Integrals of Complex Functions

Yes, that is what I was doing. I was trying to split it into real and imaginary parts. I see what you were saying about letting ##c = a + bi##, then breaking it down into real and imaginary parts. I guess I was making a mistake somewhere in the algebra, because I was not able to get the right answer.In summary, the problem asks to prove that for a function f: I → C, where I is an interval of R and f(t)=u1(t)+iv(t), if f is integral over the interval a≤t≤b, then c∫abf(t)=∫abcf(t) for any arbitrary complex constant c. The proof involves passing the factor i through
  • #1
Bashyboy
1,421
5

Homework Statement


Suppose we have the function ##f : I \rightarrow \mathbb{C}##, where ##I## is some interval of ##\mathbb{R}## the functions can be written as ##f(t) = u_1(t) + i v(t)##. Furthermore, suppose this function is integral over the interval ##a \le t \le b##, which can be found by computing

##\int_a^b f(t) = \int_a^b u(t) + i \int_a^b v(t)##

Let ##c## be some arbitrary complex constant. Proof that ## c \int_a^b f(t) = \int_a^b c f(t)##

2. Homework Equations

The Attempt at a Solution



## c \int_a^b f(t) = c \left[ \int_a^b u(t) + i \int_a^b v(t) \right] ##

## = c \int_a^b u(t) + ci \int_a^b v(t)##

Let ##c = a + bi##,

##c \int_a^b f(t) = (a+bi) \int_a^b u(t) + (a+bi) i \int_a^b v(t)##

##= a \int_a^b u(t) + bi \int_a^b u(t) + ai \int_a^b v(t) - b \int_a^b v(t)##

Here is where I had some trouble. The integrals of ##u(t)## and ##v(t)### are integrals of real-valued functions, and so I know that I can pass real-valued scalars through the integral sign; however, I do not know if I can pass ##i## through. I tried various manipulations, but all were positively unhelpful.
 
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  • #2
Passing i through is what you are asked to prove (for the case c = i), so it becomes a bit circular then, doesn't it ?

On the other hand, would you have a problem "passing i through" for a sum ? And an integral is the limit of a summation.
 
  • #3
BvU said:
Passing i through is what you are asked to prove (for the case c = i), so it becomes a bit circular then, doesn't it ?

Yes, that it what I figured the trouble to be.

BvU said:
On the other hand, would you have a problem "passing i through" for a sum ? And an integral is the limit of a summation.

Hmmm, I am not sure. We have not yet viewed the integral as the limit of a sum yet, nor has the textbook represented it as such. There is obviously something elementary that I am missing.

Here is the textbook that we are using: http://www.jiblm.org/downloads/jiblmjournal/V090515/V090515.pdf

We are presently on chapter four, which begins on page 30. Perhaps you might see some theorem I am suppose to use. I will continue to scour through the text myself.
 
  • #4
Wow, there really isn't anything about riemann sums or anything...
Anyways, there's a couple useful things (ones a definition and the other's a theorem) on the next couple pages. Try using those.
 
  • #5
Since, in the notes, it says "the proof of this theorem is very straightforward" I wouldn't worry about "passing through" the factor i.

You could split in real part and imaginary part to verify.
 

FAQ: Integrals of Complex Functions

What is a complex function?

A complex function is a function that takes in complex numbers as inputs and outputs complex numbers. It can be represented by the formula f(z) = u(x,y) + iv(x,y), where u and v are real-valued functions of the variables x and y, and i is the imaginary unit.

What is an integral of a complex function?

An integral of a complex function is a mathematical concept that represents the area under the curve of a complex function. It is denoted by the symbol ∫ and can be thought of as the sum of infinitely small rectangles under the curve.

How do you calculate integrals of complex functions?

To calculate integrals of complex functions, you can use techniques such as contour integration, substitution, and integration by parts. It is important to also consider the properties of analytic functions and Cauchy's Integral Theorem when solving integrals of complex functions.

What are some applications of integrals of complex functions?

Integrals of complex functions have many applications in physics, engineering, and mathematics. They are used to solve problems related to electric circuits, fluid mechanics, and signal processing. They are also essential in the study of complex analysis and its applications in various fields.

Are there any special types of complex functions that have simpler integrals?

Yes, there are certain special types of complex functions that have simpler integrals. These include functions that are analytic, which means they have a derivative at every point in their domain. Analytic functions have the property that their integrals along any path are equal, making them easier to integrate. Other special types of complex functions with simpler integrals include polynomials, trigonometric functions, and exponential functions.

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