Contour integrals

  • #1

Homework Statement



calculate the contour integral [itex]\oint_{C} (y^2+ix)dz[/itex] where C consists of the parabolic path z(t)=t[itex]^{2}[/itex]+it for 0≤t≤1 followed by the straight line segment from the point 1+i to the point 0

Homework Equations





The Attempt at a Solution



so the contour is in 2 parts for the first part [itex]\oint_{C} (y^2+ix)dz[/itex] = [itex]\int^{1}_{0} (t^2 + it)(2t+i)dt = \int^{1}_{0} (2t^3 + i3t^2-t)dt = i [/itex]

and second part is integral of the straight line from 1+i to 0. this can be represented by z(t) = t+ti for 1≤t≤0

[itex]\int^{0}_{1} (t+ti)(1+i)dt = \int^{0}_{1} (2ti)dt = -i[/itex]

and when i add them together i get 0. is this right? it asks what can i deduce about the function y[itex]^{2}[/itex]+ix ?
 

Answers and Replies

  • #2
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391
Why do you write the integrals that way? What is the general formula?
 
  • #3
hi voko,
Not sure what you mean but what i did was
step 1 represent curve C as z(t) in terms of t. (in this case it was given)
step 2 represent the function f(z) as f[z(t)] in terms of t
step 3 take the derivative of z(t) in terms of t [itex]\frac{dz}{dt}[/itex]
step 4 sub all the above into [itex]\int^{b}_{a} f[z(t)] \frac{dz}{dt} dt[/itex]
 
  • #4
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391
Step 2 is not done correctly in both cases. You just take z(t) instead of f(z(t)).
 
  • #5
in the question i took z(t) = t[itex]^{2}[/itex] + it and [itex]\frac{dz}{dt} = 2t + i[/itex] ans subed them in is this not right?
 
  • #6
6,054
391
Remember, z = x + iy. f(z) = y2 + ix. What does f(z) look like when z = x + iy = t2 + it?
 
  • #7
dohhhh

I see now it should be f(z) = t[itex]^{2} + it^2[/itex]

thanks a million voko
 
  • #8
just so im doing this right my integral then becomes [itex]\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} (2t^3 + i2t^3+i2t^3-2t^3)dt = \int^{1}_{0} (i4t^3)dt = i [/itex]
 
  • #9
6,054
391
just so im doing this right my integral then becomes [itex]\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} (2t^3 + i2t^3+i2t^3-2t^3)dt = \int^{1}_{0} (i4t^3)dt = i [/itex]

No, this does not look right!

[tex]\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} t^2(1 + i)2t(1 + i)dt = \int^{1}_{0} 2t^3(1 + i)^2dt = (1 + i)^2 \int^{1}_{0} 2t^3 dt \ne \int^{1}_{0} (i4t^3)dt[/tex]

And don't forget about the second part of the path.
 
  • #10
i dont understand they are equal [itex] (1 + i)^2 \int^{1}_{0} 2t^3 dt = \int^{1}_{0} (i4t^3)dt [/itex]
 
  • #11
[itex] (1 + i)^2 = 2i [/itex]
 
  • #12
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391
Indeed :) Somehow I was looking at (1 + i) as if it were (i - i). Sorry about that.

What about the second part?
 
  • #13
i havent done it out yet but when i do its going to be i as its around a simple closed curve so the have to be the same as one integral minus the other =0 am i right in my thinking on this?
 
  • #14
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391
An integral is zero over a closed curve only if the integrand is holomorphic. So do the second part, get the result and see what it means regarding this.
 
  • #15
[itex] \int^{1}_{0} (t^2 + it)(2t+i)dt [/itex]= [itex] \int^{1}_{0} 2t^3 + i3t^2 - t dt [/itex] = [itex]( \frac{1}{2} t^4 + i t^3 - \frac{1}{2} )^{1}_{0} = i[/itex] thus showing that integrand in anaylitic and holomorphic. The integrand doesnt depend on the path?
 
  • #16
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[itex] \int^{1}_{0} (t^2 + it)(2t+i)dt [/itex]

How did you get (2t + i)?
 
  • #18
or should it be [itex] \int^{1}_{0} (t^2 + it)(1+i)dt = \int^{1}_{0} t^2 + it^2 +it -t dt = \frac{1}{3} + i\frac{1}{3} -i\frac{1}{2} - \frac{1}{2} [/itex]
 
  • #19
6,054
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or should it be [itex] \int^{1}_{0} (t^2 + it)(1+i)dt = \int^{1}_{0} t^2 + it^2 +it -t dt = \frac{1}{3} + i\frac{1}{3} -i\frac{1}{2} - \frac{1}{2} [/itex]

dz in the original integral integral is replaced with z'(t)dt, which gives (1 + i)dt as you have it here correctly. So, what can be said about the holomorphicity of the integrand? Care to check the Cauchy-Riemann equation?
 
  • #20
i also had the first part wrong it should be [itex]\int^{1}_{0} (t^2 + it^2)(2t+i)dt = \int^{1}_{0} (2t^3 + it^2+i2t^3-t^2)dt = \frac{1}{2} + i\frac{1}{3} + i\frac{1}{2} - \frac{1}{3} [/itex]
 
  • #21
so the first part is [itex]\frac{1}{6} + i\frac{1}{6} [/itex] and the second part is [itex]-\frac{1}{6} - i\frac{1}{6} [/itex]
 
  • #22
which shows that the function is not holomorphic or analytic. which can be seen from applying the cauchy reimann equations to [itex] y^2 + i x [/itex]
 
  • #23
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391
For the first part I get[tex]\int_0^1 (t^2 + it^2)(2t + i)dt = (1 + i) \int_0^1 (2t^3 + it^2)dt = (1 + i)(1/2 + i/3) = 1/2 + i/3 + i/2 - 1/3 = 1/6 + 5i/6[/tex] For the second part, [tex]\int_1^0 (t^2 + it)(1 + i)dt = (1 + i) \int_1^0 (t^2 + it)dt = -(1 + i)(1/3 + i/2) = -1/3 - i/2 - i/3 + 1/2 = 1/6 - 5/6i[/tex] Which are different from your results, but still do not add up to zero
 

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