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\[\int\limits_{C}^{}dz/z\] 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

Mathematics
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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)
There's no special initial conditions, restrictions, or anything like that right?
:) yeah i remember, for now there is not :)

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Other answers:

this appears to be the same question that was posted earlier
@experimentX dude i could not do that :/
and today is the last day :/
which one??
they are all connected
i think i already did the first one.
you found sqrt2-i(pi)/4
i am not sure it is true or not
what's the answer supposed to be??
final answer should be -(pi)*i
for question no 1??
no they are connected each other book says only an answer and it is -(pi)*i
O ... let's see for 2
|dw:1343076384065:dw|
|dw:1343076480604:dw|
|dw:1343076552517:dw|
|dw:1343076614227:dw|
|dw:1343076688932:dw| sorry man ... i couldn't get any better
:) okey bro thanks again
Have you had the residue theorem?
yes i guess
but i did not learn well :/
If you know the Residue Theorem you can cut your work from computing three line integrals to one.
Your line integral on the first line is z = 1 +i y dz = i dy \[ \int_0^{-1} \frac{i}{1+i y} \, dy=\frac{1}{4} (\ln(4)-i \pi ) \]
Mimic my post above for 2 remaining lines.
hmm then will i continue with 1/4(ln4.... or i always use the i/1+iy
for the second integral z= x -I dz = dx \[\int_1^{-1} \frac{1}{x-i} \, dx=-\frac{1}{2} (i \pi ) \]
how can i understand the integral part for ex. why the second one become 1/x-i
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
On the third line z= -1 + i y dz= i dy \[ \int_{-1}^0 \frac{i}{-1+i y} \, dy \]
ok i understand i guess thank u :)
\[ \int_{-1}^0 \frac{i}{-1+i y} \, dy=-\frac{1}{4} i (\pi -2 i \log (2)) \]
log=ln
thank u so much :)
yw

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