Mathematical Methods in Physics 1-3.2 Cauchy integral formula

1. Formula
   Observation

   The board condition line reads: "The region enclosed by C is a simply connected analytic region of $f(z)$."

2. Diagram
   Observation

   A closed circular loop is drawn on the right; the instructor points to this closed curve multiple times.

C

A closed loop or contour enclosing a point inside the region.

A closed curve in the complex plane.

1. Formula
   Observation

   The function $f(z)$ appears both in the board condition line and in the numerator of the formula.

2. Audio
   Observation

   The instructor says: "this function $f(z)$" and "the $f(z)$ inside the integrand is analytic."

$f(z)$

A complex-valued function analytic in the simply connected region enclosed by C.

A function of the complex variable z.

1. Formula
   Observation

   The board condition line reads: "$z_0$ is a point inside the region," and the denominator of the formula is $z - z_0$.

2. Diagram
   Observation

   The point $z_0$ is labeled inside the closed circle.

$z_0$

A specified point located inside the region enclosed by C.

A point in the complex plane.

1. Formula
   Observation

   The formula's denominator contains $2\pi i$

i

the imaginary unit

the complex unit

1. Formula
   Observation

   The formula's denominator contains $2\pi i$

$\pi$

the constant pi

a real constant

1. Formula
   Observation

   The integral expression ends with dz

dz

the differential element when integrating along contour C

the differential element for a complex line integral

1. Formula
   Observation

   The board displays the formulas $f(z)$ and $f(z_0)$.

$f(z)$

a complex analytic function defined on a simply connected analytic domain

a simply connected region in the complex plane

1. Formula
   Observation

   The board text reads 'The region enclosed by C is a simply connected analytic domain for $f(z)$' and the symbol C appears beneath the integral sign.

C

a simple closed curve enclosing the region (the integration contour)

complex plane

1. Formula
   Observation

   The board text reads '$z_0$ is a point inside the region' and the formula contains $z_0$.

$z_0$

a fixed point inside the region (not an analytic point)

complex plane

1. Formula
   Observation

   The symbol $C_r$ appears beneath the integral sign on the board.

$C_r$

a small circle centered at $z_0$ with radius r

complex plane

1. Diagram
   Observation

   A small circle is drawn in the figure, labeled with radius r.

2. Formula
   Observation

   The parameter equation $z = z_0$ + re^{$i\theta$} contains r.

r

radius of the small circle

positive real number

1. Formula
   Observation

   The parameter equation $z = z_0$ + re^{$i\theta$} and the integration limits from 0 to $2\pi$ contain $\theta$.

$\theta$

polar angle parameter

zero to two pi

1. Formula
   Observation

   The blackboard displays the Cauchy integral formula: $f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}dz$

2. Audio
   Observation

   The instructor explains: "The value of $f(z)$ at $z_0$ … can be expressed using this integral"

If the region enclosed by C is a simply connected analytic domain of $f(z)$, and $z_0$ is a point inside that region, then the value of the function at $z_0$ can be expressed as the contour integral along C: $f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}dz$.

1. The region enclosed by C is a simply connected analytic domain of $f(z)$.

2. $z_0$ is a point inside the region enclosed by C.

1. The simply connected analytic domain condition.

1. Formula
   Observation

   The condition line on the blackboard reads: "The region enclosed by C is a simply connected analytic domain of $f(z)$."

2. Audio
   Observation

   The instructor says: "This entire C is the closed loop, and it is a simply connected analytic domain of the function $f(z)$." "Inside this enclosed region, the function is simply connected and analytic—that is, fully analytic."

The video states the prerequisite for the formula's validity as follows: the entire region enclosed by the closed contour C is a simply connected analytic domain of the function $f(z)$, meaning $f(z)$ is analytic everywhere in that region.

1. The region enclosed by C is the object under discussion.

2. $f(z)$ is simply connected and analytic in that region.

