# Exercises

## A Fourier Transform

We wish to compute for any real number $$\omega$$ the value of the integral $\hat{x}(\omega) = \int_{-\infty}^{+\infty} x(t) e^{-i \omega t}\, dt$ when $$x: \mathbb{R} \to \mathbb{R}$$ is the Gaussian function defined by $\forall \, t \in \mathbb{R}, \; x(t) = e^{-t^2/2}.$

We remind you of the value of the Gaussian integral (see e.g. Wikipedia): $\hat{x}(0) = \int_{-\infty}^{+\infty} e^{-t^2/2} \, dt = \sqrt{2\pi}$

### Questions

1. Show that for any pair of real numbers $$\tau$$ and $$\omega$$, we can compute $\int_{-\tau}^{\tau} x(t) e^{-i \omega t}\, dt$ from the line integral of a fixed holomorphic function on a path $$\gamma$$ that depends on $$\tau$$ and $$\omega$$.

2. Use Cauchy’s integral theorem to evaluate $$\hat{x}(\omega)$$.

1. We may denote $$x$$ the extension to the complex plane of the original Gaussian function $$x$$, defined by: $\forall \, z\in \mathbb{C}, \; x(z) = e^{-z^2/2}.$ It is holomorphic as a composition of holomorphic functions. Let $$\gamma = [-\tau \to \tau] + i\omega$$. The line integral of $$x$$ along $$\gamma$$ satisfies $\int_{\gamma} x(z) \, dz = \int_{0}^1 x(-\tau (1-s) + \tau s + i \omega) \, (2\tau ds)$ or, using the change of variable $$t = -\tau (1-s) + \tau s$$, $\int_{\gamma} x(z) \, dz = \int_{-\tau}^{\tau} x(t + i\omega) \, dt.$ Since $x(t + i \omega) = e^{-{(t + i\omega)^2}/{2}} = e^{-t^2/2} e^{-i\omega t} e^{\omega^2/2},$ we end up with $\int_{\gamma} x(z) \, dz = e^{\omega^2/2} \int_{-\tau}^{\tau} x(t) e^{-i2\pi f t}\, dt$

2. Let $$\nu = \tau + [0 \to i \omega]$$; on the image of this path, we have $\forall \, s \in [0,1], \; |x(\nu(s))| = \left| e^{-(\tau + i \omega s)^2/2} \right| = e^{-\tau^2/2} e^{(\omega s)^2/2} \leq e^{-\tau^2/2} e^{\omega^2/2},$ hence the M-L inequality provides $\left| \int_{\mu} x(z) \, dz \right| \leq (|\omega| e^{\omega^2/2} )e^{-\tau^2/2}$ and thus, $\forall \, \omega \in \mathbb{R}, \; \lim_{|\tau| \to +\infty} \int_{\mu} x(z) \, dz = 0.$

We may apply Cauchy’s integral theorem to the function $$x$$ on the closed polyline $[-\tau + i \omega \, \to \, \tau + i\omega \, \to \, \tau \, \to \, -\tau \, \to \, -\tau + i \omega].$ It is the concatenation of $$\gamma = [-\tau \to \tau] + i\omega$$, the reverse of $$\mu_+ = \tau + [0 \to i \omega]$$, the reverse of $$\gamma_0 = [-\tau \to \tau]$$ and finally $$\mu_- = -\tau + [0 \to i \omega]$$. The closed path used in the application of Cauchy’s integral theorem

The theorem provides $\begin{split} e^{\omega^2/2} \int_{-\tau}^{\tau} x(t) e^{-i \omega t}\, dt - \int_{\mu_+} x(z) \, dz \\ - \int_{-\tau}^{\tau} x(t) \, dt + \int_{\mu_-} x(z) \, dz = 0. \end{split}$ When $$\tau \to +\infty$$, this equality yields $\int_{-\infty}^{\infty} x(t) e^{-i2\pi f t}\, dt = \sqrt{2\pi} e^{-\omega^2/2}.$

## Cauchy’s Integral Formula for Disks

Let $$\Omega$$ be an open subset of the complex plane and $$\gamma = c + r[\circlearrowleft]$$. We assume that the closed disk $$\overline{D}(c, r)$$ is included in $$\Omega$$ (this is stronger than the requirement that $$\gamma$$ is a path of $$\Omega$$).

