a) Let f(z) = z^2 and γ(t) = 1 + it^3, t ∈ [0,1].
i) Write out the contour integral ∫γ f(z)dz as an integral with respect to t. You do not need to evaluate this integral.
ii) Evaluate the integral ∫0,1+i z^2dz
iii) What is the relationship between the integrals in (i) and (ii)? Give reasons for your answer.
b) Use Green's Theorem to evaluate the integral
∮C (5 + 10xy + y^2)dx + (6xy + 5y^2)dy
where C is the rectangle with vertices (0, 0), (0, a), (b,a) and (b, 0) traversed counterclockwise.
c) Consider the contour γR(t) = Re^it, t ∈ [0, pi]. Show that
∫γR dz/(z^2 + 4) -> 0, as R -> infinity.
so, where do you get stuck? Surely your text discusses this topic.
just wanted solutions and maybe working out as my previous year exams papers don't come with solutions. So I cant compare to see if I got it correct or not.
a)
i) To write out the contour integral ∫γ f(z)dz as an integral with respect to t, we can use the parameterization γ(t) = 1 + it^3.
First, we need to express dz in terms of dt. We do this by taking the derivative of γ(t) with respect to t:
dz/dt = d(1 + it^3)/dt = 0 + 3it^2 = 3it^2
Then, we can rewrite the integral in terms of t:
∫γ f(z)dz = ∫γ f(γ(t)) * dz/dt * dt
Since f(z) = z^2 and γ(t) = 1 + it^3, we have:
∫γ f(z)dz = ∫γ (1 + it^3)^2 * 3it^2 dt
ii) To evaluate the integral ∫0,1+i z^2dz, we can directly substitute z = x + iy and dz = dx + idy into the integral.
∫0,1+i z^2dz = ∫0,1+i (x + iy)^2(dx + idy)
Expanding the square and separating the real and imaginary parts, we get:
∫0,1+i (x^2 - y^2 + 2xyi)(dx + idy)
Rearranging the terms, we have:
∫0,1+i [(x^2 - y^2)dx - 2xydy] + i[2xydx + (x^2 - y^2)dy]
iii) The relationship between the integrals in (i) and (ii) is that they represent the same contour integral, but with different parameterizations.
In (i), we have the parameterization γ(t) = 1 + it^3, and in (ii), we have the direct parameterization of the contour in terms of x and y coordinates. The integral in (ii) can be seen as a special case of the integral in (i) where we substitute x = Re(t) and y = Im(t). The reasons for this relationship are rooted in the nature of complex analysis and parameterizations of curves.
b) To evaluate the integral ∮C (5 + 10xy + y^2)dx + (6xy + 5y^2)dy, using Green's Theorem, we can rewrite it as a double integral:
∮C (5 + 10xy + y^2)dx + (6xy + 5y^2)dy = ∫∫D (∂Q/∂x - ∂P/∂y)dA
where P = 5 + 10xy + y^2 and Q = 6xy + 5y^2, and D is the region enclosed by the rectangle C.
To evaluate the integral using Green's Theorem:
1. Calculate ∂Q/∂x and ∂P/∂y.
2. Subtract ∂P/∂y from ∂Q/∂x to get (∂Q/∂x - ∂P/∂y).
3. Set up a double integral over the region D representing the rectangle C, and evaluate (∂Q/∂x - ∂P/∂y)dA.
c) To show that ∫γR dz/(z^2 + 4) -> 0 as R -> infinity, we can use the Estimation Lemma.
First, let's parameterize the contour γR(t) = Re^(it) as z = Re^(it), where t ∈ [0, π].
Then, we can express dz in terms of dt using the chain rule:
dz = d(Re^(it)) = iRe^(it)dt
Substituting this into the integral, we have:
∫γR dz/(z^2 + 4) = ∫[0,π] iRe^(it)/(Re^(it))^2 + 4 dt
Simplifying, we get:
∫[0,π] iR/(R^2e^(2it) + 4) dt
Now, we apply the Estimation Lemma, which states that if |f(t)| <= M for all t in the contour γR and L is the length of the contour, then |∫γR f(t) dt| <= ML.
In our case, we want to show that |∫γR dz/(z^2 + 4)| approaches 0 as R approaches infinity.
The key is to show that |iR/(R^2e^(2it) + 4)| is bounded by some constant M for all t in the interval [0, π], regardless of the value of R. This would satisfy the condition of the Estimation Lemma, allowing us to conclude that the integral tends to 0.
To prove this, we can take the limit as R approaches infinity and observe that cross terms involving e^(2it) will vanish, leaving us with a constant term divided by R, which tends to 0 as R grows. Thus, the integral will tend to 0 as R approaches infinity.