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$0\leq r\leq 2, 0\leq \theta \leq \frac{\pi }{4}$

Assignment 4….1C:

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

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

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

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

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




Assignment 5 Question 1B:
Find the Taylor series and the radius of convergence for the Taylor series.

$\frac{1}{4-2z}$ at z=0

$\frac{1}{1-z}= 1+z+z^2+z^3... 4$

$\frac{1}{4-2z}= \frac{1}{4}(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+...)$


Radius of Convergence: R=1

Letter C, Homework 7
Compute the residue at the poles of each of the following functions.

$\int _{c} \frac{e^z}{z(z-2)^3} \, dz$ where C is the circle |z|=3

Find problem spots for denominator: z=0 and z=2

expand $(z-2)^3$ = $z^4-6z^3+12z^2-8z$

take derivative of denominator: $= \frac{e^z}{4z^3-18z^2+24z-8}$

Res(0)= $\frac{e^0}{4(0)^3-18(0)^2+24(0)-8}$

$\; \; \; \; \; \; \; \; \; \;$=$\frac{-1}{8}$

Res(2)= $\frac{1}{2!} \frac{d^2}{d^2 z}[(z-2)^3 \frac{e^z}{z(x-2)^3}]$
$\; \; \; \; \; \; \;$ =$\frac{1}{2} \frac{d^2}{d^2z}[\frac{e^z}{z}]$

Take derivative of function inside square brackets, then combine with remaining parts:


Plug in value (2) into function:


Multiply $2\pi i[sum \; of \; both \; residues \; found]$

$2\pi i[ \frac{-1}{8}+\frac{-e^2}{8}]$ = $\frac{-e^2+(-1) \pi i}{4}$

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