac {\varphi (z)-\varphi (a)}{z-a} & z \in D\setminus \{a\},\\[10pt] \varphi '(a) & z = a . \end {cases}\] On \(D\setminus \{a\}\) this is a quotient of analytic functions with non-vanishing denominator, hence analytic. At \(a\) it is continuous, since \[\lim _{z\to a}\frac {\varphi (z)-\varphi (a)}{z-a} = \varphi '(a)\] is precisely the definition of the derivative. So \(f\) is continuous on \(D\) and analytic on \(D\setminus \{a\}\); by the previous problem it is analytic at \(a\) too.

Alternatively and more concretely, expand \(\varphi \) about \(a\): \[\varphi (z) = \varphi (a) + \sum ^{\infty }_{n=1} c_n (z-a)^{n} \ \Longrightarrow \ f(z) = \sum ^{\infty }_{n=1} c_n (z-a)^{n-1},\] a power series with the same radius of convergence, manifestly analytic at \(a\) with value \(c_1 = \varphi '(a)\).

Uniqueness. Any two such functions agree on \(D\setminus \{a\}\), which has \(a\) as a limit point; two analytic functions agreeing on a set with a limit point in a region are identical, by the Identity Theorem. \(\blacksquare \)

This construction is the engine of the usual proof of Cauchy’s integral formula, which is why it recurs on these papers.

Solution. The hypothesis is that \(u = \operatorname {Re}f\) is bounded on the punctured disc, between \(2011\) and \(2012\). The specific numbers are decoration; boundedness of the real part is the whole content.

The origin is an isolated singularity, so exactly one of three things holds: removable, pole, or essential. Rule out the last two.

Not a pole. At a pole \(\left |f(z)\right | \rightarrow \infty \), and since \(\left |f\right |^{2} = u^{2}+v^{2}\) with \(u\) bounded, this forces \(\left |v\right | \rightarrow \infty \). But then \(f\) omits a whole half-plane near \(0\), which we rule out next in a single stroke.

Not essential. By the Casorati–Weierstrass theorem, near an essential singularity \(f\) takes values dense in \(\mathbb {C}\). In particular it would take values with real part outside \([2011,2012]\) in every punctured neighbourhood of \(0\), contradicting the hypothesis.

The pole case dies the same way: \(\left |f\right | \rightarrow \infty \) with \(2011 < u < 2012\) is impossible, because on the circle \(\left |z\right | = r\) the function \(g = e^{f}\) would satisfy \(\left |g\right | = e^{u} \in \left (e^{2011}, e^{2012}\right )\), so \(g\) is bounded and non-vanishing near \(0\); a pole of \(f\) would make \(g\) have an essential singularity, again contradicting boundedness.

Hence \(0\) is removable, and \(f\) extends analytically to \(\left |z\right | < 210\). \(\blacksquare \)

Solution. Write \(f(z) = \cot z = \frac {\cos z}{\sin z}\).

Poles. These are the zeros of \(\sin z\) not cancelled by zeros of \(\cos z\). Now \(\sin z = 0\) exactly when \(z = n\pi \), \(n \in \mathbb {Z}\) — the complex zeros of \(\sin \) are all real, as follows from \(\left |\sin z\right |^{2} = \sin ^{2}x + \sinh ^{2}y\). At those points \(\cos (n\pi ) = (-1)^{n} \neq 0\), so nothing cancels and each is a pole.

Each zero of \(\sin \) is simple, since \(\frac {d}{dz}\sin z = \cos z\) is non-zero there. Hence every pole of \(\cot z\) is simple.

Residues. For a simple pole arising as \(p/q\) with \(q(z_0) = 0\), \(q'(z_0)\neq 0\), the residue is \(p(z_0)/q'(z_0)\): \[\operatorname *{Res}_{z=n\pi }\frac {\cos z}{\sin z} = \frac {\cos (n\pi )}{\cos (n\pi )} = \boxed {1}\qquad \text {for every } n \in \mathbb {Z}.\]

So \(\cot z\) has a simple pole of residue \(1\) at every integer multiple of \(\pi \) and nowhere else — confirmed numerically at \(n = -1, 0, 1, 2\). The uniform residue is what makes \(\pi \cot (\pi z)\) the standard device for summing series such as \(\sum 1/n^{2}\).

