Complex Bootcamp Exercises: Integration II
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This set of problems focuses on the residue theorem, argument principle, and existence of branches. Some more problems on topics from the last post are also included.
Resources
Ahlfors section 4.4-4.6
Problems
Problem 1 (Prelim August 2015)
Let $f$ be a non-constant meromorphic function satisfying \(f(z) = f(z+\sqrt{2}) = f(z+i\sqrt{2})\) and assume $f$ has at most a single pole in $\mathbb{D} = \{z: |z| < 1 \}$.
-
Show that there must exist a pole in $\mathbb{D}$.
Hint
If there is no pole in $\mathbb{D}$ how many poles can $f$ have elsewhere? -
Show that this pole cannot be simple.
Hint
Try integrating along a square with side lengths $\sqrt{2}$.
Note that due to the period, the values of $f$ along $\gamma_1$ and $\gamma_2$ are identical, however we are now traversing in the opposite direction so $\gamma_1' = -\gamma_2'$. This gives us the relationship $$ \begin{align*} \int_{\gamma_2} f\,dz &= \int_{0}^{1/4} \gamma'_2(t+1/2) f(\gamma_2(t+1/2))\,dt \\&= -\int_{1/4}^0 \gamma_2'(3/4-u) f(\gamma_2(3/4 - u))\,du \\&= -\int_0^{1/4} \gamma_1'(u)f(\gamma_1(u))\,du = - \int_{\gamma_1} f\,dz\end{align*}$$ The same applies to $\gamma_3$ and $\gamma_4$. Hence, if our pole lies at $z_0 \in \mathbb{D}$, we find $$ 2\pi i\text{res}(f,z_0) = \int_\gamma f\,dz = 0 $$ allowing us to conclude that the pole at $z_0$ is not simple.
Problem 2 (Prelim January 1997)
Suppose that $f$ is a holomorphic function on a neighborhood of $\overline{\mathbb{D}}$ and on $\partial \mathbb{D}$ $f$ satisfies $|f(z)| \leq a$ when $\text{Im}(z) \geq 0$ and $|f(z)| \leq b$ when $\text{Im}(z) \leq 0$. Show that [ |f(0)| \leq \left(ab\right)^{1/2}. ]
Hint
Consider both $f(z)$ and $f(-z)$.Problem 3 (Prelim August 1996)
Suppose $f$ is analytic on an open neighborhood of $\overline{\mathbb{D}}$ and that it is injective on $|z| = 1$. Show that $f$ is injective on $\mathbb{D}$.
Hint
Apply the Jordan curve theorem. What can you deduce about solutions to $f(z) = w$ for arbitrary $w\in \mathbb{C}$?Problem 4 (Prelim January 2022)
Let $K$ be a compact connected subset of $\mathbb{C}$ containing $\pm i$. Show that there exists a holomorphic branch of $(z^2 + 1)^{-1/2}$ in $\mathbb{C} \setminus K$ and determine the possible values of its integral along closed curves in $\mathbb{C} \setminus K$.
Hint
Recall that a branch of $\log f$ exists for a function $f$ on a domain $U$ if and only if $\int_\gamma f'/f\,dz = 0$ for all closed curves $\gamma:S^1\to U$. If this this integral is nonzero, consider defining a multivalued primitive and composing with $e^z$, if the values differ by multiples of $2\pi i$ the composition is a single valued function.Finally, we must show that $\text{exp}(h(z))$ is indeed a branch of $f$. Differentiating $\text{exp}(2h(z))/(f(z))^2$ we find $$ \begin{align*} \frac{d}{dz} \text{exp}(2h(z))(z^2 + 1) &= h'(z)\text{exp}(2h(z))f(z) + 2z \text{exp}(2h(z)) \\ &= 2\text{exp}(2h(z)) \left[ \frac{-z}{z^2 + 1} + \frac{z}{z^2 + 1}\right] = 0 \end{align*} $$ showing that the functions are multiples of one another. Finally, evaluating at $z_0$ we see $\text{exp}(h(z_0)) = \text{exp}(l) = f(z_0)$ hence they are identical.
