SkykhanFalcon
\[\int\limits_{C}^{}dz/z\]
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
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agentx5
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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)
agentx5
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There's no special initial conditions, restrictions, or anything like that right?
SkykhanFalcon
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:) yeah i remember, for now there is not :)
amistre64
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this appears to be the same question that was posted earlier
SkykhanFalcon
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@experimentX dude i could not do that :/
SkykhanFalcon
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and today is the last day :/
experimentX
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which one??
SkykhanFalcon
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they are all connected
experimentX
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i think i already did the first one.
SkykhanFalcon
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you found sqrt2-i(pi)/4
SkykhanFalcon
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i am not sure it is true or not
experimentX
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what's the answer supposed to be??
SkykhanFalcon
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final answer should be -(pi)*i
experimentX
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for question no 1??
SkykhanFalcon
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no they are connected each other book says only an answer and it is -(pi)*i
experimentX
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O ... let's see for 2
experimentX
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|dw:1343076384065:dw|
experimentX
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|dw:1343076480604:dw|
experimentX
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|dw:1343076552517:dw|
experimentX
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|dw:1343076614227:dw|
experimentX
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|dw:1343076688932:dw|
sorry man ... i couldn't get any better
SkykhanFalcon
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:) okey bro thanks again
SkykhanFalcon
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@mukushla @richyw @eliassaab any idea ?
eliassaab
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Have you had the residue theorem?
SkykhanFalcon
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yes i guess
SkykhanFalcon
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but i did not learn well :/
eliassaab
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If you know the Residue Theorem you can cut your work from computing three line integrals to one.
eliassaab
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Your line integral on the first line is
z = 1 +i y
dz = i dy
\[
\int_0^{-1} \frac{i}{1+i y} \,
dy=\frac{1}{4} (\ln(4)-i
\pi )
\]
eliassaab
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Mimic my post above for 2 remaining lines.
SkykhanFalcon
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hmm then will i continue with 1/4(ln4.... or i always use the i/1+iy
SkykhanFalcon
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@eliassaab ?
eliassaab
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for the second integral
z= x -I
dz = dx
\[\int_1^{-1} \frac{1}{x-i} \,
dx=-\frac{1}{2} (i \pi )
\]
SkykhanFalcon
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how can i understand the integral part for ex. why the second one become 1/x-i
eliassaab
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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
eliassaab
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On the third line
z= -1 + i y
dz= i dy
\[
\int_{-1}^0 \frac{i}{-1+i y}
\, dy
\]
SkykhanFalcon
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ok i understand i guess thank u :)
eliassaab
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\[
\int_{-1}^0 \frac{i}{-1+i y}
\, dy=-\frac{1}{4} i (\pi
-2 i \log (2))
\]
eliassaab
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log=ln
SkykhanFalcon
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thank u so much :)
eliassaab
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yw