A complex line integral

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Homework Statement:
if ## \gamma (t):= i+3e^{2it } , t \in \left[0,4\pi \right] , then \int_0^{4\pi } \frac {dz} {z} \ ##
Relevant Equations:
complex numbers
if ## \gamma (t):= i+3e^{2it } , t \in \left[0,4\pi \right] , then \int_0^{4\pi} \frac {dz} {z} ##

in order to solve such integral i substitute z with ##\gamma(t)## and i multiply by ##\gamma'(t)##
that is:
##\int_0^{4 \pi} \frac {6e^{2it}}{i+3e^{2it}}dt=\left.log(i+3e^{2it}) \right|_0^{4 \pi}=##

##= log\left (i+3e^{i8 \pi }\right) - log\left (i+3\right)=##

##= log\left (i+3\right)-log\left (i+3\right)##

there must be something wrong but i don't see where i'm making the mistake. Because in such way the result will be zero but shouldn't be so.
 
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  • #2
anuttarasammyak
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I do not see relation between z in the integral and defined ##\gamma##. z=##\gamma##? Why (4##\pi##) in parenthesis ? Why the result should not be zero ?
 
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  • #3
fresh_42
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Homework Statement:: if γ(t):=i+3e(2it),t∈[0,4π],then∫0(4π)dzz
Relevant Equations:: complex numbers

if γ(t):=i+3e(2it),t∈[0,4π],then∫0(4π)dzz

in order to solve such integral i substitute z with γ(t) and i multiply by γ′(t)
that is:
∫0(4π)6ie(2it)i+3e(2it)dt=log(i+3e(2it))|04π=

=log(i+3e(i8π))−log(i+3)=
=log(i+3)−log(i+3)
there must be something wrong but i don't see where i'm making the mistake. Because in such way the result will be zero but shouldn't be so.
I think you should either first eliminate the complex denominator or otherwise integrate properly with ##\log| i +3e^{2it}|## and calculate the absolute value first and then insert the limits.
 
  • #4
PeroK
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Homework Statement:: if ## \gamma (t):= i+3e^\left(2it \right) , t \in \left[0,4\pi \right] , then \int_0^\left(4\pi \right) \frac {dz} {z} \ ##
Relevant Equations:: complex numbers

if ## \gamma (t):= i+3e^\left(2it \right) , t \in \left[0,4\pi \right] , then \int_0^\left(4\pi \right) \frac {dz} {z} \ ##

in order to solve such integral i substitute z with ##\gamma(t)## and i multiply by ##\gamma'(t)##
that is:
##\int_0^\left(4\pi \right) \frac {6ie^\left(2it\right)} { i+3e^\left(2it \right)} dt =\left. log(i+3e^\left(2it\right)) \right|_0^{4 \pi }=##

##= log\left (i+3e^\left(i8 \pi \right)\right) - log\left (i+3\right)=##
##= log\left (i+3\right)-log\left (i+3\right)##
there must be something wrong but i don't see where i'm making the mistake. Because in such way the result will be zero but shouldn't be so.
What about using the Residue Theorem?
 
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  • #5
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I do not see relation between z in the integral and defined ##\gamma##. z=##\gamma##? Why (4##\pi##) in parenthesis ? Why the result should not be zero ?
i am not understanding your reply...
Why (4##\pi##) in parenthesis ?
because i made a typo which i have now corrected.

I do not see relation between z in the integral and defined ##\gamma##. z=##\gamma##?
according to the theorem of complex line Integral :
##\int_{\gamma}f(z)dz= \int_a^b f(\gamma(t))\gamma'(t)dt ##

Why the result should not be zero ?

because the result given end on wolfram is not zero but ##8\pi i##
 
  • #6
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What about using the Residue Theorem?
i did not study it yet, was trying to solve it with the line integral theorem
 
  • #7
PeroK
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i did not study it yet, was trying to solve it with the line integral theorem
The least you could do is recognise you are going round the same circle four times. So, integrate from ##0## to ##\pi## and then multiply by four.

It easier once you know the residue theorem. Which gives ##8\pi i##.
 
  • #8
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The least you could do is recognise you are going round the same circle four times. So, integrate from ##0## to ##\pi## and then multiply by four.

It easier once you know the residue theorem. Which gives ##8\pi i##.
ok i will have a look to residue theorem, but just to understand, have I applied the line integral correctly?
 
  • #9
fresh_42
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ok i will have a look to residue theorem, but just to understand, have I applied the line integral correctly?
Have you read my reply? I doubt that ##\left[\log(f(t))\right]_a^b## equals ## \left[\log|f(t)|\right]_a^b##.
 
  • #10
vela
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Consider this.
$$\log e^{2\pi i} = 2\pi i$$ because ##\log## and ##e^z## are inverse functions, and
$$\log e^{2\pi i} = \log 1 = 0$$ because ##e^{2\pi i} = 1##. Which one is correct?
 