Solution. Partial fractions: \[\frac {1}{z(z-1)(z-3)} = \frac {1/3}{z} - \frac {1/2}{z-1} + \frac {1/6}{z-3}.\] On \(1 < \left |z\right | < 3\) the poles at \(0\) and \(1\) are inside and expand in \(1/z\); the pole at \(3\) is outside and expands in \(z\): \[-\frac {1/2}{z-1} = -\frac {1}{2}\sum ^{\infty }_{n=1} z^{-n}, \qquad \frac {1/6}{z-3} = -\frac {1}{18}\sum ^{\infty }_{n=0}\left (\frac z3\right )^{n} = -\sum ^{\infty }_{n=0}\frac {z^{n}}{2\cdot 3^{\,n+2}} .\] Combining the \(z^{-1}\) terms, \(\frac 13 - \frac 12 = -\frac 16\): \[\boxed {\frac {1}{z(z-1)(z-3)} = -\frac {1}{6z} \;-\;\frac {1}{2}\sum ^{\infty }_{n=2} z^{-n} \;-\;\sum ^{\infty }_{n=0}\frac {z^{n}}{2\cdot 3^{\,n+2}} .}\]

Numerically on \(\left |z\right | = 2\): \(c_{-1} = -0.1667\), \(c_{-2} = c_{-3} = -0.5\), \(c_0 = -0.0556\), \(c_1 = -0.0185\), matching \(-\frac 16\), \(-\frac 12\), \(-\frac 1{18}\) and \(-\frac 1{54}\).

Solution. Partial fractions first, since the poles are all simple and purely imaginary: \[\frac {1}{(x^{2}+4)(x^{2}+3)} = \frac {1}{x^{2}+3} - \frac {1}{x^{2}+4},\] which is checked by \(A(x^{2}+4) + B(x^{2}+3) = 1\) giving \(A = 1\), \(B = -1\).

Each piece is standard, and each may be got by the semicircle contour: for \(\alpha > 0\) the function \(\frac {1}{z^{2}+\alpha ^{2}}\) has one pole \(z = i\alpha \) in the upper half plane with residue \(\frac {1}{2i\alpha }\), so \[\int ^{\infty }_{-\infty }\frac {dx}{x^{2}+\alpha ^{2}} = 2\pi i\cdot \frac {1}{2i\alpha } = \frac {\pi }{\alpha }.\] With \(\alpha = \sqrt 3\) and \(\alpha = 2\), \[\boxed {\int ^{\infty }_{-\infty }\frac {dx}{(x^{2}+4)(x^{2}+3)} = \frac {\pi }{\sqrt 3} - \frac {\pi }{2} = \pi \left (\frac {1}{\sqrt 3}-\frac 12\right ) \approx 0.2430 .}\]

Done in one contour instead, the poles are \(2i\) and \(i\sqrt 3\) with residues \(\frac {1}{2i\cdot 2\cdot (4-3)}\cdot (-1)\)-style expressions that must be combined; splitting first is less error-prone and gives the same number.

Solution. The expansion is centred at \(\tfrac 12\), so measure everything from there. The singularities are \[z = 0:\ \left |z-\tfrac 12\right | = \tfrac 12,\qquad z = 1:\ \left |z-\tfrac 12\right | = \tfrac 12,\qquad z = \tfrac 23:\ \left |z-\tfrac 12\right | = \tfrac 16 .\] So on the annulus \(\tfrac 16 < \left |z-\tfrac 12\right | < \tfrac 12\) the function is analytic, with \(z = \tfrac 23\) inside the inner circle and \(z = 0, 1\) outside the outer one.

Is \(c_{-1}\) a residue? Yes — and this is the point of the question. For a Laurent expansion on an annulus, \[c_{-1} = \frac {1}{2\pi i}\oint _{\left |z-\frac 12\right | = r} f(z)\,dz = \sum \Big \{\text {residues of } f \text { inside the inner circle}\Big \},\] by the residue theorem applied to the circle of radius \(r\). Here exactly one singularity lies inside, namely \(z = \tfrac 23\), so \(c_{-1}\) equals that residue: \[c_{-1} = \operatorname *{Res}_{z=2/3}\frac {1}{z(z-1)(3z-2)} = \frac {1}{3z(z-1)}\bigg |_{z=2/3} = \frac {1}{3\cdot \frac 23\cdot \left (-\frac 13\right )} = -\frac {3}{2}.\]

Computing the neighbouring coefficients the same way, \[\boxed {c_{-1} = -\frac {3}{2},\qquad c_{0} = -1,\qquad c_{1} = -6 .}\]

Note the warning contained in the question: \(c_{-1}\) is a residue here only because a single pole sits inside the inner circle. On an annulus enclosing several poles, \(c_{-1}\) is their sum, and on one enclosing none it is zero while the function may still have residues elsewhere.

