The theory of functions of a complex variable belousov thunder. Complex variable functions
Complex variable functions.
Differentiation of functions of a complex variable.
This article opens a series of lessons in which I will consider typical problems related to the theory of functions of a complex variable. To successfully master the examples, you must have a basic knowledge of complex numbers. In order to consolidate and repeat the material, just visit the page. You will also need skills to find partial derivatives of the second order... Here they are, these partial derivatives ... even now I myself was a little surprised how often they occur ...
The topic that we are starting to analyze is not particularly difficult, and in functions of a complex variable, in principle, everything is clear and accessible. The main thing is to adhere to the basic rule, which I have derived empirically. Read on!
Complex variable function concept
First, let's refresh our knowledge of the school function of one variable:
Single variable function Is a rule according to which one and only one function value corresponds to each value of the independent variable (from the domain of definition). Naturally, X and Y are real numbers.
In the complex case, the functional dependence is set in the same way:
Single-valued function of a complex variable- this is the rule according to which everyone an integrated the value of the independent variable (from the domain) corresponds to one and only one complex function value. In theory, multivalued and some other types of functions are also considered, but for simplicity I will focus on one definition.
What is the difference between a complex variable function?
The main difference is that the numbers are complex. I am not being ironic. From such questions they often fall into 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 "game" can take different valid values. Roughly speaking, the function of a complex variable depends on variables and, which take on "normal" 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, consider practical examples:
Example 1
Solution: The independent variable "z", as you remember, is written as, therefore:
(1) The original function was substituted.
(2) The abbreviated multiplication formula was used for the first term. In the term - brackets have been opened.
(3) Carefully squared, not forgetting that
(4) Rearrangement of terms: first, rewrite terms in which there is no imaginary unit(first group), then the terms where they are (second group). It should be noted that it is not necessary to shuffle the terms, and this stage can be skipped (in fact, having performed it orally).
(5) For the second group, we take it out of the parentheses.
As a result, our function turned out to be represented in the form
Answer:
- the real part of the function.
- the imaginary part of the function.
What are these functions? The most ordinary functions of two variables, from which one can find such popular partial derivatives... Without mercy - we will find. But a little later.
Briefly, the algorithm of the solved problem can be written as follows: substitute into the original function, simplify and divide all terms into two groups - without an imaginary unit (real part) and with an imaginary unit (imaginary part).
Example 2
Find real and imaginary part of a function
This is an example for a do-it-yourself solution. Before throwing your checkers into battle on a complex plane, let me give you the most important advice on the topic:
BE CAREFUL! You need to be attentive everywhere, of course, but in complex numbers you should be attentive as never before! Remember, carefully open the brackets, don't lose anything. According to my observations, the most common mistake is the loss of a sign. Do not hurry!
Complete solution and answer at the end of the tutorial.
Now the cube. Using the formula for reduced multiplication, 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 a 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 the same way as in the case of a real variable function.
The bad news is that for many functions of a complex variable, the derivative does not exist at all, and you have to find out, differentiable this or that function. And "finding out" how your heart feels is associated with additional troubles.
Consider a complex variable function. In order for this function to be differentiable, it is necessary and sufficient:
1) For partial derivatives of the first order to exist. Forget about these designations right away, because in the theory of the function of a complex variable, a different notation is traditionally used: 
.
2) To carry out the so-called Cauchy-Riemann conditions:
Only in this case the derivative will exist!
Example 3
Solution decomposes into three consecutive stages:
1) Find the real and imaginary parts of the function. This task was analyzed in the previous examples, so I will write it down without comments:
Since, then:
Thus: 
- the imaginary part of the function.
I will dwell on one more technical point: in what order to write down the terms in the real and imaginary parts? Yes, in principle, no difference. For example, the real part can be written like this: 
, and imaginary - 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 fulfilled.
Undoubtedly, the good news is that partial derivatives are almost always very simple.
We check the fulfillment of the second condition: 
It turned out the same thing, but with opposite signs, that is, the condition is also satisfied.
The Cauchy-Riemann conditions are satisfied; therefore, the function is differentiable.
3) Find the derivative of the function. The derivative is also very simple and is found according to the usual rules:
The imaginary unit is considered constant when differentiating.
Answer: 
- real part, 
Is the imaginary part.