Original solution to second part
To compute the integral of this branch over a closed curve first note that any such curve $\gamma$ can be replaced by a curve on a circle of radius $> R$ with the same integral (due to Cauchy's theorem.) We can let $R$ be greater the largest modulus attained on $K$. Then note our branch restricted to $\mathbb{C} \setminus \overline{B_r(0)}$ can be continued analytically to $\mathbb{C}\setminus [-i,i]$. Finally, we can construct a dumbbell contour along the segment $[-i,i]$ like so
![Dumbbell contour](/users/jcf3275/assets/complex-ex-4/dumbbell_contour.png)
On this contour we first note that as we shrink the dumbbell the contribution from the circular regions goes to zero. Indeed, at $i$ we can let $\gamma(\theta) = re^{i\theta} + i$, then $|(\gamma(\theta)^2 + 1)^{-1/2}| \leq [r(2-r)]^{-1/2}$ after factoring. Then by the ML-lemma we find that $$ \begin{align*} \left| \int_\gamma \sqrt{z^2 + 1} \, dz \right| &\leq L(\gamma)[r(2-r)]^{-1/2} \\ &= 2\pi \frac{\sqrt{r}}{\sqrt{2-r}} \to 0 \end{align*} $$ as $r\to 0$. This shows that integrating along a simple closed circle of radius $> 1$ is equivalent to integrating up and down the segment $[-i,i]$, however when we swap directions we hop over our branch cut so the values will be negated. This negation is justified by the integral of $f'/f$ along any small circle around $\pm i$, which we know gives $\pm \pi i$ so upon exponetiating we will find the across the branch cut our brach differs by a factor of $e^{i\pi}$. Integrating in one direction we find $$ \int_{[-i,i]} \frac{1}{\sqrt{1+z^2}}\,dz = \int_{-1}^1 \frac{i}{\sqrt{1-t^2}}\,dt = \pm \pi i $$ hence the integral over any closed curve will be an integer multiple of $2\pi i$.
Toby's solution to second part
Observe that $$ \left( (1+z^2)^{-1/2} - \frac{1}{z} \right)\left( (1+z^2)^{-1/2} + \frac{1}{z} \right) = \frac{1}{1+z^2} - \frac{1}{z^2} = -\frac{1}{z^2 + z^4} $$ which, for $|z| > 1$, has absolute value below $1/|z|^4$. From this we conclude that one of the products is at most $1/|z|^2$$^\dagger$, this allows us to bound that term by $$ \left| \int_\gamma (1+z^2)^{-1/2} \,dz - \int_\gamma \frac{1}{z} \,dz \right| \leq \int_\gamma \left| \frac{1}{z^2} \right|\,dz \leq \text{Length}(\gamma) \sup |\gamma(t)|^{-2}.$$ Finally, we note that we can homotope $\gamma$ into a sequence of circular loops with arbitrarily large radius $R$, so the right hand side becomes $\leq 2\pi R\cdot R^{-2} \to 0$ as $R\to \infty$, establishing $$ \int_\gamma (1+z^2)^{-1/2} \,dz = \int_\gamma \frac{1}{z} \,dz = 2\pi n$$ for some $n\in \mathbb{Z}$.
$\dagger$ As of writing I'm unsure how to make this step rigorous.
Problem 5 (Prelim August 2020)
Assume $f:\mathbb{C}\setminus \overline{\mathbb{D}} \to \mathbb{D}$ is holomorphic. Prove that $|f’(2)| \leq 1/3$.
Hint 1
Precompose with $1/z$ and note the singularity at zero is removable to make this a map $\mathbb{D}\to \mathbb{D}$.Hint 2
Use a Mobius transformation on the disk that maps $1/2 \mapsto 0$. Then the Schwarz inequality gives a bound on the derivative at zero. This related to the "Schwarz-Pick" theorem and its very handy have memorized.Problem 6 (Prelim August 2020)
Suppose that $f$ is analytic on $\mathbb{D}$ and satisfies $|f(z)| \leq M$ for all $z\in \mathbb{D}$. Assume further that $f$ vanishes at the points $ \{z_j\}_{j=1}^N $ where $1 \leq N \leq \infty$.) Prove that
$$ |f(z)| \leq M \left| \prod_{j=1}^m \frac{z-z_j}{1-\overline{z_j}z} \right|\quad \forall z\in\mathbb{D} $$
for any $1 \leq m \leq N$ (or if $N = \infty$ then $1 \leq m < N$. )
Hint
Consider the function $$ g(z) = \frac{f(z)}{\prod_{j=1}^m \frac{z-z_j}{1-\overline{z_j}z} }. $$ Are its singularities removable? What is $|g|$ as you approach $\partial \mathbb{D}$?_Note: this is actually part (a) of a two part problem. The second part will be in a later list._
Problem 7 (Prelim January 2019)
Given $z\in \mathbb{C}$ and a closed curve in $\mathbb{C}\setminus \{z\}$ denote $n(\gamma,z)$ the index or winding number of $\gamma$ about $z$. If $\gamma$ can be written as [ \gamma(t) = \sum_{k = -N}^{N} c_ke^{ikt} ] with $c_{-N}$ and $c_N$ not both zero show that $-N \leq n(\gamma, z) \leq N$.