Solution. Statement. Let \(f\) and \(g\) be analytic inside and on a simple closed contour \(\gamma \), and suppose \[\left |g(z)\right | < \left |f(z)\right | \qquad \text {for every } z \in \gamma .\] Then \(f\) and \(f+g\) have the same number of zeros inside \(\gamma \), counted with multiplicity.

Proof. The strict inequality forces \(f \neq 0\) on \(\gamma \), and also \(f + g \neq 0\) there, since \(\left |f+g\right | \geq \left |f\right |-\left |g\right | > 0\). So the argument principle applies to both.

For \(t \in [0,1]\) put \(h_t = f + tg\). Again \(\left |h_t\right | \geq \left |f\right | - t\left |g\right | > 0\) on \(\gamma \), so \[N(t) = \frac {1}{2\pi i}\oint _{\gamma }\frac {h_t'(z)}{h_t(z)}\,dz\] is defined for every \(t\), and by the argument principle it counts the zeros of \(h_t\) inside \(\gamma \) — in particular it is an integer.

The integrand depends continuously on \(t\) and the contour is compact, so \(N\) is a continuous function of \(t\). A continuous integer-valued function on \([0,1]\) is constant, whence \(N(0) = N(1)\); that is, \(f\) and \(f+g\) have equally many zeros inside. \(\blacksquare \)

The picture usually offered — “\(g\) is too small to change how many times the image winds round \(0\)” — is exactly this continuity argument, and the “dog on a lead” image (the dog \(f+g\) cannot get round the tree a different number of times from its owner \(f\) if the lead is shorter than the distance to the tree) is the same statement again.

Solution. Apply Rouché on \(\left |z\right | = 2\), and the whole art is choosing which term dominates. On that circle, \[\left |3z^{3}\right | = 3(8) = 24,\qquad \left |z^{4}\right | = 16,\qquad \left |6\right | = 6 .\] The cubic term is the largest, so put \[f(z) = 3z^{3},\qquad g(z) = z^{4} + 6 .\] Then on \(\left |z\right | = 2\), \[\left |g(z)\right | \leq \left |z\right |^{4} + 6 = 16 + 6 = 22 < 24 = \left |f(z)\right | .\] Rouché applies, so \(z^{4}+3z^{3}+6 = f+g\) has as many zeros inside \(\left |z\right | = 2\) as \(3z^{3}\) does, namely \[\boxed {3}\] counting multiplicity (the triple zero of \(3z^{3}\) at the origin).

Two remarks. The obvious first guess \(f = z^{4}\) fails: \(\left |z^4\right | = 16\) is not larger than \(\left |3z^{3}+6\right |\), which can reach \(30\). The highest power is not automatically dominant on a circle of modest radius. And since the polynomial has degree \(4\), exactly one zero lies outside — direct computation puts the four moduli at \(1.166\), \(1.166\), \(1.640\) and \(2.693\), confirming three inside.

Solution. How many roots? On \(\left |z\right | = 1\), \[\left |e^{z}\right | = e^{\operatorname {Re}z} \leq e^{1} \approx 2.718, \qquad \left |3z^{2}\right | = 3 .\] Since \(e < 3\), take \(f(z) = -3z^{2}\) and \(g(z) = e^{z}\), so that \(\left |g\right | \leq e < 3 = \left |f\right |\) on the circle. By Rouché, \(e^{z}-3z^{2}\) has as many zeros in the unit disc as \(-3z^{2}\): exactly \[2 \qquad \text {(counted with multiplicity).}\]

Both are real. Consider the real function \(h(x) = e^{x} - 3x^{2}\) on \([-1,1]\): \[h(-1) = e^{-1} - 3 \approx -2.632 < 0,\qquad h(0) = 1 > 0,\qquad h(1) = e - 3 \approx -0.282 < 0 .\] By the intermediate value theorem \(h\) has a root in \((-1,0)\) and another in \((0,1)\). These are two distinct roots of \(e^{z}=3z^{2}\) lying in the unit disc, and we have just shown there are only two altogether. Hence all the roots in the disc are real. \(\blacksquare \)

Numerically they are \(x \approx -0.4590\) and \(x \approx 0.9100\), both comfortably inside. Note the shape of the argument: counting by Rouché and exhibiting by the intermediate value theorem, then matching the two counts. Neither half alone would do.