1. Audio
   Observation

   The instructor says: "The integrand is not merely $f(z)$; rather, the integrand is $f(z)$ divided by z minus $z_0$."

2. Formula
   Observation

   The integrand in the formula is $\frac{f(z)}{z-z_0}$.

1. The clip ends at the phrase "So we see that, for this large integrand," and the subsequent reasoning is not included in this clip.

When proving Cauchy's integral formula, the video first stresses that one must not focus solely on the analyticity of the numerator $f(z)$, but rather analyze the entire integrand $\frac{f(z)}{z-z_0}$ as an object.

1. The object of study is the integrand of the integral $\oint_C \frac{f(z)}{z-z_0}dz$

1. Cauchy integral formula

1. Formula
   Observation

   Blackboard title "Cauchy integral formula" and the formula $f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} dz$.

If the region enclosed by C is a simply connected domain where $f(z)$ is analytic, and $z_0$ is a point inside that region, then $f(z_0)$ can be expressed via a contour integral along the boundary C.

1. the region enclosed by C is a simply connected domain where $f(z)$ is analytic

2. $z_0$ is a point inside the region

1. Audio
   Observation

   The instructor mentions, "The integral over a large loop can actually be reduced to the integral over a small loop encircling the non-analytic point $z_0$."

Using Cauchy's integral theorem for multiply connected domains, the integral along the large contour C can be transformed into the integral along the small contour $C_r$ encircling the singularity $z_0$, thereby simplifying computation.

1. The integrand is analytic in the annular region between C and $C_r$

1. Cauchy integral formula

1. Formula
   Observation

   the board title reads "Cauchy's integral formula", followed by $f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}dz$

2. Audio
   Observation

   the instructor concludes, "So this is the Cauchy integral formula we just discussed."

the video writes Cauchy's integral formula as: if the region enclosed by C is a simply connected domain where $f(z)$ is analytic, and $z_0$ is a point inside that region, then $f(z_0)$ can be expressed as a contour integral along C. This formula relates the value of the function at an interior point to its contour integral along the boundary loop.

1. the region enclosed by C is a simply connected domain where $f(z)$ is analytic

2. $z_0$ is a point inside the region enclosed by C

1. Audio
   Observation

   the instructor says, "Convert the large-loop integral into a small-loop integral," "as long as this loop encloses the non-analytic region," "all results are the same."

2. Diagram
   Observation

   The right-hand diagram simultaneously plots the outer contour C and the inner contour $C_r$, both encircling the same point $z_0$

The video emphasizes that, in the proof, the integral along the large contour C can be converted into the integral along the small contour $C_r$. As long as the new contour still encloses the non-analytic region and winds around the same point $z_0$, the integral values obtained from contours of different sizes are identical. Here, the key is not the shape itself, but whether the contour encloses the same singularity region.

1. The new contour still encloses the original non-analytic region

2. What is being compared is the contour integral of the same integrand $\frac{f(z)}{z-z_0}$

1. Cauchy's integral formula

1. Formula
   Observation

   Below the chalkboard, $\lim_{r\to 0}\oint_{C_r}\frac{f(z)}{z-z_0}dz=i\lim_{r\to 0}\int_0^{2\pi}f$($z_0$+re^{$i\theta$})$d\theta$ is written

2. Audio
   Observation

   The instructor says, "I can let my r approach zero, and the result remains unchanged."

The video further explains that, since the integral value remains unchanged as long as the contour encloses the point $z_0$, we may take the limit $r\to 0$, shrinking the small circle toward its center. The purpose of this step is to exploit the continuity of the analytic function, replacing f($z_0$+re^{$i\theta$}) with $f(z_0)$ in the limit.

1. $C_r$ is centered at $z_0$

2. The integrand is analytic everywhere in the region except at $z_0$

1. The method of converting the integral over the large contour into the integral over the small contour

1. Formula
   Observation

   the value of f at z zero equals one over two pi i times the integral along the closed curve C of f evaluated at z divided by z minus z zero, with respect to z

If the region enclosed by $C$ is a simply connected analytic region of $f(z)$, and $z_0$ is a point in that region, then Cauchy's integral formula holds.