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  • #11
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Have you read my reply? I doubt that ##\left[\log(f(t))\right]_a^b## equals ## \left[\log|f(t)|\right]_a^b##.
it is not the same, but according to the boundaries of integrations the argument is always positive, no?
 
  • #12
fresh_42
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it is not the same, but according to the boundaries of integrations the argument is always positive, no?
What do you mean by positive if you didn't get rid of the complex numbers? As I see it you have three possibilities:
  1. Integrate ##\dfrac{6e^{2it}}{ i + 3e^{2it}}=\dfrac{(6 e^{2it}) \cdot ( -i + 3e^{2it})}{1+9e^{4it}}##.
  2. Apply the correct formula for ##\displaystyle{\int \dfrac{f'(t)}{f(t)}}\,dt = \log\,|f(t)|\,## with the absolute value in the logarithm before inserting the limits.
  3. Residue theorem.
 
  • #13
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Sorry but(not considering the boundaries of integration) isn't this like this?
##\int \frac {6ie^{2it}}{i+3e^{2it}}dt=log|i+3e^{2it}|= log|i+3(cos(2t)+i sin(2t))| ##
 
  • #14
fresh_42
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I haven't done the math, but this looks ok. You only have to compute the (real valued) length at some stage. All three ways should of course result in the same number.
 
  • #15
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I haven't done the math, but this looks ok. You only have to compute the (real valued) length at some stage. All three ways should of course result in the same number.
ok, so if the above is true, then also this holds. Correct?
##\int_0^{4\pi} \frac {6ie^{2it}}{i+3e^{2it}}dt=\left.log|i+3e^{2it}|\right|_0^{4\pi}=\left. log|i+3(cos(2t)+i sin(2t))|\right|_0^{4\pi} ##
 
  • #16
fresh_42
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ok, so if the above is true, then also this holds. Correct?
##\int_0^{4\pi} \frac {6ie^{2it}}{i+3e^{2it}}dt=\left.log|i+3e^{2it}|\right|_0^{4\pi}=\left. log|i+3(cos(2t)+i sin(2t))|\right|_0^{4\pi} ##
I think so. But why don't you calculate the absolute value? This has to be done prior to the evaluation at the end points.
 
  • #17
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because i don't understand what should i do with that absolute value.
 
  • #18
fresh_42
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It is the norm of a complex number: ##|z|=\sqrt{|z|\cdot|\bar{z}|}=\sqrt{\Re(z)^2+\Im(z)^2}##, and the integral of ##\dfrac{y'}{y}## is ##\log|y|##. Hence you cannot avoid the norm.

You can also proceed by possibility #1 where you only have ##c\cdot e^{nit}## terms to integrate.

I would solve the integral in all three ways for practicing, starting from #1 to #3. You can e.g. look up the residue theorem on Wikipedia.
 
  • #19
vela
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Apply the correct formula for ##\displaystyle{\int \dfrac{f'(t)}{f(t)}}\,dt = \log\,|f(t)|\,## with the absolute value in the logarithm before inserting the limits.
I don't think this is correct for complex integration. It'll just lead to the same result that confused the OP in the first place.
 
  • #20
fresh_42
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I don't think this is correct for complex integration. It'll just lead to the same result that confused the OP in the first place.
Yeah, maybe. I haven't checked. But the idea of complex path integrals is a real parameterization, so that possible complex numbers become constants and the integral a real one.
 
  • #21
vela
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ok i will have a look to residue theorem, but just to understand, have I applied the line integral correctly?
Kind of, but not really. How's that for confusing?

You used the idea that ##\int (dz/z) = \log z##, but you have to be careful. You're assuming ##\log z## is an antiderivative of ##1/z##, but under what conditions can you say this? There are ways to work around this difficulty. My question above, which you seem to have completely ignored, was intended to get you to consider what you're doing more carefully.
 
  • #22
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Kind of, but not really. How's that for confusing?

You used the idea that ##\int (dz/z) = \log z##, but you have to be careful. You're assuming ##\log z## is an antiderivative of ##1/z##, but under what conditions can you say this? There are ways to work around this difficulty. My question above, which you seem to have completely ignored, was intended to get you to consider what you're doing more carefully.
i did not ignore it, was just trying to see the connection with the exercise and your statement
 
  • #23
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Consider this.
$$\log e^{2\pi i} = 2\pi i$$ because ##\log## and ##e^z## are inverse functions, and
$$\log e^{2\pi i} = \log 1 = 0$$ because ##e^{2\pi i} = 1##. Which one is correct?
both are correct
 
  • #24
vela
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So we must conclude that ##2\pi i = 0##?

By the way, how do you know the original integral shouldn't evaluate to 0?
 

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