Solution. Statement. Let \(f\) be meromorphic inside and on a positively oriented simple closed contour \(\gamma \), with no zeros or poles on \(\gamma \). Then \[\frac {1}{2\pi i}\oint _{\gamma }\frac {f'(z)}{f(z)}\,dz = Z - P,\] where \(Z\) and \(P\) are the numbers of zeros and poles inside \(\gamma \), each counted with multiplicity.

Proof. The integrand \(f'/f\) is analytic inside \(\gamma \) except at the zeros and poles of \(f\), so by the residue theorem it suffices to compute the residue at each.

If \(f\) has a zero of order \(m\) at \(z_0\), write \(f(z) = (z-z_0)^{m}g(z)\) with \(g\) analytic and \(g(z_0) \neq 0\). Then \[\frac {f'(z)}{f(z)} = \frac {m(z-z_0)^{m-1}g + (z-z_0)^{m}g'}{(z-z_0)^{m}g} = \frac {m}{z-z_0} + \frac {g'(z)}{g(z)},\] the last term being analytic at \(z_0\). So the residue is \(+m\).

If \(f\) has a pole of order \(p\) at \(z_1\), the same computation with \(f(z) = (z-z_1)^{-p}g(z)\) gives residue \(-p\).

Summing over all zeros and poles and multiplying by \(2\pi i\) gives \(Z - P\). \(\blacksquare \)

Why “argument”. Formally \(\frac {f'}{f} = \frac {d}{dz}\log f\), so the integral measures the total change in \(\log f\) round \(\gamma \). The modulus returns to its starting value, so only \(\arg f\) contributes, and \[Z - P = \frac {1}{2\pi }\Delta _{\gamma }\arg f(z)\] is the number of times the image curve \(f\circ \gamma \) winds about the origin.

Solution. Apply Rouché twice, on each boundary circle, and subtract.

On \(\left |z\right | = 3\): \(\left |z^{5}\right | = 243\), while \[\left |6z^{3} - 10\right | \leq 6(27) + 10 = 172 < 243 .\] So with \(f = z^{5}\) the polynomial has \(5\) zeros in \(\left |z\right | < 3\).

On \(\left |z\right | = 2\): now \(\left |z^{5}\right | = 32\) but \(\left |6z^{3}\right | = 48\), and \[\left |z^{5} - 10\right | \leq 32 + 10 = 42 < 48 .\] So with \(f = 6z^{3}\) the polynomial has \(3\) zeros in \(\left |z\right | < 2\).

Both inequalities are strict, so there are no zeros on either circle and the counts are unambiguous. Subtracting, \[\boxed {5 - 3 = 2 \text { zeros in } 2 < \left |z\right | < 3 .}\]

The dominant term changes between the two circles — \(z^{5}\) at radius \(3\), \(6z^{3}\) at radius \(2\) — and that change is the whole reason zeros lie between them. Directly computing the roots gives moduli \(1.114,\ 1.210,\ 1.210,\ 2.478,\ 2.478\): three inside radius \(2\), two in the annulus, exactly as counted.

Solution. Write \(\Omega = \mathbb {C}\setminus \{\gamma \}\) and \(n(z) = \frac {1}{2\pi i}\displaystyle \int _{\gamma }\frac {dw}{w-z}\).

