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Let \(f(t)=u(t)+i v(t)\) be a (piecewise) continuous complex-valued function of a real variable \(t\) defined in the interval \(a \leq t \leq b\). Show that if \(F(t)=U(t)+i V(t)\) is a function such that \(d F / d t=f(t)\), then $$\int_{a}^{b} f(t) d t=F(b)-F(a) .$$ This is the fundamental theorem of calculus for complex variables.

Short Answer

Expert verified
Using the properties of complex-valued functions and the Fundamental Theorem of Calculus, we have shown that \(\int_{a}^{b} f(t) dt = F(b) - F(a)\). This is the stated Fundamental Theorem of Calculus for complex variables.

Step by step solution

01

Recognize the forms of f(t) and F(t)

We have \(f(t) = u(t) + iv(t)\) and it's integral \(F(t) = U(t) + iV(t)\) where \(U(t)\) and \(V(t)\) are antiderivatives of \(u(t)\) and \(v(t)\) respectively.
02

Express integral of f(t)

We express the integral of \(f(t)\) from \(a\) to \(b\) as follows, \(\int_{a}^{b} f(t) dt = \int_{a}^{b} (u(t) + iv(t)) dt\). Using the property of integral of the sum is the sum of integrals, it becomes \(\int_{a}^{b} u(t)dt + i \int_{a}^{b} v(t)dt\).
03

Apply Fundamental Theorem of Calculus

We can now apply the Fundamental Theorem of Calculus to each integral separately, which gives us \(U(b) - U(a) + i[V(b) - V(a)]\). It is the same as \(F(b) - F(a)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex-Valued Function
A complex-valued function is a type of function that maps real variables to complex numbers. In simpler terms, it takes a real input and gives you an output that has both a real part and an imaginary part. The imaginary number is usually represented by the letter "i". So, if you have a function like \(f(t) = u(t) + iv(t)\), this means that \(u(t)\) is the real part, and \(v(t)\) is the imaginary part.

Understanding these can be easier if you think of them as two separate functions. One only deals with real numbers while the other handles the imaginary part. When working with complex-valued functions, it is crucial to keep these two components in mind. Using calculus on these functions involves working separately with the real and imaginary parts, as we often integrate or differentiate each component on its own.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a bridge between differentiation and integration. For complex-valued functions, it works similarly to how it does for real-valued functions, but we consider both the real and imaginary parts.
  • In essence, if you have an antiderivative for your function, integrating over a specific interval will give the difference in this antiderivative at the endpoints.
  • In terms of the provided example, if \(F(t) = U(t) + iV(t)\) is an antiderivative of \(f(t) = u(t) + iv(t)\), then using this theorem gives us \[\int_{a}^{b} f(t) dt = F(b) - F(a)\].

This equation essentially simplifies our computation, neatly linking the values of our antiderivative at two points to the integral of the function over that interval.
Piecewise Continuous Function
A piecewise continuous function is a function that may not be continuous in the traditional sense but can be broken down into segments where it is continuous. Picture it as a train of different continuous "pieces."

These functions are important because they allow us to still use calculus over intervals that include points of discontinuity. For complex-valued functions like \(f(t) = u(t) + iv(t)\), even if \(u(t)\) and \(v(t)\) aren't completely smooth or continuous over an interval, they can be split into parts where they are continuous. Each piece can be handled using the basics of calculus, including differentiation and integration, with results like those provided by the fundamental theorem of calculus.
  • For example, even if there is a jump or "break" at a certain point, if the function is continuous in between, we can apply our integrals part by part.
Understanding this concept can help make the analysis of more complicated functions manageable.

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Most popular questions from this chapter

Show that the following functions are entire. (a) \(f(z)=\left\\{\begin{array}{ll}\frac{e^{2 z}-1}{z^{2}}-\frac{2}{z} & \text { for } z \neq 0, \\ 2 & \text { for } z=0 .\end{array}\right.\) (b) \(f(z)=\left\\{\begin{array}{ll}\frac{\sin z}{2} & \text { for } z \neq 0, \\\ 1 & \text { for } z=0 .\end{array}\right.\) (c) \(f(z)=\left\\{\begin{array}{ll}\frac{\cos z}{z^{2}-\pi^{2} / 4} & \text { for } z \neq \pm \pi / 2, \\ -1 / \pi & \text { for } z=\pm \pi / 2 .\end{array}\right.\)

Show that (a) the sum and the product of two entire functions are entire, and (b) the ratio of two entire functions is analytic everywhere except at the zeros of the denominator.

Show that \(\frac{d}{d z}(\ln z)=\frac{1}{z} .\) Hint: Find \(u(x, y)\) and \(v(x, y)\) for \(\ln z\) and differentiate them.

Let \(C\) be the circle \(|z-i|=3\) integrated in the positive sense. Find the value of each of the following integrals. (a) \(\oint_{C} \frac{e^{z}}{z^{2}+\pi^{2}} d z\), (b) \(\oint_{C} \frac{\sinh z}{\left(z^{2}+\pi^{2}\right)^{2}} d z\), (c) \(\oint_{C} \frac{d z}{z^{2}+9}\), (d) \(\oint_{C} \frac{d z}{\left(z^{2}+9\right)^{2}}\), (e) \(\oint_{C} \frac{\cosh z}{\left(z^{2}+\pi^{2}\right)^{3}} d z\) (f) \(\oint_{C} \frac{z^{2}-3 z+4}{z^{2}-4 z+3} d z\).

Prove the following identities. (a) \(\cos ^{-1} z=-i \ln \left(z \pm \sqrt{z^{2}-1}\right)\), (b) \(\sin ^{-1} z=-i \ln \left[i z \pm \sqrt{1-z^{2}}\right]\), (c) \(\tan ^{-1} z=\frac{1}{2 i} \ln \left(\frac{i-z}{i+z}\right)\), (d) \(\quad \cosh ^{-1} z=\ln \left(z \pm \sqrt{z^{2}-1}\right)\), (e) \(\sinh ^{-1} z=\ln \left(z \pm \sqrt{z^{2}+1}\right)\), (f) \(\tanh ^{-1} z=\frac{1}{2} \ln \left(\frac{1+z}{1-z}\right)\).

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