Cauchy’s Inequality and Liouville’s Theorem

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Suppose that a function $f$ is analytic inside and on a positively oriented circle $C_R$, centered at $z_0$ and with radius $R$. Then by Cauchy Integral Formula
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}dz.$$
Since $C_R$ is compact (i.e. closed and bounded) and $f(z)$ is continuous, $|f(z)|$ assumes a maximum (and also a minimum) on $C_R$. Let us call the maximum value of $|f(z)|$ on $C_R$ $N_R$. On $C_R$, $|z-z_0|=R$. Hence, we have the upper bound for $|f^{(n)}(z_0)|$ as
\begin{align*}
|f^{(n)}(z_0)|&=\frac{n!}{2\pi}\left|\oint_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}dz\right|\\
&\leq\frac{n!M_R}{R^n}
\end{align*}
for $n=1,2,\cdots$. This inequality is called Cauchy’s Inequality.

Using Cauchy’s Inequality, we can prove the following Liouville’s Theorem.

Theorem. If $f$ is entire and bounded in the complex plane $\mathbb{C}$, then $f(z)$ is constant.

Proof. Since $f$ is bounded, there exists $M>0$ such that $|f(z)|\leq M$ for all $z\in\mathbb{C}$. Since $f$ is entire, for any $R>0$,
$$|f'(z)|\leq\frac{M}{R}$$
by Cauchy’s Inequality. As $R\to\infty$, $\frac{M}{R}\to 0$, so we obtain $|f'(z)|=0$ for all $z\in\mathbb{C}$. This implies that $f'(z)=0$ for all $z\in\mathbb{C}$ and hence, $f(z)$ is constant.

Morera’s Theorem

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Cauchy’s Integral Theorem says that if a function $f(z)$ is analytic throughout some simply connected domain $D$, then for any contour $C$ in $D$, $\oint_C f(z)dz=0$. It turns out that the converse of Cauchy’s Theorem is also true, namely

Theorem. If a function $f(z)$ is continuous in a simply connected domain $D$ and $\oint_C f(z)dz=0$ for every closed contour $C$ within $D$, then $f(z)$ is analytic throughout $D$.

This theorem is called Morera’s Theorem. Let us prove the theorem.

First we show that under the assumption the line integral $\int_\gamma f(z)dz$ is independent on the path $\gamma$. Let $\gamma$ be a path from $z_1$ to $z_2$ and choose $\gamma_1$ another path from $z_1$ to $z_2$ as shown in the figure.


Then $C: \gamma\cup (-\gamma_1)$ is a positively oriented contour. So, by assumption,
\begin{align*}
0&=\oint_C f(z)dz\\
&=\int_\gamma f(z)dz+\int_{-\gamma_1}f(z)dz\\
&=\int_\gamma f(z)dz-\int_{\gamma_1}f(z)dz.
\end{align*}
That is,
$$\int_\gamma f(z)dz=\int_{\gamma_1}f(z)dz.$$
Hence, the line integral $\int_\gamma f(z)dz$ does not depend on the path $\gamma$ but only on the endpoints $z_1$ and $z_2$.

Let us define $F(z)=\int_{z_0}^z f(w)dw$. Then
\begin{align*}
F(z+\Delta z)-F(z)&=\int_{z_0}^{z+\Delta z} f(w)dw-\int_{z_0}^z f(w)dw\\
&=\int_z^{z+\Delta z} f(w)dw.
\end{align*}
Since $\int_z^{z+\Delta z}dw=\Delta z$,
$$f(z)=\frac{1}{\Delta z}\int_z^{z+\Delta z}f(z) dw$$
and so, we have
$$\frac{F(z+\Delta z)-F(z)}{\Delta z}-f(z)=\frac{1}{\Delta z}\int_z^{z+\Delta z}[f(w)-f(z)]dw.$$
Since $f$ is continuous at $z$, given $\epsilon>0$ there exists $\delta>0$ such that $|f(w)-f(z)|<\epsilon$ whenever $|w-z|<\delta$. If $z+\Delta z$ is close enough to $z$ so that $|\Delta z|<\delta$, then
$$\left|\frac{F(z+\Delta z)-F(z)}{\Delta z}-f(z)\right|<\frac{1}{|\Delta z|}\epsilon|\Delta z|=\epsilon.$$
Therefore,
$$F'(z)=\lim_{\Delta z\to 0}\frac{F(z+\Delta z)-F(z)}{\Delta z}=f(z).$$
That is, $F(z)$ analytic in $D$. This means that $f(z)$ is analytic by Cauchy Integral Formula.