\(n\) is continuous on \(\Omega \). This is the estimate of the Cauchy-kernel problem: fixing \(z_0 \in \Omega \) and \(d = \operatorname {dist}(z_0,\{\gamma \}) > 0\), for \(\left |z-z_0\right | < d/2\) we get \[\left |n(z)-n(z_0)\right | = \frac {1}{2\pi }\left |\left (z-z_0\right )\int _{\gamma }\frac {dw}{(w-z)(w-z_0)}\right | \leq \frac {\left |z-z_0\right | L}{2\pi (d/2)(d)} \longrightarrow 0 .\]

\(n\) is integer-valued. Parametrise \(\gamma \) on \([0,1]\) and set \[\varphi (t) = \int ^{t}_{0}\frac {\gamma '(s)}{\gamma (s)-z}\,ds .\] Then \(\frac {d}{dt}\left [e^{-\varphi (t)}\big (\gamma (t)-z\big )\right ] = 0\), so \(e^{-\varphi (t)}(\gamma (t)-z)\) is constant. Since \(\gamma \) is closed, \(\gamma (1) = \gamma (0)\), and comparing \(t=0\) with \(t=1\) gives \(e^{\varphi (1)} = 1\). Hence \(\varphi (1) = 2\pi i k\) for some integer \(k\), and \(n(z) = k\).

Conclusion. A continuous integer-valued function on a connected set is constant. Each connected component of \(\Omega \) is connected, so \(n\) is constant on it. \(\blacksquare \)

The same “continuous and integer-valued, hence constant” step powers the proof of Rouché; it is the workhorse of the subject.

Solution. (a). Rouché’s Theorem. If \(f\) and \(g\) are analytic inside and on a simple closed contour \(\gamma \) and \(\left |g\right | < \left |f\right |\) everywhere on \(\gamma \), then \(f\) and \(f+g\) have the same number of zeros inside \(\gamma \), counting multiplicity.

(b). The Fundamental Theorem of Algebra. Let \[P(z) = a_n z^{n} + a_{n-1}z^{n-1} + \cdots + a_0, \qquad a_n \neq 0,\ n \geq 1 .\] Put \(f(z) = a_n z^{n}\) and \(g(z) = P(z) - f(z)\), the lower-order part.

On the circle \(\left |z\right | = R\), \[\frac {\left |g(z)\right |}{\left |f(z)\right |} \leq \frac {\left |a_{n-1}\right |R^{n-1} + \cdots + \left |a_0\right |} {\left |a_n\right | R^{n}} = \frac {1}{\left |a_n\right |}\left (\frac {\left |a_{n-1}\right |}{R} + \frac {\left |a_{n-2}\right |}{R^{2}} + \cdots + \frac {\left |a_{0}\right |}{R^{n}}\right ),\] and every term tends to \(0\) as \(R \rightarrow \infty \). Choose \(R\) large enough that the bracket is less than \(\left |a_n\right |\); then \(\left |g\right | < \left |f\right |\) on \(\left |z\right | = R\).

By Rouché, \(P = f + g\) has as many zeros in \(\left |z\right | < R\) as \(a_n z^{n}\), namely \(n\) counting multiplicity. In particular \(n \geq 1\) guarantees at least one root. \(\blacksquare \)

The estimate is the precise form of “a polynomial looks like its leading term far out”. Note that all \(n\) roots are captured: no root can lie outside \(\left |z\right | = R\), since the count inside is already \(n\) and the degree is \(n\).

Solution. (a). Continuity. Let \(a_0 \notin \{\gamma \}\) and \(d = \operatorname {dist}(a_0,\{\gamma \}) > 0\), which is positive because \(\{\gamma \}\) is compact. For \(\left |a - a_0\right | < d/2\) every \(z\) on \(\gamma \) satisfies \(\left |z-a\right | \geq d/2\), and \[n(\gamma ;a) - n(\gamma ;a_0) = \frac {a-a_0}{2\pi i}\int _{\gamma }\frac {dz}{(z-a)(z-a_0)},\] so with \(L\) the length of \(\gamma \), \[\left |n(\gamma ;a) - n(\gamma ;a_0)\right | \leq \frac {\left |a-a_0\right |}{2\pi }\cdot \frac {L}{(d/2)\,d} = \frac {L\left |a-a_0\right |}{\pi d^{2}} \xrightarrow [a \to a_0]{} 0 .\] Hence \(n(\gamma ;\cdot )\) is continuous, and being integer-valued it is locally constant, so constant on each connected component.

