Question

Complex Functions: Cauchy's Integral Formula
compute:
integral of (dz/(z-3)^4 ) dz where c: /z-2/=4

Answer 1

i cant make sense of how you notated it

Answer 2

c is a contour im assuming

Answer 3

is the dz iterated twice by accident?

Answer 4

ill attach a file of the question more clearly

Answer 5

that is clearer, thanx :)

Answer 6

no thank you for taking the time to look at my problem

Answer 7

im reading up on what i might have forgotten,

c: defines a circle such that |z-2|=4

Answer 8

yes thats right, and i know theres another part to it to find the answer using : 2Pi/n! multiplied by f^n (z0)

Answer 9

this might take me a few ...

Answer 10

its okay

Answer 11

i think that n! is related to this right?

Answer 12

yes i know that the n in the problem has to do with the exponent 4 . its something like this :

Answer 13

yes thats right thats part of solving it

Answer 14

is this what |z-2| = 3 represents?

I spose I never really did get a grasp on this ....

Answer 15

= 4 that is

Answer 16

\[\frac{1}{(z-3)^4}\]

the 3 there tells us that theres is a point "3" inside the circle?

Answer 17

i think that part is to make sure that the z was in the circle , being g that -2 is a point in the circle. I just get quite confused with this section in complex/

Answer 18

\[f^{(n)}(\alpha)=\frac{1}{2\pi\ i\ }\int_{C}\frac{f(z)}{(z-\alpha)^{n+1}}\]
is what im looking at

Answer 19

http://ndp.jct.ac.il/tutorials/complex/node36.html

Answer 20

as far as I can tell; f(z) = 1 according to that ...

Answer 21

does:\[\int_C\to\int_{0}^{2\pi}\]
??

Answer 22

yes i tried working on it myself a bit just just now.(z0)= -3, f(z)=1, f(z0) = -3 not quite sure if this is correct.

Answer 23

\[\frac{1}{(z-3)^4}=\frac{a}{(z-3)}+\frac{b}{(z-3)^2}+\frac{c}{(z-3)^3}+\frac{d}{(z-3)^4}\]

\[1=a(z-3)^3+b(z-3)^2+c(z-3)+d\]
d=1
\[1=a(z-3)^3+b(z-3)^2+c(z-3)+1\]

\[0=a(z^3-9z^2+27z-27)+b(z^2-6z+9)+c(z-3)\]

\[0=(a)z^3+(-9a+b)z^2+(27a-6b+c)z -(27a+9b-3c)\]
a=0; b=0;

\[0=(c)z -(27a+9b-3c)\]

c=0

sooo, this seems to be alot of working just to get :)

\[2\pi\ i\left(\frac{d^3}{dz^3}(1)_{z=3}\right) \]

Answer 24

i must say it is a lot to grasp. But I thank you anyways for spending a long time on it!! lol

Answer 25

well, i do need the practice ;)

Answer 26

i need lots of it. lol