Functions of a complex variable.
Differentiation of functions of a complex variable.

This article opens a series of lessons in which I will look at typical tasks associated with 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, it is enough to visit the page. You will also need skills to find second order partial derivatives. Here they are, these partial derivatives ... even now I was a little surprised how often they occur ...

The topic that we are starting to analyze is not particularly difficult, 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 is derived by me empirically. Read on!

The concept of a function of a complex variable

First, let's refresh our knowledge about the school function of one variable:

Function of one variable 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 given similarly:

Single-valued function of a complex variable is the rule that everyone integrated the value of the independent variable (from the domain) corresponds to one and only one comprehensive 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 function of a complex variable?

The main difference is that numbers are complex. I'm not being ironic. From such questions they often fall into a stupor, at the end of the article I will tell a cool story. On 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 values. Roughly speaking, the function of a complex variable depends on the variables and , which take "usual" values. From this fact the following point follows logically:

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

Decision: The independent variable "z", as you remember, is written as , therefore:

(1) Substituted into the original function.

(2) For the first term, the reduced multiplication formula was used. In the term, the brackets were opened.

(3) Carefully squared, not forgetting that

(4) Rearrangement of terms: first rewrite terms , in which there is no imaginary unit(first group), then terms, where there is (second group). It should be noted that it is not necessary to shuffle the terms, and this stage can be skipped (actually doing it verbally).

(5) The second group is taken out of brackets.

As a result, our function turned out to be represented in the form

Answer:
is the real part of the function .
is 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: 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 a function

This is a do-it-yourself example. Before you throw yourself into battle on the complex plane with drafts, 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 careful more than ever! Remember that, carefully expand the brackets, don't lose anything. According to my observations, the most common mistake is the loss of sign. Do not hurry!

Complete Solution and the answer at the end of the lesson.

Now cube. Using the abbreviated multiplication formula, we derive:
.

Formulas are very convenient to use in practice, as they greatly 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 exactly the same way as in the case of a function of a real variable.

The bad news is that for many functions of a complex variable, there is no derivative at all, and you have to figure out is differentiable one function or another. And “figuring out” how your heart feels is associated with additional troubles.

Consider a function of a complex variable . For this function to be differentiable, it is necessary and sufficient that:

1) For there to be partial derivatives of the first order. Forget about these notations right away, since in the theory of the function of a complex variable, another version of the notation is traditionally used: .

2) To carry out the so-called Cauchy-Riemann conditions:

Only in this case will the derivative exist!

Example 3

Decision decomposed into three successive stages:

1) Find the real and imaginary parts of the function. This task was analyzed in previous examples, so I will write it down without comment:

Since , then:

Thus:

is the imaginary part of the function .

I'll stop at one more technical point: in what order write terms in real and imaginary parts? Yes, basically it doesn't matter. 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 fulfilled.

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 in differentiation is considered a constant.

Answer: - real part is the imaginary part.
The Cauchy-Riemann conditions are met, .

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 to find the function of a complex variable?

The derivative can be found using the formula:

In this case:

Thus

It is necessary to solve the inverse problem - in the resulting expression, you need to isolate . In order to do this, it is necessary in terms and to take out of brackets:

The reverse action, as many have noticed, is somewhat more difficult to perform, for verification it is always better to take the expression and on the draft or verbally open the brackets back, making sure that it will turn out 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 exemplary sample finishing touches at the end of the lesson.

Are the Cauchy-Riemann conditions always satisfied? Theoretically, they are more often not fulfilled than they are. But in practical examples I don’t remember a case where they were not executed =) Thus, if your partial derivatives “did not converge”, then with a very high probability we 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

Decision: The solution algorithm is completely preserved, but at the end a new fad is added: finding the derivative at a point. For the cube, the required formula has already been derived:

Let's define the real and imaginary parts of this function:

Attention and again attention!

Since , then:


Thus:
is the real part of the function ;
is 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 fulfilled.

The Cauchy-Riemann conditions are satisfied, therefore, the function is differentiable:

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 consolidate:

Example 6

Determine the real and imaginary parts of a function . Check the fulfillment of the Cauchy-Riemann conditions. Calculate .

Decision and sample finishing at the end of the lesson.

In the theory of complex analysis, other functions of a complex argument are also defined: exponential, sine, cosine, etc. These functions have unusual and even bizarre properties - and it's really interesting! I really want to tell you, but here, it just so happened, not a reference book or a textbook, but a solution, so I will consider the same task with some common functions.

First about the so-called Euler formulas:

For anyone valid numbers, the following formulas are valid:

You can also copy it into your notebook as a reference.

Strictly speaking, there is only one formula, but usually, for convenience, they also write special case with a minus indicator. The parameter does not have to be a single letter, it can be a complex expression, a function, it is only important that they take only valid values. Actually, we will see it right now:

Example 7

Find derivative.

Decision: The general line of the party remains unshakable - it is necessary to single out the real and imaginary parts of the function. I will give a detailed solution, and comment on each step below:

Since , then:

(1) Substitute for "z".

(2) After substitution, it is necessary to separate the real and imaginary parts first in exponent exhibitors. To do this, open the brackets.

(3) We group the imaginary part of the indicator, putting the imaginary unit out of brackets.

(4) Use school action with degrees.

(5) For the multiplier, we use the Euler formula , while .

(6) We open the brackets, as a result:

is the real part of the function ;
is the imaginary part of the function .

Further actions are standard, let's 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 will not find the derivative.

Decision: 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) We substitute instead of "z".

(2) First, select the real and imaginary parts inside the sinus. For this purpose, open the brackets.

(3) We use the formula , while .

(4) Use parity of hyperbolic cosine: and hyperbolic sine oddness: . Hyperbolics, although not of this world, but in many ways resemble similar trigonometric functions.

Eventually:
is the real part of the function ;
is the imaginary part of the function .

Attention! The minus sign refers to the imaginary part, and in no case should we lose it! For a visual 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 met.

Answer:, , the Cauchy-Riemann conditions are satisfied.

With cosine, ladies and gentlemen, we understand 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 complicated examples, since everyone can handle something like peeled peanuts. At the same time, train your attention! Nutcracker at the end of the lesson.

Well, in conclusion, I will consider one more interesting example when the complex argument is in the denominator. We met a couple of times in practice, let's analyze something simple. Oh, I'm getting old...

Example 11

Determine the real and imaginary parts of the function. Check the fulfillment of the Cauchy-Riemann conditions.

Decision: 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 simple - the standard 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 the school formula. In the denominator we already have , so the conjugate expression will be . Thus, you need to multiply the numerator and denominator by: