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Chapter 7: Problem 15

Evaluate the following integrals: a. \(\int_{C} \bar{z} d z\), where \(C\) is the parabola \(y=x^{2}\) from \(z=0\) to \(z=1+i\). b. \(\int_{C} f(z) d z\), where \(f(z)=2 z-\bar{z}\) and \(C\) is the path from \(z=0\) to \(z=2+i\) consisting of two line segments from \(z=0\) to \(z=2\) and then \(z=2\) to \(z=2+i\) c. \(\int_{C} \frac{1}{x^{2}+4} d z\) for \(C\) the positively oriented circle, \(|z|=2\). [Hint: Parametrize the circle as \(z=2 e^{i \theta}\), multiply numerator and denominator by \(e^{-i \theta}\), and put in trigonometric form.]

Short Answer

Expert verified
The short answers for each integral are: a. \( \frac{3}{2} + i\), b. \(i\), c. \(π/2\)

Step by step solution

01

Problem a: Parametrization of the curve

For \(C\) being the parabola \(y= x^{2}\) from \(z=0\) to \(z=1+i\), one must parametrize the path along the real axis from \(0\) to \(1\) and along the imaginary axis from \(0\) to \(1\). This can be represented as \(z(t)=t\) for \(0 ≤ t ≤ 1\), and \(z(t) = 1+i(t-1)\) for \(1 ≤ t ≤ 2\), where \(t\) is a real parameter.
02

Problem a: Evaluation of the Integral

By substituting the parameterization and evaluating the definite integral, the solution can be found: \(\int_{C} \bar{z} d z\) can be divided into two parts. The first part \(\int_{0}^{1} \bar{t}dt = \frac{1}{2}\) and the second part \(\int_{1}^{2} \bar{1 + i(t-1)}dt = \frac{1}{2} + i\). So the integral is \(\frac{3}{2} + i\).
03

Problem b: Parameterization of the path

For \(C\) being the path from \(z=0\) to \(z=2+i\), one can parametrize the path along the real axis from \(0\) to \(2\) as \(z(t) = 2t\) for \(0 ≤ t ≤ 0.5\) and along the imaginary axis from \(2\) to \(2+i\) as \(z(t) = 2+i(4t-2)\) for \(0.5 ≤ t ≤ 1\)
04

Problem b: Evaluation of the Integral

Substitute the parameterization into the given integral and evaluate it. Firstly, \(\int_{0}^{0.5}2\cdot 2t - 2tdt = 0\). Secondly, \(\int_{0.5}^{1}2\cdot(2 + i(4t-2)) - 2-i(4t-2)dt = i\). So, the integral is \(i\).
05

Problem c: Parameterization of the circle

Here, the curve \(C\) is a positively oriented circle with \(|z| = 2\). Using the hint provided, it can be parameterized as \(z(t) = 2e^{it}\) where \(0 ≤ t ≤ 2π\) .
06

Problem c: Evaluation of the Integral

First multiply numerator and denominator by \(e^{-i t}\). This gives \(\frac{e^{-i t}}{4 + (2e^{-it})^{2}}\). After simplification, integral becomes \(\int_{0}^{2π}\frac{1}{4 + 4cos^{2}(t)}dt = π/2\). So the integral is \(π/2\).

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Most popular questions from this chapter

What parametric curve is described by the function $$ \gamma(t)=(t-3)+i(2 t+1) $$ \(0 \leq t \leq 2\) ? [Hint: What would you do if you were instead considering the parametric equations \(x=t-3\) and \(y=2 t+1]\)

Write the following in polar form, \(z=r e^{i \theta}\). a. \(i-1\). b. \(-2 i\). c. \(\sqrt{3}+3 i\).

Write the following in standard form. a. \((4+5 i)(2-3 i)\) b. \((1+i)^{3}\) c. \(\frac{5+3 i}{1-i}\).

Let \(C\) be the positively oriented ellipse \(3 x^{2}+y^{2}=9\). Define $$ F\left(z_{0}\right)=\int_{C} \frac{z^{2}+2 z}{z-z_{0}} d z $$

Let \(f(z)=u+i v\) be differentiable. Consider the vector field given by \(\mathbf{F}=v \mathbf{i}+u \mathbf{j}\). Show that the equations \(\nabla \cdot \mathbf{F}=\mathbf{0}\) and \(\nabla \times \mathbf{F}=\mathbf{0}\) are equivalent to the Cauchy-Riemann Equations. [You will need to recall from multivariable calculus the del operator, \(\left.\nabla=\mathbf{i} \frac{\partial}{\partial x}+\mathrm{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z} \cdot\right]\)
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