We wish to prove that for any holomorphic function $$f:\Omega \to \mathbb{C}$$, $\forall \, z \in \Omega, \; |z - c| < r \, \Rightarrow \, f(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw.$

### Questions

1. What is the value of the line integral above when $$|z - c| > r$$?

2. Compute the line integral above when $$z=c$$ as an integral with respect to a real variable. When happens in this case when $$r \to 0$$ ?

3. Let $$\epsilon>0$$ be such that $$|z| + \epsilon < r$$ and let $$\lambda = z + \epsilon[\circlearrowleft]$$. Provide two paths $$\mu$$ and $$\nu$$ whose images belong to (different) star-shaped subsets of $$\Omega \setminus \{z\}$$ and such that for any continuous function $$g:\Omega \setminus \{z\}\to \mathbb{C},$$ $\int_{\gamma} g(w) \, dw = \int_{\lambda} g(w) dw + \int_{\mu} g(w) \, dw + \int_{\nu} g(w) dw.$

4. Prove Cauchy’s integral formula for disks.

5. Show that $$f'(z)$$ can be computed as a line integral on $$\gamma$$ of an expression that depends on $$f(w)$$ and not on $$f'(w)$$. What property of $$f'$$ does this expression shows?

1. If $$|z - c| > r$$, the function $$w \mapsto {f(w)} / (w-z)$$ is defined and holomorphic in $$\Omega \setminus \{z\}$$. Let $$\rho$$ be the minimum between $$|z-c|$$ and the distance between $$c$$ and $$\mathbb{C} \setminus \Omega$$. By construction, the open disk $$D(c, \rho)$$ is a star-shaped subset of $$\Omega \setminus \{z\}$$ and it contains the image of $$\gamma$$. Thus, Cauchy’s integral theorem provides $\frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw = 0.$

2. When $$z=c$$, we have $\begin{split} \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw =&\, \frac{1}{i2\pi} \int_0^1 \frac{f(c+re^{i2\pi t})}{c+re^{i2\pi t} - c} (i2\pi re^{i2\pi t}dt) \\ =&\, \int_0^1 f(c+re^{i2\pi t}) dt. \end{split}$ By continuity of $$f$$ at $$c$$, the limit of this integral when $$r \to 0$$ is $$f(c)$$.

3. Assume for the sake of simplicity that $$z = c + x$$ for some real number $$x \in \left[0,r \right[$$. Let $$\alpha = \arccos x/r$$; define $$\mu$$ as the concatenation $\begin{array}{lll} \mu \; = & c+ [x + i\epsilon \to re^{i\alpha}] &\, | \\ & c + r e^{i[\alpha \to 2\pi-\alpha]} &\, | \\ & c + [re^{-i\alpha} \to x - i\epsilon] &\, | \\ & c + x + \epsilon e^{i [-\pi/2 \to -3\pi/2]} & \, . \end{array}$ and $$\nu$$ as the concatenation $\begin{array}{lll} \nu \; = & c+ [x - i\epsilon \to re^{-i\alpha}] &\, | \\ & c + r e^{i[-\alpha \to \alpha]} &\, | \\ & c + [re^{i \alpha} \to x + i\epsilon] &\, | \\ & c + x + \epsilon e^{i [\pi/2 \to -\pi/2]} & \, . \end{array}$

Since the closure of $$D(c,r)$$ is included in $$\Omega$$, there is a $$\rho > r$$ such that $$D(c,\rho) \subset \Omega$$. The image of $$\mu$$ belongs to the set $D(c, \rho) \setminus \{z + t \;|\; t \geq 0\}$ while the image of $$\nu$$ belongs to the set $D(c, \rho) \setminus \{z + t \;|\; t \leq 0\}.$ Both sets are star-shaped and included in $$\Omega$$.

Additionally, for any continuous function $$g:\Omega \setminus \{z\}\to \mathbb{C}$$

• the integral of $$g$$ on $$c+ [x + i\epsilon \to re^{i\alpha}]$$ and its reverse path on one hand, the integral of $$g$$ on $$c+[re^{-i\alpha} \to x - i\epsilon]$$ and its reverse path on the other hand are opposite numbers.

• the sum of the integral of $$g$$ on $$c + r e^{i[\alpha \to 2\pi-\alpha]}$$ and $$c + r e^{i [-\alpha \to \alpha]}$$ is equal to its integral on $$\gamma = c + r [\circlearrowleft]$$.

• the sum of the integral of $$g$$ on $$c + x + \epsilon e^{i [-\pi/2 \to -3\pi/2]}$$ and $$c + x + \epsilon e^{i [\pi/2 \to -\pi/2]}$$ is equal to the opposite of its integral on $$\lambda = c + x + \epsilon[\circlearrowleft]$$.

