Find the value of a function of a complex variable. Functions of a complex variable
Functions of a complex variable.
Differentiation of functions of a complex variable.
This article begins a series of lessons in which I will look at typical tasks, related to the theory of functions of a complex variable. To successfully master the examples, you must have basic knowledge about complex numbers. In order to consolidate and repeat the material, just visit the page. You will also need the skills to find second order partial derivatives. Here they are, these partial derivatives... even now I was a little surprised at how often they occur...
The topic that we are beginning to examine does not present any particular difficulties, and in the functions of a complex variable, in principle, everything is clear and accessible. The main thing is to adhere to the basic rule, which I derived experimentally. Read on!
Concept of a function of a complex variable
First, let's refresh our knowledge about the school function of one variable:
Single variable function is a rule according to which each value of the independent variable (from the domain of definition) corresponds to one and only one value of the function. Naturally, “x” and “y” are real numbers.
In the complex case, the functional dependence is specified similarly:
Single-valued function of a complex variable- this is the rule according to which everyone comprehensive the value of the independent variable (from the domain of definition) corresponds to one and only one comprehensive function value. The theory also considers multi-valued and some other types of functions, but for simplicity I will focus on one definition.
What is the difference between a complex variable function?
The main difference: complex numbers. I'm not being ironic. Such questions often leave people in a stupor; at the end of the article I’ll tell you a funny story. At the lesson Complex numbers for dummies we considered a complex number in the form . Since now the letter “z” has become variable, then we will denote it as follows: , while “x” and “y” can take different valid meanings. Roughly speaking, the function of a complex variable depends on the variables and , which take on “ordinary” values. The following point logically follows from this fact:
The function of a complex variable can be written as:
, where and are two functions of two valid variables.
The function is called real part functions
The function is called imaginary part functions
That is, the function of a complex variable depends on two real functions and . To finally clarify everything, let’s look at practical examples:
Example 1
Solution: The independent variable “zet”, as you remember, is written in the form , therefore:
(1) We substituted .
(2) For the first term, the abbreviated multiplication formula was used. In the term, the parentheses have been opened.
(3) Carefully squared, not forgetting that
(4) Rearrangement of terms: first we rewrite the terms , in which there is no imaginary unit(first group), then the terms where there are (second group). It should be noted that shuffling the terms is not necessary, and this step can be skipped (by actually doing it orally).
(5) For the second group we take it out of brackets.
As a result, our function turned out to be represented in the form
Answer:
– real part of the function.
– imaginary part of the function.
What kind of functions did these turn out to be? The most common functions of two variables from which you can find such popular partial derivatives. Without mercy, we will find it. But a little later.
Briefly, the algorithm for the solved problem can be written as follows: we substitute , into the original function, carry out simplifications and divide all terms into two groups - without an imaginary unit (real part) and with an imaginary unit (imaginary part).
Example 2
Find the real and imaginary part of the function
This is an example for independent decision. Before you rush into battle on the complex plane with your checkers drawn, let me give you the most important advice on this topic:
BE CAREFUL! You need to be careful, of course, everywhere, but in complex numbers you should be more careful than ever! Remember that, carefully open the brackets, do not lose anything. According to my observations, the most common mistake is the loss of a sign. Do not hurry!
Complete solution and the answer at the end of the lesson.
Now the cube. Using the abbreviated multiplication formula, we derive:
.
Formulas are very convenient to use in practice, since they significantly speed up the solution process.
Differentiation of functions of a complex variable.
I have two news: good and bad. I'll start with the good one. For a function of a complex variable, the rules of differentiation and the table of derivatives of elementary functions are valid. Thus, the derivative is taken in exactly the same way as in the case of a function of a real variable.
The bad news is that for many complex variable functions there is no derivative at all, and you have to figure out is it differentiable one function or another. And “figuring out” how your heart feels is associated with additional problems.
Let's consider the function of a complex variable. In order to this function was differentiable necessary and sufficient:
1) So that first-order partial derivatives exist. Forget about these notations right away, since in the theory of functions of a complex variable a different notation is traditionally used: 
.
2) To carry out the so-called Cauchy-Riemann conditions:
Only in this case will the derivative exist!
Example 3
Solution is divided into three successive stages:
1) Let's find the real and imaginary parts of the function. This task was discussed in previous examples, so I’ll write it down without comment:
Since then:
Thus: 
– imaginary part of the function.
Let me touch on one more technical point: in what order write the terms in the real and imaginary parts? Yes, in principle, it doesn’t matter. For example, the real part can be written like this: 
, and the imaginary one – like this: .
2) Let us check the fulfillment of the Cauchy-Riemann conditions. There are two of them.
Let's start by checking the condition. We find partial derivatives:
Thus, the condition is satisfied.
Of course, the good news is that partial derivatives are almost always very simple.
We check the fulfillment of the second condition: 
The result is the same, but with opposite signs, that is, the condition is also fulfilled.
The Cauchy-Riemann conditions are satisfied, therefore the function is differentiable.
3) Let's find the derivative of the function. The derivative is also very simple and is found by normal rules:
The imaginary unit is considered a constant during differentiation.
Answer: 
– real part, 
– imaginary part.
The Cauchy-Riemann conditions are satisfied, .
There are two more ways to find the derivative, they are, of course, used less frequently, but the information will be useful for understanding the second lesson - How to find a function of a complex variable?
