Restore the analytic function from the imaginary part. Recovery of an analytic function from its real or imaginary part

Let us consider a complex-valued function of the real variable z(t) differentiable at the point t and some of its neighborhood.

The presence of a non-zero derivative
means the presence of a tangent to the graph of the function with an angle of inclination to the real axis equal to
.

Consider now the complex-valued analytic function of a complex variable
. Let be
, where is a real number. Then
is a complex-valued function of the real variable z(t), differentiable at the point t and some of its neighborhood.

The tangent to the graph of the function, as discussed above, has an angle of inclination to the real axis equal to
.

By the complex function theorem
, that's why

.Consequently,
- argument of the derivative of an analytic function
. has the meaning of the angle of rotation of the tangent to the curve at the point when it is displayed through the function
.

Because
,
, then
-the modulus of the derivative of an analytic function has the meaning of a stretch factor when displayed by a function
. All this is true at the points where derivative is non-zero.

If two curves are displayed by means of an analytic function

, then the angle of inclination of the tangent to each curve changes at the point z by the same angle
, so the angles between the curves are preserved when displayed by the analytic function(at the points where derivative is non-zero).

A mapping that preserves the angles between curves is called conformal. That's why mapping by analytic function(at the points where derivative is non-zero)is conformal.

Example. Linear display
(
), as shown above, is reduced to rotation by an angle
and stretching in once.

Reconstruction of an analytic function from its real or imaginary part.

Let the function
, it is required to determine whether it can be a real part of some analytic function
,

The same problem can be posed with respect to the imaginary part. Let the function
, it is required to determine whether it can be the imaginary part of some analytic function
, and if so, restore this function.

When solving these problems, we first need to check whether such an analytic function exists
.

fair theorem.The real and imaginary parts of an analytic function are harmonic functions (that is, they satisfy the Laplace equation).

Proof. If
- is an analytic function, then the Cauchy-Riemann conditions are satisfied
. Differentiate in a particular way the first equality in x, the second in y and add. Get
, so the function
- harmonic. Differentiate in a particular way the first equality in y, the second in x, and subtract the second from the first equality. Get
, so the function
- harmonic.

Therefore, if the function
or function
are not harmonic, then it is impossible to construct an analytic function.

Let the function
and function
- harmonic functions. Let us show how it is possible to restore an analytic function from the known real part
.

Restore function by
similarly.

1 way.

Comparing both expressions, we determine
. Now.

Comment. When recovering by
the function is restored up to a real constant, not an imaginary one.

2 way.
(as in the first method). If problems arise when integrating the second Cauchy-Riemann condition, then it is possible to differentiate the resulting relation in x and equate to a known function.

. Solving this differential equation, we get
,
+C,.

3 way. In the first two ways, the function is restored as a function of x,y. It is much nicer to get it in the form f(z). The third method uses the formula for the derivative
. Since the function
known, then
defined as a function of (x,y). The function is determined by the formula

.

Example. Function is set
=
. Check if it is possible to recover an analytic function with such a real part. If possible, then restore.

Check for yourself that the given function is harmonic.

Comparing these expressions, we have ,

. Therefore + Сi=
.

.
,

Therefore + Сi =
.

Here C is a complex number.

Let us consider a complex-valued function of the real variable z(t) differentiable at the point t and some of its neighborhood.

The presence of a non-zero derivative
means the presence of a tangent to the graph of the function with an angle of inclination to the real axis equal to
.

Consider now the complex-valued analytic function of a complex variable
. Let be
, where is a real number. Then
is a complex-valued function of the real variable z(t), differentiable at the point t and some of its neighborhood.

The tangent to the graph of the function, as discussed above, has an angle of inclination to the real axis equal to
.

By the complex function theorem
, that's why

.Consequently,
- argument of the derivative of an analytic function
. has the meaning of the angle of rotation of the tangent to the curve at the point when it is displayed through the function
.

Because
,
, then
-the modulus of the derivative of an analytic function has the meaning of a stretch factor when displayed by a function
. All this is true at the points where derivative is non-zero.

If two curves are displayed by means of an analytic function

, then the angle of inclination of the tangent to each curve changes at the point z by the same angle
, so the angles between the curves are preserved when displayed by the analytic function(at the points where derivative is non-zero).

A mapping that preserves the angles between curves is called conformal. That's why mapping by analytic function(at the points where derivative is non-zero)is conformal.

Example. Linear display
(
), as shown above, is reduced to rotation by an angle
and stretching in once.