Cauchy Integral Formula

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Suppose that $f(z)$ is analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is a point exterior to $C$, then by Cauchy’s Integral Theorem,
$$\oint_C\frac{f(z)}{z-z_0}dz=0.$$
Now, the question is what would be the value of the integral
$$\oint_C\frac{f(z)}{z-z_0}dz$$
if $z_0$ is a point interior to $C$? To answer this question, let us consider a tiny circle centerd at $z_0$ and with radius $r$, also oriented counterclockwise as shown in the figure.

Then using Cauchy’s Integral Theorem, we obatin
$$\oint_C\frac{f(z)}{z-z_0}dz-\oint_{C_1}\frac{f(z)}{z-z_0}dz=0.$$
Let $z=z_0+re^{i\theta}$. Then
\begin{align*}
\oint_{C_1}\frac{f(z)}{z-z_0}dz&=\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}rie^{i\theta}d\theta\\
&=i\int_0^{2\pi}f(z_0+re^{i\theta})d\theta\\
&\to 2\pi if(z_0)
\end{align*}
as $r\to 0$. Therefore, we obtain
$$f(z_0)=\frac{1}{2\pi}\oint_C\frac{f(z)}{z-z_0}dz.$$
This formula is called Cauchy Integral Formula.

Example. Let $C$ be the positively oriented circle $|z|=2$. The function $f(z)=\frac{z}{9-z^2}$ is analytic within and on $C$. Since $z_0=-i$ is interior to $C$, by Cauchy Integral Formula,
\begin{align*}
\oint_C\frac{z}{(9-z^2)(z+i)}dz&=\oint_C\frac{\frac{z}{9-z^2}}{z+i}dz\\
&=2\pi\frac{-i}{9-(-i)^2}\\
&=\frac{\pi}{5}.
\end{align*}

Derivatives

Suppose that $|\Delta z_0|$ is small enough so that $z_0+\Delta z_0$ is still interior to the contour $C$. Then by Cauchy Integral Formula, we obtain
$$\frac{f(z_0+\Delta z_0)-f(z_0)}{\Delta z_0}=\frac{1}{2\pi i\Delta z_0}\left\{\oint_C\frac{f(z)}{z-z_0-\Delta z_0}dz-\oint_C\frac{f(z)}{z-z_0}dz\right\}.$$
So,
\begin{align*}
f'(z_0)&=\lim_{\Delta z_0\to 0}\frac{f(z_0+\Delta z_0)-f(z_0)}{\Delta z_0}\\
&=\lim_{\Delta z_0\to 0}\frac{1}{2\pi i\Delta z_0}\left\{\oint_C\frac{f(z)}{z-z_0-\Delta z_0}dz-\oint_C\frac{f(z)}{z-z_0}dz\right\}\\
&=\lim_{\Delta z_0\to 0}\frac{1}{2\pi i\Delta z_0}\oint_C\frac{\Delta z_0f(z)}{(z-z_0-\Delta z_0)(z-z_0)}dz\\
&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz.
\end{align*}
The same process can be repeated for $f'(z_0)$ and we obtain
$$f^{\prime\prime}(z_0)=\frac{2}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^3}dz.$$
Continuing, we get
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}$$
for $n=0,1,2,\cdots$ where $f^{(0)}(z)=f(z)$. This formula may also be called Cauchy Integral Formula which includes the Cauchy Integral Formula we derived earlier.

Example. If $C$ is the positively oriented unit circle $|z|=1$ and $f(z)=\exp(2z)$, then
\begin{align*}
\oint_C\frac{\exp(2z)}{z^4}dz&=\oint_C\frac{f(z)}{(z-0)^{3+1}}\\
&=\frac{2\pi i}{3!}f^{\prime\prime\prime}(0)\\
&=\frac{8\pi i}{3}
\end{align*}
where $f(z)=\exp(2z)$.