Hint
Can you relate this integral to an application of the argument principle on a suitable function?Problem 8 (Prelim January 2017)
Prove that the range of the entire function $z\mapsto z^2 + \cos(z)$ is all of $\mathbb{C}$
Hint
Picard's theorem tells us that this functions range can exclude at most one value. Use this to deduce that if this value exists it is in $\mathbb{R}$ then apply Rouche's theorem to show it does not exist.By some simple calculus we can find that the minimum of $z^2+\cos(z)$ on $\mathbb{R}$ is $1$, hence $w < 1$. For the remaining cases we can use symmetric Rouche's theorem on the functions $f(z) = z^2 + \cos(z) - w$ and $g(z) = \cos(z) - w$ to show both have the same number of roots in $\mathbb{C}$, and since $\cos$ is surjective this will complete the proof.
To do this, define the path $\Lambda = \partial \{x + i y: |x| < 2n\pi, |y| < n\}$. On horizontal segments we note that $$|cos(z) - w| \geq \left| \frac{e^{\pm n} + e^{\mp n}}{2} - w \right| \geq \frac{1}{2} e^n - |w| $$ while $|f(z) - g(z)| = |z^2| = 4\pi^2n^2 + n^2$, hence if we take $n$ suitably large we find $$|f(z) - g(z)| < |f(z)| + |g(z)|$$ on the horizontal sides of $\Lambda$. On the vertical sides we note that $\cos(\pm 2\pi n + iy) = (e^y+e^{-y})/2 \geq 1 > w$, so $g(z) > 0$. The triangle inequality yields $|z^2| \leq |z^2 + \cos(z) - w| + |\cos(z) -w|$, and we know equality occurs if and only if both $f$ and $-g$ point in the same direction at $z$. Since $\cos(z) - w > 0 $ we see that for equality to occur must require $z^2 \in \mathbb{R}$ which can only occur if $z\in \mathbb{R}$ or $i\mathbb{R}$. The only points on the vertical edges satisfying this condition are $\pm 2n \pi$, but at these points we find $$ |4n^2 \pi^2| <| 4n^2 \pi^2 + 1 - w| + |1-w| $$ hence showing $|f-g| < |f|+|g|$ on $\Lambda$. Finally, since $z\mapsto \cos z$ is surjective it follows $z\mapsto z^2 + \cos z$ is a surjective as well.
The main part of the argument using Rouche's theorem is modified from this answer.
Problem 9 (Prelim January 2011)
Let $f$ be analytic in an anulus $ U = \{z: r < |z| < R\}$ where $0 < r < R$. Assume $f$ has no zeros in $U$. Show there exists an integer $n$ and a holomorphic function $g:U\to \mathbb{C}$ such that $f(z) = z^n e^{g(z)}$
Hint
What conditions must $f$ satisfy for this to be true with $n = 0$? If $\gamma$ is a simple closed curve in $U$ then what is the index $n(f\circ \gamma,0)$? Does it depend on $\gamma$?Problem 10 (Prelim January 2016)
Show that the equation $\sin(f(z)) = z$ has a solution $f$ that is analytic on the region $U = \{z\in \mathbb{C}:|z| < 1\text{ or }\text{Im}(z) \ne 0\}$.
Hint
Write $\sin(w)$ in terms of $e^{iw}$ and solve for it. You will need to prove the existence of branches of two functions on $U$ to produce $f$.Fix $z \in U$, we wish to solve $$ \frac{e^{iw} - e^{-iw}}{2i} = \sin(w) = z.$$ We see that $e^{iw}$ must solve the quadratic $$ \left(e^{iw}\right)^2 -2iz\left(e^{iw} \right) - 1 = 0 $$ which gives $$ e^{iw} = \frac{2iz +\sqrt{4-4z^2}}{2} = iz + \sqrt{1-z^2}$$ for some branch of $z\mapsto \sqrt{1-z^2}$ defined on $U$. To observe that such a branch exists note that its logarithmic derivative is $$ - \frac{z}{1-z^2} $$ which is meromorphic with two poles at $z = \pm 1$. By the above note about $n(\gamma,\pm 1) = 0$ for all $\gamma$ in $U$ the residue theorem gives us $$ \int_\gamma - \frac{z}{1-z^2}\,dz = 0 $$ hence a primitive $P$ exists, and we can take our branch to be $\sqrt{1-z^2} = e^{P(z)}$, given the primitive has correct initial value.
