Contour Integrals: Calculate \oint_{C} (y^2+ix)dz

[gtfitzpatrick]

Homework Statement

calculate the contour integral $\oint_{C} (y^2+ix)dz$ where C consists of the parabolic path z(t)=t$^{2}$+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 $\oint_{C} (y^2+ix)dz$ = $\int^{1}_{0} (t^2 + it)(2t+i)dt = \int^{1}_{0} (2t^3 + i3t^2-t)dt = i$

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

$\int^{0}_{1} (t+ti)(1+i)dt = \int^{0}_{1} (2ti)dt = -i$

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

 

[voko]

Why do you write the integrals that way? What is the general formula?

 

[gtfitzpatrick]

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 $\frac{dz}{dt}$
step 4 sub all the above into $\int^{b}_{a} f[z(t)] \frac{dz}{dt} dt$

 

[voko]

Step 2 is not done correctly in both cases. You just take z(t) instead of f(z(t)).

 

[gtfitzpatrick]

in the question i took z(t) = t$^{2}$ + it and $\frac{dz}{dt} = 2t + i$ ans subed them in is this not right?

 

[voko]

Remember, z = x + iy. f(z) = y² + ix. What does f(z) look like when z = x + iy = t² + it?

 

[gtfitzpatrick]

dohhhh

I see now it should be f(z) = t$^{2} + it^2$

thanks a million voko

 

[gtfitzpatrick]

just so I am doing this right my integral then becomes $\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} (2t^3 + i2t^3+i2t^3-2t^3)dt = \int^{1}_{0} (i4t^3)dt = i$

 

[voko]

just so I am doing this right my integral then becomes $\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} (2t^3 + i2t^3+i2t^3-2t^3)dt = \int^{1}_{0} (i4t^3)dt = i$

No, this does not look right!

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

And don't forget about the second part of the path.

 

[gtfitzpatrick]

i don't understand they are equal $(1 + i)^2 \int^{1}_{0} 2t^3 dt = \int^{1}_{0} (i4t^3)dt$

 

[gtfitzpatrick]

$(1 + i)^2 = 2i$

 

[voko]

Indeed :) Somehow I was looking at (1 + i) as if it were (i - i). Sorry about that.

What about the second part?

 

[gtfitzpatrick]

i haven't 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?

 

[voko]

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.

 

[gtfitzpatrick]

$\int^{1}_{0} (t^2 + it)(2t+i)dt$= $\int^{1}_{0} 2t^3 + i3t^2 - t dt$ = $( \frac{1}{2} t^4 + i t^3 - \frac{1}{2} )^{1}_{0} = i$ thus showing that integrand in anaylitic and holomorphic. The integrand doesn't depend on the path?

 

[voko]

$\int^{1}_{0} (t^2 + it)(2t+i)dt$

How did you get (2t + i)?

 

[gtfitzpatrick]

f'[z(t)]?

 

[gtfitzpatrick]

or should it be $\int^{1}_{0} (t^2 + it)(1+i)dt = \int^{1}_{0} t^2 + it^2 +it -t dt = \frac{1}{3} + i\frac{1}{3} -i\frac{1}{2} - \frac{1}{2}$

 

[voko]

or should it be $\int^{1}_{0} (t^2 + it)(1+i)dt = \int^{1}_{0} t^2 + it^2 +it -t dt = \frac{1}{3} + i\frac{1}{3} -i\frac{1}{2} - \frac{1}{2}$

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?

 

[gtfitzpatrick]

i also had the first part wrong it should be $\int^{1}_{0} (t^2 + it^2)(2t+i)dt = \int^{1}_{0} (2t^3 + it^2+i2t^3-t^2)dt = \frac{1}{2} + i\frac{1}{3} + i\frac{1}{2} - \frac{1}{3}$

 

[gtfitzpatrick]

so the first part is $\frac{1}{6} + i\frac{1}{6}$ and the second part is $-\frac{1}{6} - i\frac{1}{6}$

 

[gtfitzpatrick]

which shows that the function is not holomorphic or analytic. which can be seen from applying the cauchy reimann equations to $y^2 + i x$

 

[voko]

For the first part I get$$\int_0^1 (t^2 + it^2)(2t + i)dt = (1 + i) \int_0^1 (2t^3 + it^2)dt = (1 + i)(1/2 + i/3) = 1/2 + i/3 + i/2 - 1/3 = 1/6 + 5i/6$$ For the second part, $$\int_1^0 (t^2 + it)(1 + i)dt = (1 + i) \int_1^0 (t^2 + it)dt = -(1 + i)(1/3 + i/2) = -1/3 - i/2 - i/3 + 1/2 = 1/6 - 5/6i$$ Which are different from your results, but still do not add up to zero