Question

Let C denote the positively oriented boundary of the square whose sides lie along the lines x=(+/-)2 and y=(+/-)2. Evaluate the integral, over C of (e^(-z))/(z-((i*pi)/2))

Answer 1

rewrite it again

Answer 2

what are the lines

Answer 3

ohh its +- 2 , man so stupid

Answer 4

so the singularity is at \[\frac{i \pi}{2}\]

Answer 5

and pi/2 is about 1.7 , so its inside the contour

Answer 6

so all you have to do is sub \[\frac{i \pi }{2}\] into the function on top

Answer 7

so you get -1

Answer 8

and thats the answer

Answer 9

complex analysis for the win

Answer 10

I am told that the answer is pi?

Answer 11

I mean 2pi

Answer 12

ohh yeh , I forgot to multiply by the 2pi i fudge factor

Answer 13

no its actually -2 pi i

Answer 14

ur confusing me :|

Answer 15

so I plug (i*Pi)/2 into e^(-z) and then multiply by 2*pi*i?

Answer 16

yes, thats the cauchy integral formula isnt it

Answer 17

yes, thank you!!!!!

Answer 18

if you have \[\int\limits \frac{f(z)}{(z-a)^{n+1}} = 2 \pi i \frac{ f ^{(n)} (a) } {n!}\]

Answer 19

so you differentite it zero times ( which does nothing ) , and divide by 0! (which is 1 )

Answer 20

how do I mark this as a good answer?

Answer 21

Thats the formula I was looking for. It really is that simple. However, that specific form only works for problems in the form. If you have multiple poles in the denominator it can make the problem different.

Answer 22

I personally prefer the more general questions where you actually have to find the residues

Answer 23

or the ones like (1/sinz) , where you come up with a general odd series and multiply both sides by sinz , compare coeffiecents etc

Answer 24

but they are a bit longer.

Answer 25

most singularities are simple poles though , can get the resides in like 30seconds by L hopitals rule

Answer 26

wait

Answer 27

so stupid lols

Answer 28

well I have been on hollidays for like 3weeks

Answer 29

\[e^{ \frac{-\pi i }{2}} = -i \]

Answer 30

=\[2 \pi i \times -i = 2\pi \]

Answer 31

so many mistakes, not really my day, lucky dont have a test today

Answer 32

The Cauchy Integral Formula is
\[f(s) = (i/(2 \pi i))\int\limits_{C}^{?}f(z) dz/(z - s).\]Here
\[f(z) = e^{-z}, s = \pi i/2,\]
\[f(s) = e^{-\pi i/2} = -i,\]
so
\[-i = \left( 1/(2 \pi i \right)\int\limits_{c}^{?}e^{-z}dz / (z - \pi i / 2),\]
since pi * i/2 is in the interior of C. Hence,\[\int\limits_{C}^{?}e^{-z} dz/(z - \pi i/2) = 2 \pi.\]