Now to fully solve the equation we must find a logarithm of $iz + \sqrt{1-z^2} = iz + e^{P(z)}$. Proceeding as usual, we compute its logarithmic derivative as $$ \frac{i + P'(z)e^{P(z)}}{iz + e^{P(z)}} $$ which has poles at $z$ for which $e^{P(z)} = iz$. This is equivalent to $1-z^2 = -z^2$ for any choice of our branch $e^{P(z)}$, hence the function is holomorphic in $U$ and since $U$ is simply connected this allows us to construct a primitive. This primitive is then our branch of $\log(iz + \sqrt{1-z^2})$. Multiplying it by $-i$ then gives the function $f$ satisfying $f(z) = w$ in $U$.
Problem 11 (Prelim August 2016)
Evaluate the integrals [ I = \int_C \sqrt{1-z^2}\,dz \quad J = \int_{0}^\infty \frac{x\sin \pi x}{1-x^2}\,dx ] for some branch of the square root function (indicate which) and $C$ is the positively oriented circle $|z| = 2$.
Hint
Note that for $I$ there are no residues, but instead branch points.For $J$ consider integrating $$f(z) = \frac{ze^{i\pi z}}{1-z^2}$$ over a large semicircle with side on $\mathbb{R}$, closely avoiding the poles at $z = \pm 1$.
Another strategy is to use the substitution $u = z^{-1}$, which transforms integral to be over $|z| = 1/2$ with a single pole inside, allowing us to use the residue theorem. Note that finding the residues can be difficult due to choice of branches.
For $J$ consider the contour that is a large semicircle of radius $R > 0$ with smaller semicircles of radius $r$ cut out at the points $z = \pm 1$.
![Circular contour avoiding poles on real axis](/users/jcf3275/assets/complex-ex-4/circle_contour.png)
Over this contour we will integrate $$ f(z) = \frac{z e^{\pi i z}}{1-z^2}.$$ First, note that on the largest semicircle we have $$ \begin{align*} \left| \int_{\gamma_R} f(x)\,dx \right| &\leq 2\int_0^{\pi/2} \frac{R^2}{R^2-1} e^{-\pi R\sin \theta}\,d\theta \\&\leq 2\int_0^{\pi/2} \frac{R^2}{R^2-1} e^{-2 R\theta}\,d\theta \\&= \frac{R^2}{R^2 - 1} \left[ \frac{1}{2R} - \frac{1}{2R}e^{-\pi R} \right] \end{align*}$$ which converges to zero as $R \to \infty$. Now, taking the limit as $r\to 0$ of the small circles around the poles at $z = \pm 1$ we see that each circular arc's integral approaches $-\pi i\text{res}_{z = \pm 1}(f)$ (this is because we can write $f(z) = R/(z-p) + g(z)$ where $R$ is the residue at $z = p$ and $g(z)$ is holomorphic near $z = p$. Integrating on the half arc will arrive at $-\pi i R.$ See Ahlfors 4.5.3.) Quick calculations show that $$ \text{res}_{z=1}(f) = \text{res}_{z= -1}(f) = \frac{1}{2} $$ hence by applying Cauchy's theorem and letting $R\to \infty, r\to 0$ we find $$\int_\mathbb{R} f(z)\,dz = \pi i $$ and finally, equating imaginary parts and dividing by 2 (due to evenness of the integrand) we arrive at $$ J = \frac{\pi}{2} $$
Problem 12 (Prelim January 2014)
Suppose $f$ is an entire function with the property that $f(z)$ is real if and only if $z$ is real. Show that $f’(z) \ne 0$ for all real $z$.
Hint
If $f'(w) = 0$ consider applying the argument principle to $f(w) - w$.Problem 13 (Riccardo’s Quandary)
Consider the following statement: Any function $f$ continuous on the annulus $A = {z\in \mathbb{C}: 1/2 \leq |z| \leq 2}$ such that $f_{\partial A}$ is injective and $f_{A^o}$ is holomorphic is also injective on the interior $A^o$.
