\[\int\limits_{C}^{}dz/z\]
where c consists of three line segments:
From z = 1 to z = 1-i
From z = 1-i to z = -1-i
From z = -1-i to z = -1

- anonymous

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- jamiebookeater

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- anonymous

At first I though this wasn't going to be so hard, but then I read the sub-instructions >_<
@amistre64 , @TuringTest , any insight on this one guys?
PS: You did post ALL the contextual information for this one this time right @SkykhanFalcon ? ;-) (/me remembers your last super-challenging differentials question)

- anonymous

There's no special initial conditions, restrictions, or anything like that right?

- anonymous

:) yeah i remember, for now there is not :)

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## More answers

- amistre64

this appears to be the same question that was posted earlier

- anonymous

@experimentX dude i could not do that :/

- anonymous

and today is the last day :/

- experimentX

which one??

- anonymous

they are all connected

- experimentX

i think i already did the first one.

- anonymous

you found sqrt2-i(pi)/4

- anonymous

i am not sure it is true or not

- experimentX

what's the answer supposed to be??

- anonymous

final answer should be -(pi)*i

- experimentX

for question no 1??

- anonymous

no they are connected each other book says only an answer and it is -(pi)*i

- experimentX

O ... let's see for 2

- experimentX

|dw:1343076384065:dw|

- experimentX

|dw:1343076480604:dw|

- experimentX

|dw:1343076552517:dw|

- experimentX

|dw:1343076614227:dw|

- experimentX

|dw:1343076688932:dw|
sorry man ... i couldn't get any better

- anonymous

:) okey bro thanks again

- anonymous

@mukushla @richyw @eliassaab any idea ?

- anonymous

Have you had the residue theorem?

- anonymous

yes i guess

- anonymous

but i did not learn well :/

- anonymous

If you know the Residue Theorem you can cut your work from computing three line integrals to one.

- anonymous

Your line integral on the first line is
z = 1 +i y
dz = i dy
\[
\int_0^{-1} \frac{i}{1+i y} \,
dy=\frac{1}{4} (\ln(4)-i
\pi )
\]

- anonymous

Mimic my post above for 2 remaining lines.

- anonymous

hmm then will i continue with 1/4(ln4.... or i always use the i/1+iy

- anonymous

@eliassaab ?

- anonymous

for the second integral
z= x -I
dz = dx
\[\int_1^{-1} \frac{1}{x-i} \,
dx=-\frac{1}{2} (i \pi )
\]

- anonymous

how can i understand the integral part for ex. why the second one become 1/x-i

- anonymous

How much is y on the second line? -1
How much is x on the seconfd line? x from 1 to -1
z= x -i

- anonymous

On the third line
z= -1 + i y
dz= i dy
\[
\int_{-1}^0 \frac{i}{-1+i y}
\, dy
\]

- anonymous

ok i understand i guess thank u :)

- anonymous

\[
\int_{-1}^0 \frac{i}{-1+i y}
\, dy=-\frac{1}{4} i (\pi
-2 i \log (2))
\]

- anonymous

log=ln

- anonymous

thank u so much :)

- anonymous

yw

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