Question

I need help about complex analysis

Answer 1

I might be able to help. Whats up?

Answer 2

Residues I'm assuming?

Answer 3

yes, I have huge problem with that :/ can u write and attach it?

Answer 4

Yeah, let me look at my complex analysis book really fast to make sure I remember how to do this :P

Answer 5

ok, u can take as much time as u need :) thanks :)

Answer 6

Alright let me write it up. I'll probably type it on here after I get it written down since the Latex is easier to type in.

Answer 7

type it wherever u want, just type :D

Answer 8

i can look it up too if you like. memory tells me it is
\[2\pi i \sum_k Res\frac{P}{Q}\] yes?

Answer 9

yes

Answer 10

but I don't know how to find Res, my solution never matches with solution in book :/

Answer 11

refresh my memory. do we use pfd or just find the zeros of the denominator?

Answer 12

what is pfd?

Answer 13

partial fractions

Answer 14

as I know, we have just to find zeros of denominator.

Answer 15

The zeros of the denominator are AWFUL.

Answer 16

and for this integral, it's i and 3i ( because -i and -3i are not in C )

Answer 17

ok i cheated and got \[\frac{\pi}{3\sqrt{7}}\] now lets see if we can do it. find the zeros of the denominator yes?

Answer 18

i, -i, 3i and -3i, but we use just zeros with non-negative imaginary part.

Answer 19

oh no, those are not the zeros

Answer 20

how they aren't?

Answer 21

maybe they are something nice in polar for hold on let me check

Answer 22

oh, sorry, I was looking at some other integral. wait.

Answer 23

I forgot to write 10 before x^2 :)

Answer 24

ok i see the gimmick. these are the imaginary 6th roots of

Answer 25

hold the phone before we go any further, the one you gave me had a denominator of
\[x^4+x^2+9\]

Answer 26

is that the right denominator?

Answer 27

no, it's the second one I sent u

Answer 28

second one says the same thing. maybe you forgot to save before you sent?

Answer 29

grrrr, wait :D

Answer 30

that one :)

Answer 31

lorda mercy. ok answer is
\[\frac{\pi}{12}\] by cheating. now lets see what we can do

Answer 32

factors nicely, get
\[x^2=-1\] or
\[x^2=-9\] zeros are what you said. ok

Answer 33

That 10 makes it 1359135183509232 times easier. lol.

Answer 34

fine now lets see if we can recall how to find the residues...

Answer 35

is it not
\[\frac{f(z)}{g'(z)}\]?

Answer 36

these being simple poles. i think that is what you compute.

Answer 37

I did it the same, but solution in book is \[5\pi/12\]

Answer 38

I got just pi/12

Answer 39

me too :/ I hope we did it good :) maybe it's typing mistake in book ( I hope ) :)

Answer 40

@malevolence. how did you compute the residues? is it what i wrote?

Answer 41

Yeah. Simple poles.

Answer 42

i mean it is
\[\frac{f(i)}{g'(i)}\]

Answer 43

and
\[\frac{f(3i)}{g'(3i)}\] yes

Answer 44

Yeah

Answer 45

ok whew. it has been years. then add, multiply by
\[2\pi\]

Answer 46

ok, guys/girls, u both got \[\pi/12\] as a solution?

Answer 47

i got the sum of the residues as as
\[\frac{-i}{24}\]

Answer 48

and therefore an answer of
\[\frac{\pi}{12}\]

Answer 49

Yup.

Answer 50

i checked in wolfram alpha originally and got the same thing, and malevolence did too yes?

Answer 51

so to hell with the back of the book

Answer 52

Hahahahaha

Answer 53

than it must be correct :)
hehe , I'm happy now because I did it like u 2 :) thank u :)

Answer 54

thanks for refreshing my memory. hadn't had my copy of bak newman out for years

Answer 55

I'm gonna burn this book :D

Answer 56

http://www.wolframalpha.com/input/?i=integral+of+%28x^2-x-2%29%2F%28%28x^2%2B9%29%28x^2%2B1%29%29%2Cx%2C-infinity%2Cinfinity

Answer 57

(can't find my alfors)

Answer 58

Wolfram agrees lol. I would trust that >.> You're welcome :D

Answer 59

do I have to download wolfram to check some other problems? where can I find wolfram?

Answer 60

not that i want to do this again but partial fractions actually makes the computation easier

Answer 61

wolframalpha.com

Answer 62

It does indeed lol.

Answer 63

it is web based and free!

Answer 64

nothing to download, no muss no fuss never do annoying algebra again

Answer 65

lol^

Answer 66

thank u a lot :)
well, I am Gorica ( female ), I'm from Serbia, I study technomathematics. You 2?

Answer 67

James. I'm from Virginia and I'm a physics/mathematics double major. :D

Answer 68

nice to meet you :) as I see, you will help me about termodinamics next year ;)

Answer 69

Haha xP I'll do my best :P

Answer 70

how can I help you? :D do you want ice-cream? :D

Answer 71

I would love some :D

Answer 72

You have it when you come to Serbia :D