Question

integral in C; In=1/(z^2+2z+4) with C={|z|=1}

Answer 1

\[I_{n}= \int\limits_{C}^{}1/(z ^{2}+2z+4) with C={|z|=1}\]

Answer 2

\[I _{n}=\int\limits_{C}^{}=1/(z ^{2}+2z+4)z^{n}\]

Answer 3

help me

Answer 4

parametrize \(|z|=1\) as \(z=e^{it}\) \(0\leq t <2\pi\) then
\[\int_0^{2\pi}f(e^{it})z'(t)dt\]

Answer 5

\[\int_0^{2\pi}\frac{ie^{it}}{e^{2it}+2e^{it}+4}\]

Answer 6

I tried
but it equal a integral very difficult

Answer 7

use residue theory

Answer 8

I trying

Answer 9

also...I can't fully tell what you integral is

Answer 10

is it

\[\oint\limits_{|z|=1}\frac{1}{z ^{2}+2z+4}dz\]
?

Answer 11

miss z^n

Answer 12

\[I_n=\oint\limits_{|z|=1}\frac{1}{z^n(z ^{2}+2z+4)}dz\]

Answer 13

ok

Answer 14

hmmm none of the residues are inside the unit circle though...

Answer 15

the only pole you have inside \(|z|=1\) is zero. find the residue there

Answer 16

?

Answer 17

z=0 is a pole of order n

Answer 18

oh i see i there is a \(z^n\) there let me be quiet

Answer 19

ok z=0 is singular point

Answer 20

yeah i am a moron
integral i wrote would be zero because function has no poles inside unit circle

Answer 21

thanks @satellite73 and @Zarkon I trying use residuees theory

Answer 22

@Zarkon you think residues =??

Answer 23

what do you think it is

Answer 24

I think it is [1/(z1-z2)]*[1/z1^n]*[1/z2^n]
with z1,z2 are solution of z^2+2z+4

Answer 25

doesn't your formula depend in the order of z1-z2. that seems weird

Answer 26

I get

\[I_n=2\pi i \times \left\{\begin{matrix}0 & n\equiv 0 (\bmod3) \\ \frac{1}{2^{n+1}} & n\equiv 1(\bmod 3)\\\frac{-1}{2^{n+1}} & n\equiv 2 (\bmod3) \end{matrix}\right.\]

Answer 27

you calculation \[\lim_{z \rightarrow 0}zf(z)\]

Answer 28

I looked at the Laurent expansion

Answer 29

if you take that limit you get 1/4

Answer 30

at least for n=1

Answer 31

example of the Laurent expansion?
we have
\[\int\limits_{C}^{}f(z)=2piiRes[f(z),z=z _{0} with Resf(z)=\lim_{z \rightarrow z _{0}}(z-z _{0})f(z)\]

Answer 32

that is only for a pole of order 1

Answer 33

your pole is of order n

Answer 34

the pole point is 0

Answer 35

yes....and it is of order n