1. The region enclosed by $C$ is a simply connected analytic region of $f(z)$

2. $z_0$ is a point in the region

1. Formula
   Observation

   One over two pi i times the integral along the closed curve C of one divided by z minus z zero, with respect to z, equals one if z zero lies inside the region bounded by C, and zero if z zero lies outside the region bounded by C

When $f(z) = 1$, Cauchy's integral formula reduces to a very useful special case for determining whether a point lies inside the integration contour.

1. f evaluated at z is identically equal to one

1. Cauchy's integral formula

1. Formula
   Observation

   The derivative of f at z equals one over two π i times the integral along the closed curve C of f evaluated at sigma divided by sigma minus z, integrated with respect to sigma, then differentiated with respect to z; this derivative also equals one over two π i times the integral along the closed curve C of f evaluated at sigma divided by sigma minus z squared, integrated with respect to sigma.

Differentiating both sides of the Cauchy integral formula with respect to $z$ yields the formula for the first derivative of an analytic function.

1. $f(z)$ is analytic inside $C$

1. Cauchy's integral formula

1. Formula
   Observation

   The second derivative of f at z equals two factorial divided by two π i, multiplied by the integral along the closed curve C of f evaluated at sigma divided by sigma minus z cubed, integrated with respect to sigma.

Differentiating the first derivative formula further yields the formula for the second derivative of an analytic function.

1. $f(z)$ is analytic inside $C$

1. first derivative of an analytic function

1. Formula
   Observation

   The blackboard fully writes "Cauchy's integral formula: $f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} dz$"

2. Audio
   Observation

   The instructor verbally explains that this formula relates the function value at a point $z_0$ inside the region to a contour integral around that point.

If the region enclosed by C is a simply connected domain of analyticity for $f(z)$, and $z_0$ is a point inside that region, then $f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}dz$.

1. The region enclosed by C is a simply connected domain of analyticity for $f(z)$

2. $z_0$ is a point inside the region enclosed by C

Holds for C, $f(z)$, and the interior point $z_0$ satisfying the above conditions

1. Audio
   Observation

   The instructor mentions, "We will need to apply Cauchy's integral theorem for multiply connected domains."

In a multiply connected domain, if a function is analytic between all boundaries, then the integral along the outer boundary equals the sum of the integrals along the inner boundaries.

1. The function is analytic within the multiply connected domain

For any multiply connected domain and analytic function satisfying the conditions

1. Audio
   Observation

   The instructor says, "The entire proof of this Cauchy integral formula relies solely on the Cauchy integral theorem for multiply connected domains, which we discussed earlier."

The video points out that the proof of Cauchy's integral formula shown in this segment is fundamentally based on Cauchy's integral theorem for multiply connected domains, discussed earlier.

1. Cauchy's integral theorem for multiply connected domains has already been established.

2. The integrand satisfies the analyticity condition within the corresponding region.

A general explanation of the theorem used as the basis for this segment's proof.

1. Audio
   Observation

   The instructor says, 'As long as it is an analytic function... within the region enclosed by C, which is simply connected and analytic, then we know that this analytic function's derivatives—yes, you can compute derivatives of arbitrarily high order; any order of derivative can be taken.'

2. Caption evidence
   Observation

   Subtitle: "Analytic functions are infinitely differentiable!"

If a function $f(z)$ is analytic on a simple closed curve C and in the simply connected region enclosed by C, then $f(z)$ has derivatives of all orders in that region, and each derivative can be expressed via a contour integral formula.

1. $f(z)$ is analytic on and inside C

2. C is a simple closed curve

For any point z in the region and any nonnegative integer n

1. Audio
   Observation

   The instructor says, "Let us prove this fact. How do we prove it? Well, look at the integrand of this integral."

2. Formula
   Observation

   The blackboard displays "Proof:" below the previous content

3. Audio
   Observation

   The instructor continues, 'Although the function $f(z)$ inside your integrand is analytic... the integrand is not just $f(z)$; this integrand is $f(z)$ divided by z minus $z_0$.'

1. This segment presents only the initial observation of the proof, without the full derivation chain.

1. Expression

   $$$$
   Explanation

   After entering the proof, focus first on the integrand of integral $\oint_C \frac{f(z)}{z-z_0}dz$.

   Justification

   The video explicitly states orally, 'We will examine the integrand of this integral.'

   Shown in the video
2. Expression

   $$\frac{f(z)}{z-z_0}$$
   Explanation

   It is pointed out that the integrand is not merely $f(z)$, but the quotient $\frac{f(z)}{z-z_0}$.

   Justification

   The instructor says, 'The integrand is not just $f(z)$; this integrand is $f(z)$ divided by z minus $z_0$.'

   Shown in the video

The derivation in this segment proceeds only as far as 'the proof should begin with the overall integrand $\frac{f(z)}{z-z_0}$'; subsequent reasoning does not appear.

1. Formula
   Observation

   The blackboard step-by-step demonstrates the derivation from $\oint_C$ to $\oint_{C_r}$ and then to the parametrized integral.

1. Expression

   $$\oint_C \frac{f(z)}{z - z_0} dz = \oint_{C_r} \frac{f(z)}{z - z_0} dz$$
   Explanation

   Using Cauchy's integral theorem for multiply connected domains, the integral over the large contour C is transformed into the integral over the small contour $C_r$.

   Justification

   Cauchy's Integral Theorem on Multiply Connected Domains

   Shown in the video
2. Expression

   $$z \in C_r, \quad z = z_0 + re^{i\theta}$$
   Explanation

   Parameterize the small loop $C_r$ with center $z_0$, radius r, and polar angle $\theta$.

   Justification

   Parametric equations of a circle

   Shown in the video
3. Expression

   $$dz = ire^{i\theta} d\theta$$
   Explanation

   Differentiate both sides of the parametric equations to obtain the expression for dz.

   Justification

   Differentiation rules for complex functions

   Shown in the video
4. Expression

   $$\oint_{C_r} \frac{f(z)}{z - z_0} dz = \int_0^{2\pi} \frac{f(z_0 + re^{i\theta})}{re^{i\theta}} ire^{i\theta} d\theta$$
   Explanation

   Substitute the parametric expressions for z and dz into the integral, and change the limits of integration from the contour to theta ranging from 0 to 2 pi.

   Justification

   Parametric substitution in integrals

   Shown in the video

By parameterizing the integration path, the contour integral in the complex plane is successfully converted into a definite integral in the real variable $\theta$, laying the groundwork for taking the limit $r \to 0$ to prove the formula.

1. Formula
   Observation

   The first line of the board work gives $\oint_C \frac{f(z)}{z-z_0}dz=\oint_{C_r}\frac{f(z)}{z-z_0}dz$, where $z\in C_r$ and $z=z_0$+re^{$i\theta$}.

2. Formula
   Observation

   The next line writes =$\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}$ire^{$i\theta$}$d\theta$.

3. Audio
   Observation

   The instructor says, "The r and r cancel, and e to the i theta and e to the i theta cancel, so the result emerges immediately."

1. Expression

   $$\oint_C \frac{f(z)}{z-z_0}\,dz=\oint_{C_r}\frac{f(z)}{z-z_0}\,dz$$
   Explanation

   First rewrite the integral along the large contour C as an integral along the small circle $C_r$.

   Justification

   The video verbally explains that the large contour can be replaced by a small contour, provided the non-analytic region remains enclosed.

   Shown in the video
2. Expression

   $$z=z_0+re^{i\theta},\quad dz=ire^{i\theta}\,d\theta$$
   Explanation

   Apply polar-coordinate parameterization to the small circle $C_r$ and express the differential in terms of $\theta$.

   Justification

   The blackboard directly gives the parametric equations and their corresponding differentials.

   Shown in the video
3. Expression

   $$\oint_{C_r}\frac{f(z)}{z-z_0}\,dz=\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}\,ire^{i\theta}\,d\theta$$
   Explanation

   Substitute z and dz to obtain a definite integral in terms of $\theta$.

   Justification

   Parametric substitution.

   Shown in the video
4. Expression

   $$\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}\,ire^{i\theta}\,d\theta=i\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta$$
   Explanation

   The r and $e^{i\theta}$ in numerator and denominator cancel, leaving only the coefficient i and the integral with respect to $\theta$.

   Justification

   Algebraic simplification; the instructor explicitly states, "r cancels with r, and e to the i theta cancels with e to the i theta."

   Shown in the video

Obtain the intermediate result $\oint_C \frac{f(z)}{z-z_0}$\,$dz=i\int_0^{2\pi}f$($z_0$+re^{$i\theta$})\,$d\theta$.

1. Formula
   Observation

   Below the blackboard, write $\oint_C \frac{f(z)}{z-z_0}dz=\lim_{r\to 0}\oint_{C_r}\frac{f(z)}{z-z_0}dz=i\lim_{r\to 0}\int_0^{2\pi}f$($z_0$+re^{$i\theta$})$d\theta=2\pi i f(z_0)$.

2. Audio
   Observation

   The instructor says, "When my r tends to zero, due to the continuity of this analytic function..., we can write it as $f(z_0)$, and then factor it out."

3. Audio
   Observation

   The instructor continues, "Inside, only the integral with respect to theta from 0 to $2\pi$ remains, which yields the result $2\pi$."

1. Expression

   $$\oint_C \frac{f(z)}{z-z_0}\,dz=\lim_{r\to 0}\oint_{C_r}\frac{f(z)}{z-z_0}\,dz$$
   Explanation

   Write the original integral as the limit as the small circle's radius tends to zero.

   Justification

   The video explains that as long as the contour encloses $z_0$, the integral value does not change with the contour's size, so we may take $r\to 0$.

   Shown in the video
2. Expression

   $$\lim_{r\to 0}\oint_{C_r}\frac{f(z)}{z-z_0}\,dz=i\lim_{r\to 0}\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta$$
   Explanation

   Use the previously parameterized expression and then take the overall limit.

   Justification

   Substitution from the previous step's derivation.

   Shown in the video
3. Expression

   $$i\lim_{r\to 0}\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta=i\int_0^{2\pi}\lim_{r\to 0}f(z_0+re^{i\theta})\,d\theta$$
   Explanation

   Move the limit inside the integral sign.

   Justification

   The video's verbal justification is 'continuity of analytic functions'; the stricter conditions for interchanging limit and integral are not fully elaborated in this segment.

   Shown in the video
4. Expression

   $$\lim_{r\to 0}f(z_0+re^{i\theta})=f(z_0)$$
   Explanation

   As $r \to 0$, the point $z_0$+re^{$i\theta$} approaches $z_0$, and the function value approaches $f(z_0)$.

   Justification

   Continuity.

   Shown in the video
5. Expression

   $$i\int_0^{2\pi}f(z_0)\,d\theta=i f(z_0)\int_0^{2\pi}d\theta$$
   Explanation

   $f(z_0)$ is independent of $\theta$ and can be moved outside the integral sign.

   Justification

   A constant function can be factored out of the integral.

   Shown in the video
6. Expression

   $$i f(z_0)\int_0^{2\pi}d\theta=2\pi i f(z_0)$$
   Explanation

   Evaluate the remaining angular integral to obtain length $2\pi$.

   Justification

   Integrate $\theta$ from 0 to $2\pi$, yielding $2\pi$.

   Shown in the video