The derivative can be found using the formula: 
In this case: 
Thus
We have to solve the inverse problem - in the resulting expression we need to isolate . In order to do this, it is necessary in the terms and outside the brackets:
The reverse action, as many have noticed, is somewhat more difficult to perform; to check, it is always better to take the expression on a draft or orally open the brackets back, making sure that the result is exactly
Mirror formula for finding the derivative: 
In this case: 
, That's why:
Example 4
Determine the real and imaginary parts of a function 
. Check the fulfillment of the Cauchy-Riemann conditions. If the Cauchy-Riemann conditions are met, find the derivative of the function.
Quick Solution and an approximate sample of the final design at the end of the lesson.
Are the Cauchy-Riemann conditions always satisfied? Theoretically, they are not fulfilled more often than they are fulfilled. But in practical examples I don’t remember a case where they were not fulfilled =) Thus, if your partial derivatives “do not converge”, then with a very high probability you can say that you made a mistake somewhere.
Let's complicate our functions:
Example 5
Determine the real and imaginary parts of a function 
. Check the fulfillment of the Cauchy-Riemann conditions. Calculate
Solution: The solution algorithm is completely preserved, but at the end a new point will be added: finding the derivative at a point. For cube required formula already printed:
Let's define the real and imaginary parts of this function:
Attention and attention again!
Since then:
Thus:
– real part of the function;
– imaginary part of the function.
Checking the second condition: 
The result is the same, but with opposite signs, that is, the condition is also fulfilled.
The Cauchy-Riemann conditions are satisfied, therefore the function is differentiable:
Let's calculate the value of the derivative at the required point:
Answer:, , the Cauchy-Riemann conditions are satisfied,
Functions with cubes are common, so here’s an example to reinforce:
Example 6
Determine the real and imaginary parts of a function 
. Check the fulfillment of the Cauchy-Riemann conditions. Calculate.
Solution and example of finishing at the end of the lesson.
In theory comprehensive analysis Other functions of a complex argument are also defined: exponent, sine, cosine, etc. These functions have unusual and even bizarre properties - and this is really interesting! I really want to tell you, but here, as it happens, is not a reference book or textbook, but a solution book, so I will consider the same problem with some common functions.
First about the so-called Euler's formulas:
For anyone valid numbers, the following formulas are valid: 
You can also copy it into your notebook as reference material.
Strictly speaking, there is only one formula, but usually for convenience they also write a special case with a minus in the exponent. The parameter does not have to be a single letter; it can be a complex expression or function, it is only important that they accept only valid meanings. Actually, we will see this right now:
Example 7
Find the derivative.
Solution: The general line of the party remains unshakable - it is necessary to distinguish the real and imaginary parts of the function. I will give a detailed solution and comment on each step below:
Since then:
(1) Substitute “z” instead.
(2) After substitution, you need to select the real and imaginary parts first in the indicator exhibitors. To do this, open the brackets.
(3) We group the imaginary part of the indicator, placing the imaginary unit out of brackets.
(4) We use the school action with degrees.
(5) For the multiplier we use Euler’s formula, and .
(6) Open the brackets, resulting in:
– real part of the function;
– imaginary part of the function.
Further actions are standard, let us check the fulfillment of the Cauchy-Riemann conditions: 
Example 9
Determine the real and imaginary parts of a function 
. Check the fulfillment of the Cauchy-Riemann conditions. So be it, we won’t find the derivative.
Solution: The solution algorithm is very similar to the previous two examples, but there are very important points, That's why First stage I will comment again step by step:
Since then:
1) Substitute “z” instead.
(2) First, we select the real and imaginary parts inside the sinus. For these purposes, we open the brackets.
(3) We use the formula, and 
.
(4) Use parity of hyperbolic cosine: And oddity of hyperbolic sine: . Hyperbolics, although out of this world, are in many ways reminiscent of similar trigonometric functions.
Eventually:
– real part of the function;
– imaginary part of the function.
Attention! The minus sign refers to the imaginary part, and under no circumstances should we lose it! For a clear illustration, the result obtained above can be rewritten as follows:
Let's check the fulfillment of the Cauchy-Riemann conditions: 
The Cauchy-Riemann conditions are satisfied.
Answer:, , the Cauchy-Riemann conditions are satisfied.
Ladies and gentlemen, let’s figure it out on our own:
Example 10
Determine the real and imaginary parts of the function. Check the fulfillment of the Cauchy-Riemann conditions.
I deliberately chose more difficult examples, because everyone seems to be able to cope with something, like shelled peanuts. At the same time, you will train your attention! Nut cracker at the end of the lesson.
Well, in conclusion, I’ll consider one more interesting example, when the complex argument is in the denominator. It’s happened a couple of times in practice, let’s look at something simple. Eh, I'm getting old...
Example 11
Determine the real and imaginary parts of the function. Check the fulfillment of the Cauchy-Riemann conditions.
Solution: Again it is necessary to distinguish the real and imaginary parts of the function.
If , then
The question arises, what to do when “Z” is in the denominator?
Everything is simple - the standard one will help method of multiplying the numerator and denominator by the conjugate expression, it has already been used in the examples of the lesson Complex numbers for dummies. Let's remember school formula. We already have in the denominator, which means the conjugate expression will be . Thus, you need to multiply the numerator and denominator by: