Integrals

Enter homework on integrals here.

## Question 1B

Evaluate the following contour integrals by parameterizing the contouring:

$\int _{c} \frac{z+2}{z}dz\; \;,$ where C is the upper half of circle of radius 2 centered at origin

$z=z0+re^{iθ}, θ->[0,pi]$
$z=2e^{it}, t->[0,pi]$
$dz=2ie^{it}dt$
$\int_{0}^{pi} \frac{2e^{it}+2}{2e^{it}}\, 2ie^{it} dt$
$\int_{0}^{pi}\frac{2ei^{2it}=2ie^{it}}{e^{it}} dt$
$\int_{0}^{pi} 2ie^{it} + 2i dt$
$2\int_{0}^{pi} ie^{it}+2i dt$
$2[e^{it}+2it]$ from 0 to pi
$2[(e^{ipi+2ipi})-(e^{0}+0)]$
$2[cos(pi)+isin(pi)+2ipi-1]$
$2[2ipi-2]$
$4ipi-4$

## Question 1C

Evaluate the following contour integrals by parameterizing the contouring:

$\int _{c} \left | x \right |^2dz\; \;, z=t+it$, $0\leq t\leq 1$

$(\sqrt{t^2+t^4})^2 \rightarrow t^2+t^4 \; \; \; \; dz=1+2it$

$=\int (t^2+t^4)*(1+2it)\, dt$

$=\int (t^2+2it^3+t^4+2it^5)\,dt$

$=\frac{t^3}{3}+\frac{2it^4}{4}+\frac{t^5}{5}+\frac{2it^6}{6} |_0^1$

$=\frac{1}{3}+\frac{i}{2}+\frac{1}{5}+\frac{i}{3}$

$=\frac{5}{15}+\frac{3}{15}+\frac{3i}{6}+\frac{2i}{6}$

$=\frac{8}{15}+\frac{5i}{6}$

## Question 2

Evaluate the following contour integrals using the fundamental theorem of calculus for contour integrals

a)$\int _C 3z^2-5z+i\; dz$where C is the line segment from $z=i$to $z=1$

(1)
\begin{align} \int_{i}^{1}3z^2-5z+i\; dz = \frac{3z^3}{3}-\frac{5z^2}{2} + iz|_{i}^{1} \\=1-\frac{5}{2}+ i -(\frac{3i^3}{3} - \frac{5i^2}{2} +i^2) \\=1-\frac{5}{2}+ i -(-i - \frac{5}{2} -1) \\=1-\frac{5}{2}+ i -(-i - \frac{5}{2} -1) \\=1-\frac{5}{2}+ i + i - \frac{5}{2} +1) \\=2+ 2i - \frac{10}{2} \\=2+ 2i - 5 \\= 2i - 3 \end{align}

b)
$\int _C\frac{1}{z}\; dz$where C is any contour in the right-half plane $z=-3i$to $z=3i$

(2)
\begin{align} \int_{-3i}^{3i}\frac{1}{z}\; dz=ln{z}|_{-3i}^{3i}=\left [ log\left ( 3i \right ) \right ]-\left [ log\left ( -3i \right ) \right ]=\left [ log\left ( 3 \right )+i\frac{\pi}{2} \right ]-\left [ log\left ( 3 \right )+i\frac{\pi}{2} \right ] \end{align}
(3)
\begin{align} =log\left ( 3 \right )+i\frac{\pi}{2}-log\left ( 3 \right )-i\frac{\pi}{2}=log\left ( 3 \right )-log\left ( -3 \right )=i\pi \end{align}

## Question 3

Establish the estimate $\left | \int_{C} \frac{1}{z^{2}-1}dz \right | \leq \frac{\pi}{3}\\$where C is the first quadrant portion of the circle $\left | z \right | = 2$.

(4)
\begin{align} \left | \int_{C} \frac{1}{z^{2}-1}dz \right | \\ \end{align}

Use the formula for finding the estimate.

(5)
\begin{align} \int_{C} (f(z))dz \leq m*length(r)\\ \end{align}

Calculate length r.

(6)
\begin{align} \frac{1}{4}\pi (2)^2 = \pi\\ \end{align}

Calculate $max_{z \euro C} \left | f(z) \right |$.

(7)
\begin{align} f(z)=\frac{1}{4e^{2}-1}\\ \left | z_1+z_2 \right | \geq \left | \left|z_1 \right | - \left| z_2 \right | \right | \leq 3\\ \left|z_1 \right |^{2}=2^{2}=4\\ \left|z_2\right|=1\\ \left|z_1 \right | - \left| z_2 \right |=3\\ max_{z \euro C} \left | f(z) \right |= \frac{1}{3}\\ \end{align}

Multiply the max with the length to get the answer.

(8)
\begin{align} \left | \int_{C} \frac{1}{z^{2}-1}dz \right | \leq \frac{\pi}{3}\\ \end{align}