Example. Let $z_0$ be any point interior to a positively oriented simple closed contour $C$. When $f(z)=1$, Cauchy Integral Formula tells that
$$\oint_C\frac{dz}{z-z_0}=2\pi i\ \mbox{and}\ \oint_C\frac{dz}{(z-z_0)^{n+1}}=0,\ n=1,2,\cdots.$$

Determinants II: Determinants of Order $n$

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A determinant of order $n$ can be calculated by expanding it in terms of determinants of order $n-1$. Let $A=(a_{ij})$ be an $n\times n$ matrix and let us denote by $A_{ij}$ the $(n-1)\times (n-1)$ matrix obtained by deleting the $i$-th row and the $j$-th column from $A$:

Then $\det A$ is given by the Laplace expansion
\begin{align*}
\det A&=(-1)^{i+1}a_{i1}\det A_{i1}+\cdots+(-1)^{i+n}a_{in}\det A_{in}\\
&=(-1)^{1+j}a_{1j}\det A_{1j}+\cdots+(-1)^{n+j}a_{nj}\det A_{nj}.
\end{align*}

All the properties of the determinants of order 2 we studied here still hold in general for the determinants of order $n$. In particular,

Theorem. Let $A^1,\cdots,A^n$ be column vectors of dimension $n$. They are linearly dependent if and only if
$$\det(A^1,\cdots,A^n)=0.$$

Example. Let us calculate the determinant of the following $3\times 3$ matrix
$$A=\begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}.$$
You may use any column or row to calculate $\det A$ using the Laplace expansion. In this example, we use the first row to calculate $\det A$. By the Laplace expansion,
\begin{align*}
\det A&=a_{11}\det A_{11}-a_{12}\det A_{12}+a_{13}\det A_{13}\\
&=a_{11}\left|\begin{array}{cc}
a_{22} & a_{23}\\
a_{32} & a_{33}
\end{array}\right|-a_{12}\left|\begin{array}{cc}
a_{21} & a_{23}\\
a_{31} & a_{33}
\end{array}\right|+a_{13}\left|\begin{array}{cc}
a_{21} & a_{22}\\
a_{31} & a_{32}
\end{array}\right|.
\end{align*}
Replay this with
$$A=\begin{pmatrix}
2 & 1 & 0\\
1 & 1 & 4\\
-3 & 2 & 5
\end{pmatrix}.$$
Since the first row or the third column contains 0, you may want to use the first row or the third column to do the Laplace expansion.

For $3\times 3$ matrices, there is a quicker way to calculate the determinant as shown in the following figure. You multiply three entries along each indicated arrow. When you multiply three entries along each red arrow, you also multiply by $-1$. This is called the Rule of Sarrus named after a French mathematician Pierre Frédéric Sarrus. Please be warned that the rule of Sarrus works only for $3\times 3$ matrices.

The Rule of Sarrus

Example. [Cross Product] Let $v=v_1E_1+v_2E_2+v_3E_3$ and $w=w_1E_1+w_2E_2+w_3E_3$ be two vectors in Euclidean 3-space $\mathbb{R}^3$. The cross product is defined by
$$v\times w=\left|\begin{array}{ccc}
E_1 & E_2 & E_3\\
v_1 & v_2 & v_3\\
w_1 & w_2 & w_3
\end{array}\right|.$$
Note that the cross product is perpendicular to both $v$ and $w$.

Clearly, if there are many 0 entries in a given determinant, it would be easier to calculate the determinant since you will have a lesser than usual number of terms that you actually have to calculate in the Laplace expansion. For any given determinant, we can indeed make it happen. Recall the theorem we studied here:

Theorem. If one adds a scalar multiple of one column (row) to another column (row), then the value of the determinant does not change.

Using the particular column (row) operation in the Theorem, we can turn a given determinant into one with more 0 entries.

Example. Find
$$\left|\begin{array}{cccc}
1 & 3 & 1 & 1\\
2 & 1 & 5 & 2\\
1 & -1 & 2 & 3\\
4 & 1 & -3 & 7
\end{array}\right|.$$

Solution. By the above Theorem,
\begin{align*}
\left|\begin{array}{cccc}
1 & 3 & 1 & 1\\
2 & 1 & 5 & 2\\
1 & -1 & 2 & 3\\
4 & 1 & -3 & 7
\end{array}\right|&=\left|\begin{array}{cccc}
0 & 4 & -1 & -2\\
2 & 1 & 5 & 2\\
1 & -1 & 2 & 3\\
4 & 1 & -3 & 7
\end{array}\right|\ (\mbox{add $-1$ times row 3 to row 1})\\
&=\left|\begin{array}{cccc}
0 & 4 & -1 & -2\\
0 & 3 & 1 & -4\\
1 & -1 & 2 & 3\\
4 & 1 & -3 & 7
\end{array}\right|\ (\mbox{add $-2$ times row 3 to row 2})\\
&=\left|\begin{array}{cccc}
0 & 4 & -1 & -2\\
0 & 3 & 1 & -4\\
1 & -1 & 2 & 3\\
0 & 5 & -11 & -5
\end{array}\right|\ (\mbox{add $-4$ times row 3 to row 4})\\
&=\left|\begin{array}{ccc}
4 & -1 & -2\\
3 & 1 & -4\\
5 & -11 & -5
\end{array}\right|\ (\mbox{Laplace expansion along column 1})
\end{align*}
You may compute the resulting determinant of order 3 using the rule of Sarrus or you may further simpify it. For instance, you may do:
\begin{align*}
\left|\begin{array}{ccc}
4 & -1 & -2\\
3 & 1 & -4\\
5 & -11 & -5
\end{array}\right|&=\left|\begin{array}{ccc}
7 & 0 & -6\\
3 & 1 & -4\\
5 & -11 & -5
\end{array}\right|\ (\mbox{add row 2 to row 1})\\
&=\left|\begin{array}{ccc}
7 & 0 & -6\\
3 & 1 & -4\\
38 & 0 & -49
\end{array}\right|\ (\mbox{add 11 times row 2 to row 3})\\
&=\left|\begin{array}{cc}
7 & -6\\
38 & -49
\end{array}\right|\ (\mbox{Laplace expansion along column 2})\\
&=-115.
\end{align*}

Determinants I: Determinants of Order 2

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Let $A=\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}$. Then we define the determinant $\det A$ by
$$\det A=ad-bc.$$
$\det A$ is also denoted by $|A|$ or $\left|\begin{array}{ccc}
a & b\\
c & d
\end{array}\right|$. In terms of the column vectors $A^1,A^2$, the determinant of $A$ may also be written as $\det(A^1,A^2)$.

Example. If $A=\begin{pmatrix}
2 & 1\\
1 & 4
\end{pmatrix}$, then $\det A=8-1=7$.

Property 1. The determinant $\det (A^1,A^2)$ may be considered as a bilinear map of the column vectors. As a bilinear map $\det (A^1,A^2)$ is linear in each slot. For example, if $A^1=C+C’$ then
$$\det(A^1,A^2)=\det(C,A^2)+\det(C’,A^2).$$
If $x$ is a number,
$$\det(xA^1,A^2)=x\det(A^1,A^2).$$

Property 2. If the two columns are equal, the determinant is 0.

Property 3. $\det I=\det (E^1,E^2)=1$.

Combining Properties 1-3, we can show that:

Theorem. If one adds a scalar multiple of one column to another, then the value of the determinant does not change.

Proof. We prove the theorem for a particular case.
\begin{align*}
\det(A^1+xA^2,A^2)&=\det(A^1,A^2)+x\det(A^2,A^2)\\
&=\det(A^1,A^2).
\end{align*}

Theorem. If the two columns are interchanged, the determinant changes by a sign i.e.
$$\det(A^2,A^1)=-\det(A^1,A^2).$$

Proof. \begin{align*}
0&=\det(A^1+A^2,A^1+A^2)\\
&=\det(A^1,A^2)+\det(A^2,A^1).
\end{align*}

Theorem. $\det A=\det {}^tA$.

Proof. This theorem can be proved directly from the definition of $\det A$.

Remark. Because of this theorem, we can also say that if one adds a scalar multiple of one row to another row, then the value of the determinant does not change.

Theorem. The column vectors $A^1,A^2$ are linearly dependent if and only if $\det(A^1,A^2)=0$.

Proof. Suppose that $A^1,A^2$ are linearly dependent. Then there exists numbers $x,y$, not all equal to 0 such that $xA^1+yA^2=0$. Let us say $x\ne 0$. Then $A^1=-\frac{y}{x}A^2$. So,
\begin{align*}
\det(A^1,A^2)&=\det\left(-\frac{y}{x}A^2,A^2\right)\\
&=-\frac{y}{x}\det(A^2,A^2)\\
&=0.
\end{align*}
To prove the converse, suppose that $A^1,A^2$ are linearly independent. Then $E^1,E^2$ can be written as linear combinations of $A^1,A^2$:
\begin{align*}
E^1&=xA^1+yA^2,\\
E^2&=zA^1+wA^2.
\end{align*}
Now,
\begin{align*}
1&=\det(E^1,E^1)\\
&=xw\det(A^1,A^2)+yz\det(A^2,A^1)\\
&=(xw-yz)\det(A^1,A^2).
\end{align*}
Hence, $\det(A^1,A^2)\ne 0$.