(b). Vanishing on the unbounded component. Let \(M = \max _{z\in \gamma }\left |z\right |\). For \(\left |a\right | > M\), \[\left |n(\gamma ;a)\right | \leq \frac {1}{2\pi }\cdot \frac {L}{\left |a\right | - M} \xrightarrow [\left |a\right |\to \infty ]{} 0 .\] So \(n(\gamma ;a) \rightarrow 0\); but \(n\) takes only integer values and is constant on the unbounded component, so that constant must be \[\boxed {n(\gamma ;a) = 0 \qquad \text {for } a \text { in the unbounded component.}}\] \(\blacksquare \)

The two parts together say the winding number is a locally constant integer that dies at infinity — which is what licenses “count only the singularities you have enclosed”.

Solution. On the unit circle, estimate each piece: \[\left |5z^{4}\right | = 5,\qquad \left |e^{z} + 2\right | \leq \left |e^{z}\right | + 2 \leq e^{\operatorname {Re}z} + 2 \leq e + 2 \approx 4.718 .\] Since \(4.718 < 5\), take \[f(z) = -5z^{4},\qquad g(z) = e^{z}+2,\] so that \(\left |g\right | < \left |f\right |\) on \(\left |z\right | = 1\) and Rouché applies: \(e^{z}-5z^{4}+2 = f+g\) has as many zeros in the unit disc as \(-5z^{4}\), namely \[\boxed {4}\] counted with multiplicity.

The margin is thin — \(4.718\) against \(5\) — so the bound \(\left |e^{z}\right | \leq e\) on \(\left |z\right |=1\) has to be used, not the crude \(\left |e^{z}\right | \leq e^{\left |z\right |}\), which gives the same number here but would fail on a slightly larger circle. Independent computation of the winding number of the image confirms \(4\).

Solution. Two applications of Rouché, one on each boundary circle.

On \(\left |z\right | = 2\): \(\left |2z^{5}\right | = 2(32) = 64\), while \[\left |-6z^{2} + z + 1\right | \leq 6(4) + 2 + 1 = 27 < 64 .\] So \(2z^{5}-6z^{2}+z+1\) has \(5\) zeros in \(\left |z\right | < 2\) — all of them, as it must, the polynomial having degree \(5\).

On \(\left |z\right | = 1\): now \(\left |2z^{5}\right | = 2\) but \(\left |6z^{2}\right | = 6\), and \[\left |2z^{5} + z + 1\right | \leq 2 + 1 + 1 = 4 < 6 .\] Taking \(f = -6z^{2}\), the polynomial has \(2\) zeros in \(\left |z\right | < 1\).

Both inequalities being strict, there are no zeros on either circle, so \[\boxed {5 - 2 = 3 \text { zeros in } 1 \leq \left |z\right | \leq 2 .}\]

Computing the roots directly gives moduli \(0.332,\ 0.514,\ 1.328,\ 1.486,\ 1.486\) — two inside the unit circle and three in the annulus, as counted. As in the earlier annulus problem, the dominant term switches between the two radii, and that switch is precisely why zeros lie in between.

Solution. (a). Definition. For a closed rectifiable curve \(\gamma \) and a point \(a\) not on \(\gamma \), the winding number (or index) of \(\gamma \) about \(a\) is \[n(\gamma ; a) = \frac {1}{2\pi i}\int _{\gamma }\frac {dz}{z-a} .\] It is an integer, and it counts the net number of times \(\gamma \) travels anticlockwise around \(a\).

(b). Let \(a\) and \(b\) lie in the same path-connected component \(U\) of \(\mathbb {C}\setminus \{\gamma \}\), and let \(\sigma : [0,1] \rightarrow U\) be a path from \(a\) to \(b\).

The function \(t \mapsto n\big (\gamma ; \sigma (t)\big )\) is well defined, since \(\sigma (t)\) never meets \(\gamma \). It is continuous, by the estimate of part (a) of the previous problem applied at each point of the compact set \(\sigma ([0,1])\). And it takes only integer values.

A continuous integer-valued function on the connected interval \([0,1]\) is constant. Evaluating at the endpoints, \[n(\gamma ; a) = n\big (\gamma ;\sigma (0)\big ) = n\big (\gamma ;\sigma (1)\big ) = n(\gamma ; b). \qquad \blacksquare \]

For an open subset of \(\mathbb {C}\), connected and path-connected coincide, so this is the same statement as “\(n\) is constant on each component”. What makes the argument work is not analysis but topology: continuity plus discreteness of the value set.