- Find a counter example to this statement.
Hint
Try to apply the proof of Problem 3 to this problem, now with a contour of two concentric circles connected by a path. Where does the proof fail in this case? - Find an additional hypothesis such that this statement is true. Prove it.
Hint
Attempt the simpler case of $f$ having a holomorphic extension in some neighborhood of $A$. To extend this to the case of $A$ just continuous as the boundary prove the *dog on a leash* lemma (if $\gamma,\gamma'$ are two paths in $\C$ such that for all $t$ we have $|\gamma(t) - \gamma'(t)| < |\gamma(t)| then $n(\gamma,0) = n(\gamma',0)$.)
In the case of $f$ extending holomorphically to a neighborhood of $A$ we fix any point $z \in f(A)^o$. Without loss of generality we suppose that $z \not\in [1/2,2].$ We select the contour $\Gamma = \gamma_1 \ast \gamma_2 \ast \gamma_3 \ast \gamma^{-1}_2$ where $\gamma_1$ parametrizes the circle of radius 2 counter clockwise, $\gamma_2$ is a segment connecting $2$ to $1/2$, and $\gamma_3$ is parametrizes the circle of radius 1/2 clockwise.
![Annulus contour](/users/jcf3275/assets/complex-ex-4/annulus_contour.png)
By argument principle we see that the number of solutions $f(\xi) = z$ is exactly $$ \begin{align*} \frac{1}{2\pi i} \int_\Gamma \frac{f'(\xi)}{f(\xi) - f(z)}\,d\xi &= \frac{1}{2\pi i} \left(\int_{f\circ \gamma_1} \frac{1}{\xi - f(z)}\,d\xi + \int_{f\circ \gamma_3} \frac{1}{\xi - f(z)}\,d\xi\right) \\ &= n(f\circ \gamma_1,f(z)) + n(f\circ \gamma_3,f(z)) = \pm 1\text{ or }0. \end{align*}$$ We know that both of these winding numbers are 1, 0, or -1 due to the injectivity of $f$ along $\partial A$, but is it possible for both to be equal to $\pm 1$? The parameterizations of the circles $\gamma_1,\gamma_3$ are oriented in opposite directions so we shall seek an $f$ that reverses the orientation of one and keep the second the same. Such a function is $$ f(z) = z + \frac{2}{z}, $$ because for small $z$ the function resembles $z^{-1}$ which reverses orientation of $\gamma_3$ while for large $z$ it resembles the identity preserving the orientation of $\gamma_1$. From here it is easy to verify $f$ is injective on $\partial A$ and that $f(1) = f(2) = 3$ giving our counterexample for part (a).
This exploration suggests that requiring $f$ to either preserve or reverse orientation for the two circles comprising $\partial A$. For our purposes, we will define this as $n(\gamma_1, z) \ne n(\gamma_3,z)$ for all $z \in \mathbb{C}$. Returning to the attempt at a proof our new assumption guarantees $n(f\circ \gamma_1,f(z)), n(f\circ \gamma_3,f(z))$ cannot both be $1$ or $-1$, hence $n(f\circ \Gamma,z) = \pm 1\text{ or }0$ giving injectivity by the argument principle.
At this point what remains is to extend this result to $f$ analytic in $A^o$ and continuous on $A$. To do this, define $$A_\delta = \{z\in \mathbb{C}: 1/2 + \delta \leq |z| \leq 2 - \delta\}$$ with corresponding contour $\Gamma^\delta$. Due to $A$ being compact $f$ is uniformly continuous on $A$. Given $z\in f(A)^o$ let $\epsilon = d(z, \partial f(A)).$ By uniform continuity there exists $\hat\delta > 0$ such that if $|x-y| < \hat\delta \implies |f(x)-f(y)| < \epsilon$ for all $x,y \in A$. Therefore for any $0 < \delta < \hat\delta$ we have $$ |f\circ \Gamma(t) - f\circ \Gamma^\delta(t)| < \epsilon \leq |f\circ \Gamma(t) - z|.$$ This shows that $\Gamma,\Gamma^\delta$ satisfy the hypothesis of the dog on a leash lemma, hence $$ n(f\circ \Gamma^\delta, z) = n(f\circ \Gamma, z) = \pm 1\text{ or } 0$$ for all sufficiently small $\delta$, and now we are able to apply the argument principle on $\Gamma^\delta$ proving injectivity.