Reconstruction of an analytic function from its real or imaginary part.

Let the function
, it is required to determine whether it can be a real part of some analytic function
,

The same problem can be posed with respect to the imaginary part. Let the function
, it is required to determine whether it can be the imaginary part of some analytic function
, and if so, restore this function.

When solving these problems, we first need to check whether such an analytic function exists
.

fair theorem.The real and imaginary parts of an analytic function are harmonic functions (that is, they satisfy the Laplace equation).

Proof. If
- is an analytic function, then the Cauchy-Riemann conditions are satisfied
. Differentiate in a particular way the first equality in x, the second in y and add. Get
, so the function
- harmonic. Differentiate in a particular way the first equality in y, the second in x, and subtract the second from the first equality. Get
, so the function
- harmonic.

Therefore, if the function
or function
are not harmonic, then it is impossible to construct an analytic function.

Let the function
and function
- harmonic functions. Let us show how it is possible to restore an analytic function from the known real part
.

Restore function by
similarly.

1 way.

Comparing both expressions, we determine
. Now.

Comment. When recovering by
the function is restored up to a real constant, not an imaginary one.

2 way.
(as in the first method). If problems arise when integrating the second Cauchy-Riemann condition, then it is possible to differentiate the resulting relation in x and equate to a known function.

. Solving this differential equation, we get
,
+C,.

3 way. In the first two ways, the function is restored as a function of x,y. It is much nicer to get it in the form f(z). The third method uses the formula for the derivative
. Since the function
known, then
defined as a function of (x,y). The function is determined by the formula

.

Example. Function is set
=
. Check if it is possible to recover an analytic function with such a real part. If possible, then restore.

Check for yourself that the given function is harmonic.

Comparing these expressions, we have ,

. Therefore + Сi=
.

.
,

Therefore + Сi =
.

Here C is a complex number.

Recovery of an analytic function from its real or imaginary part

Example.Find Analytic Function f(z), if

u(x,y) = Re f(z) = and f(i) = 2.

Solution

1. Find the partial derivatives of the function u(x,y)

2. From the 2nd Cauchy-Riemann condition (1)

Differentiating with respect to y, we get

To find the function j(y), we use the 1st Cauchy–Riemann condition (1). Equating = derivative

we obtain an ordinary differential equation of the 1st order

from which we determine j(y)

j(y) = = – +C.

Thus, we get the function

3. We write the desired function f(z) in the form

We transform the resulting expression to a function of the variable z using the equalities

z = x+iy and = = .

We get

f(z) = = +iC = or

f(z) = +C, where C is an arbitrary complex constant.

4. Find the value of the constant C using the condition f(i) = 2:

We get C = i and

f(z) = – 2iz+i

Answer: f(z) = – 2iz + i.

STEM Plus application

Let the function u= be given (the point between the variables x and y is obligatory).

1. "Remember" u=(highlight and press Alt+Enter)

2. Calculate the derivative ux by highlighting and pressing Alt+=(or by selecting the given expression and using the menu Extra® Function® Find derivative).

Type vy and press Alt+Ins to insert the result of the ux calculation. Get the string

This is the Cauchy–Riemann condition vy= ux.

3. Select and using the menu Extra® Function® Find an antiderivative calculate the antiderivative in y. Type v and press Alt+Ins. We get

4. Open this formula ( Shift+F9) and replace C with f(x). We get

5. Remember f(x)= (thus we make it clear that f(x) is an unknown function).

6. Calculate the derivative of vx and uy, and write the results as an equation vx= uy (2nd Cauchy-Riemann condition).

7. We simplify the resulting equation (terms containing y will be reduced).

We get an ordinary differential equation

Select -2 and use the DERIVE menu to calculate the antiderivative with respect to x.

We insert the result into the expression for v instead of f(x). We get

8. Get the desired function

f(z) = +i (– 2x + C).

9. Find C from the initial condition f(i)=2. To do this, "remember" x=0, y=1 and solve the equation

I (– 2x + C)=2

relative to C (highlight and press Alt+?). The result C=1 is inserted instead of C into the expression for f(z). We get

f(z) = +i (– 2x + 1).

As a result, the desired function is expressed in terms of x and y. 10. To find the expression f(z) in terms of z, “remember” that

Here w denotes the conjugate of z.

If we now select the right side of the equality

f(z) = +i (– 2x + 1)

and press Alt+=, then the expression will be simplified taking into account the fact that

It will so happen that