The Cauchy-Riemann conditions are satisfied,.
There are two more ways to find the derivative, they are, of course, used less often, but the information will be useful for understanding the second lesson - How do I find the function of a complex variable?
The derivative can be found by the formula: 
In this case: 
Thus
We have to solve the inverse problem - in the resulting expression, you need to isolate. In order to do this, it is necessary in the terms and put out of the bracket:
The reverse action, as many have noticed, is somewhat more difficult to perform, for verification it is always better to take an expression and on a draft or orally open back the parentheses, making sure that it will turn out exactly
Mirror formula for finding the derivative: 
In this case: 
, therefore:
Example 4
Determine 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.
A short solution and a rough sample of finishing at the end of the tutorial.
Are the Cauchy-Riemann conditions always satisfied? In theory, they are more often not executed than they are. But in practical examples I don’t remember a case that they didn’t work =) Thus, if your partial derivatives “did not agree”, then with a very high probability you can say that you have made a mistake somewhere.
Let's complicate our functions:
Example 5
Determine 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 fad will be added: finding the derivative at the point. For the cube, the required formula has already been deduced:
Let's define the real and imaginary parts of this function:
Attention and attention again!
Since, then:
Thus:
- the real part of the function;
- the imaginary part of the function.
Checking the second condition: 
It turned out the same thing, but with opposite signs, that is, the condition is also satisfied.
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 an example to pinpoint:
Example 6
Determine real and imaginary parts of a function 
... Check the fulfillment of the Cauchy-Riemann conditions. Calculate.
Solution and sample finishing at the end of the lesson.
In the theory of complex 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'd like to tell you very much, but here, it just so happened, not a reference book or a textbook, but a solver, so I will consider the same problem with some common functions.
First, about the so-called Euler's formulas:
For anyone actual number the following formulas are valid: 
You can also rewrite it in a notebook as a reference material.
Strictly speaking, there is only one formula, but usually, for convenience, they also write a special case with a minus in the indicator. The parameter does not have to be a lonely letter, a complex expression, function can act as a function, it is only important that they accept only valid values. Actually, we will see it right now:
Example 7
Find the derivative.
Solution: The party's general line remains unshakable - it is necessary to single out the real and imaginary parts of the function. I will give a detailed solution, and below I will comment out each step:
Since, then:
(1) Substitute for "z".
(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, taking the imaginary unit out of the brackets.
(4) We use school action with degrees.
(5) For the factor, we use the Euler formula, while.
(6) Expand the parentheses, resulting in:
- the real part of the function;
- the imaginary part of the function.
Further actions are standard, check the fulfillment of the Cauchy-Riemann conditions: 
Example 9
Determine real and imaginary parts of a function 
... Check the fulfillment of the Cauchy-Riemann conditions. We will not find the derivative.
Solution: The solution algorithm is very similar to the previous two examples, but there are very important points, so I will again comment out the initial stage step by step:
Since, then:
1) Substitute for "z".
(2) First, select the real and imaginary parts inside sine... For this purpose, we open the brackets.
(3) We use the formula, while 
.
(4) We use parity of hyperbolic cosine: and odd hyperbolic sine:. Hyperbolics, although not of this world, in many ways resemble similar trigonometric functions.
Eventually:
- the real part of the function;
- the imaginary part of the function.
Attention! The minus sign refers to the imaginary part, and we don't lose it in any case! For a visual illustration, the result obtained above can be rewritten as follows:
Let us check the fulfillment of the Cauchy-Riemann conditions: 
The Cauchy-Riemann conditions are fulfilled.
Answer:,, the Cauchy-Riemann conditions are satisfied.
With the cosine, ladies and gentlemen, we 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 picked up examples that are more complicated, because everyone will cope with something like peeled peanuts. At the same time, you will train your attention! Nutcracker at the end of the lesson.
Well, in conclusion, I will consider another interesting example when a complex argument is in the denominator. I've met a couple of times in practice, let's analyze 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 separate the real and imaginary parts of the function.
If, then
The question arises, what to do when “z” is in the denominator?
Everything is ingenuous - the standard will help trick 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... We remember the school formula. We already have it in the denominator, which means it will be a conjugate expression. Thus, you need to multiply the numerator and denominator by: