Let 0<r<R and f be the function f: U̇R(0) ⟶ℂ z ⟼(R+z)/((R-z) z) Show that f(z)=(1)/(

Let 0<r<R and f be the function f: U̇R(0) ⟶ℂ z ...

Let $0<r<R$ and $f$ be the function $$ \begin{aligned} f: \dot{U}_{R}(0) & \longrightarrow \mathbb{C} \\ z & \longmapsto \frac{R+z}{(R-z) z} \end{aligned} $$ Show that $f(z)=\frac{1}{z}+\frac{2}{R-z}$, and, by integrating over the curve $\alpha$, $$ \alpha:[0,2 \pi] \longrightarrow \mathbb{C}, \quad \alpha(t)=r \exp (\mathrm{i} t) $$ that $$ \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{R^{2}-r^{2}}{R^{2}-2 R r \cos t+r^{2}} d t=1 $$ Show in a similar manner: $$ \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{R \cos t}{R^{2}-2 R r \cos t+r^{2}} d t=\frac{r}{R^{2}-r^{2}}, \text { if } 0 \leq r<R $$

Let $0<r<R$ and $f$ be the function
$$
\begin{aligned}
f: \dot{U}_{R}(0) & \longrightarrow \mathbb{C} \\
z & \longmapsto \frac{R+z}{(R-z) z}
\end{aligned}
$$
Show that $f(z)=\frac{1}{z}+\frac{2}{R-z}$, and, by integrating over the curve $\alpha$,
$$
\alpha:[0,2 \pi] \longrightarrow \mathbb{C}, \quad \alpha(t)=r \exp (\mathrm{i} t)
$$
that
$$
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{R^{2}-r^{2}}{R^{2}-2 R r \cos t+r^{2}} d t=1
$$
Show in a similar manner:
$$
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{R \cos t}{R^{2}-2 R r \cos t+r^{2}} d t=\frac{r}{R^{2}-r^{2}}, \text { if } 0 \leq r<R
$$

Key Concepts

Partial Fraction Decomposition

This technique involves expressing a complicated rational function as a sum of simpler fractions. By breaking the given function into parts that depend on the singularities separately, it becomes much easier to handle, particularly when integrating over contours. In many cases, such decomposition enables the application of standard integration formulas or residue calculation methods.

Residue Theorem

A central result in complex analysis, the residue theorem allows the evaluation of contour integrals by summing the residues (the coefficients of the 1/(z?z?) term in a function's Laurent series) of the function’s isolated singularities inside the contour. This theorem is especially powerful when dealing with rational functions over closed curves, as it converts an integral into a sum over a discrete set of points.

Contour Integration

Contour integration is the process of integrating a complex-valued function along a directed path or curve in the complex plane. By choosing an appropriate contour, such as a circle, and parameterizing it, one can simplify the evaluation of integrals, particularly when combined with techniques like partial fractions or the residue theorem. This method is essential for solving integrals that are otherwise difficult to approach using real-variable methods.

Parametrization of Curves in the Complex Plane

This concept involves representing a contour or path in the complex plane using a real parameter. For instance, a circular path can be described in terms of an angle parameter, which allows the conversion of a complex integral into an ordinary integral over a real interval. This transformation is crucial in making the evaluation of complex integrals more tractable by applying known real analysis techniques.

Transcript

00:03 We are given a line integral over a curve c, and we are given the parameterization for this curve, and we are asked to evaluate this line integral.

00:21 So the line integral is over the curve c of y plus sine x dx, plus z squared plus cosine y, d, y, plus x cubed d z.

00:32 The curve c has parametric equation r of t equals the vector sine t, cosine t sine 2 t from t equals 0 to 2 pi so first we'll notice that this line integral can also be written as the line integral over c of f a vector field f where f is the vector field y plus sine x i plus z squared plus cosine y j plus y cubed sorry not y cubed x cubed x cubed x k and therefore it follows that the curl of f is equal to negative 2 z i minus 3x squared j minus 1 k now notice that by trigger then it is sine of 2 t is equal to 2 sine t cosine of t and so looking at the formula, the vector equation for our curve, we see that curve c lies on the surface z, which is sine of 2t, equals 2 times x, which is sine t, times y, which is cosine t.

03:12 Now we'll let s be the part of this surface that is bounded by c.

03:34 Then the projection of s onto the xy plane.

03:45 This is a region d, and this is, in fact, a unit disk for x squared plus y squared is less than or equal to one.

04:05 Now, looking at the way our curve is set up, our curve c is traversed clockwise when it is viewed from above.

04:31 Therefore, this induces a downward orientation, on our surface s.

04:49 And we're going to use a previous result that simplifies calculating surface integrals.

04:55 So we have that z, our surface, is a function of x and y.

05:01 So we can say z equals g of xy equals 2xy.

05:08 And we have that the function p, this will be the first component to the curl of f, which was negative 2 z, substituting from our surface c, this is negative 4xy...

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