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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

Show that $$ \int_{C} \frac{d z}{(z-1-i)^{n+1}}=\left\\{\begin{array}{cl} 0, & n \neq 0 \\ 2 \pi i, & n=0 \end{array}\right. $$ for \(C\) the boundary of the square \(0 \leq x \leq 2,0 \leq y \leq 2\) taken counterclockwise. [Hint: Use the fact that contours can be deformed into simpler shapes (like a circle) as long as the integrand is analytic in the region between them. After picking a simpler contour, integrate using parametrization.].

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, \(\left.\nabla=\mathbf{i} \frac{\partial}{\partial x}+\mathrm{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z} \cdot\right]\)

Find all \(z\) such that \(z^{4}=16 i\). Write the solutions in rectangular form, \(z=a+i b\), with no decimal approximation or trig functions.

Find the principal value of \(i^{i}\). Rewrite the base, \(i\), as an exponential first.

Show that \(\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\) using trigonometric identities and the exponential forms of these functions.
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