Therefore, the equality $\int_{\gamma} g(w) \, dw = \int_{\lambda} g(w) dw + \int_{\mu} g(w) \, dw + \int_{\nu} g(w) dw.$ holds.

4. We may apply the result of the previous question to the function $$w \mapsto f(w)/(w-z)$$. As it is holomorphic on $$\Omega \setminus \{z\}$$, the star-shaped version of Cauchy’s integral theorem provides $\int_{\mu} \frac{f(w)}{w-z} dw = \int_{\nu} \frac{f(w)}{w-z} dw = 0,$ hence $\frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw = \frac{1}{i2\pi} \int_{\lambda} \frac{f(w)}{w-z} dw.$ We proved in question 2. that the right-hand side of this equation tends to $$f(z)$$ when $$\epsilon \to 0$$, thus $\frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} dw = f(z),$ which is Cauchy’s integral formula for disks.

5. Let $$z \in \Omega$$. There are some $$c\in \Omega$$ and $$r> 0$$ such that $$z \in D(c, r)$$ and $$\overline{D}(z,r) \subset \Omega$$; let $$\gamma = c + r[\circlearrowleft]$$. For any complex number $$h$$ such that $$|z+h - c| < r$$, we have by Cauchy’s formula $\begin{split} \frac{f(z+h) - f(z)}{h} = &\, \frac{1}{i2\pi} \int_{\gamma} \frac{1}{h} \left(\frac{1}{w - z - h} -\frac{1}{w - z} \right) f(w) \, dw \\ = &\, \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{(w-z-h)(w-z)} \, dw \end{split}$ To met the condition on $$h$$, assume that $$|h| \leq \epsilon$$ where $0 < \epsilon < \min_{t \in [0,1]} |z - \gamma(t)|.$ Write the line integral above as an integral with respect to the real parameter $$t \in [0,1]$$; its integrand is dominated by a constant: $\forall \, t \in [0,1], \; \left| \frac{1}{i2\pi}\frac{f(\gamma(t))}{(\gamma(t)-z-h)(\gamma(t)-z)} \gamma'(t) \right| \leq \frac{1}{\epsilon^2} \max_{t \in [0,1]} |f(\gamma(t))|.$ Thus, Lebesgue’s dominated convergence theorem provides the existence of the derivative of $$f$$ at $$z$$ as well at its value: $f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{(w-z)^2} \, dw.$ Now it’s pretty clear that we can iterate the previous argument: consider the right-hand side of the above equations as a function of $$z$$, build its difference quotient and pass to the limit. The process provides $f''(z) = \lim_{h \to 0} \frac{f'(z+h) - f'(z)}{h} = \frac{1}{i2\pi} \int_{\gamma} \frac{2f(w)}{(w-z)^3} \, dw.$ The argument is valid for any $$z \in \Omega$$: the function $$f'$$ is also holomorphic.

## The Fundamental Theorem of Algebra

### Question

Prove that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

Let $$p: \mathbb{C} \to \mathbb{C}$$ be a polynomial with no complex root. The function $$f= 1/p$$ is defined and holomorphic on $$\mathbb{C}$$. Additionally, as $$|p(z)| \to +\infty$$ when $$|z| \to + \infty$$, the modulus of $$f$$ is bounded. By Liouville’s theorem, $$f$$ is constant, hence $$p$$ is constant too.
Show that any non-constant holomorphic function $$f:\mathbb{C} \to \mathbb{C}$$ has an image which is dense in $$\mathbb{C}$$: $\forall \, w \in \mathbb{C}, \; \forall \, \epsilon > 0, \; \exists \, z \in \mathbb{C}, \; |f(z) - w| < \epsilon.$
Assume that the image of $$f$$ is not dense in $$\mathbb{C}$$: there is a $$w\in\mathbb{C}$$ and a $$\epsilon > 0$$ such that for any $$z\in \mathbb{C}$$, $$|f(z) - w| \geq \epsilon$$. Now consider the function $$z \mapsto 1/(f(z) - w)$$; it is defined and holomorphic in $$\mathbb{C}$$. Additionally, $\forall \, z \in \mathbb{C},\; \left| \frac{1}{f(z) - w}\right| \leq \frac{1}{\epsilon}.$ By Liouville’s theorem, this function is constant, hence $$f$$ is constant too.