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Cauchy-Riemann equations are fundamental in complex analysis, providing essential conditions for a function of a complex variable to be complex differentiable, or analytic. Named after Augustin-Louis Cauchy and Bernhard Riemann, these equations connect the partial derivatives of the real and imaginary parts of a complex function.
Cauchy-Riemann equations are a set of partial differential equations which provide the necessary conditions for a complex function to have complex differentiability or, equivalently, be holomorphic. Take a function like this where the functions u and v are real-valued. Then, these equations read:
- ∂u/∂x = ∂v/∂y
- ∂u/∂y =-∂v/∂x
Let’s learn how to solve Cauchy-Riemann Equations in detail:
What is Cauchy-Riemann Equation?
Cauchy-Riemann equations are therefore necessary conditions such that a complex function may be differentiable. In other words, the Cauchy-Riemann equations join real and complex analysis into the definition of holomorphic functions. The applications of these equations have gone a long way in fluid dynamics, electromagnetism, and thermodynamics. These have enabled powerful techniques of complex analysis and give insight into the behavior of complex functions by guaranteeing the independence of the direction of a function’s derivative in the complex plane.
Cauchy-Riemann Equation
For any analytic function, f(z) = u(x,y) + iv(x,y), where z = x + iy, the following are the Cauchy-Riemann equations:
- ∂u/∂x = ∂v/∂y
- ∂u/∂y =-∂v/∂x
These given above two equations are used to relate the partial derivatives of the real part (u) and imaginary part (v) of the function.
Assume that these two equations are satisfied, such that their partial derivatives exist and in this case and if they are continuous, then the function is complex differentiable at that point. The equations will then ensure the limit defining the derivative exists and is independent of the direction of approach in the complex plane.
Derivation of Cauchy-Riemann Equations
Let’s us consider a complex function f(z) = u(x,y) + iv(x,y), where z = x + iy, and u and v are real-valued functions.
For a f(z) to be complex differentiable at a point, z₀ say, the following limit must exist and be independent of the direction of approach:
f'(z₀) = limh→0 [f(z₀ + h) – f(z₀)] / h where ‘h’ is a complex number approaching zero.
Now, Let us approach this limit from two different directions:
Along the real axis (h = Δx):
f'(z₀) = limΔx→0 [f(x₀ + Δx, y₀) – f(x₀, y₀)]/Δx
= ∂u/∂x + i∂v/∂x . . .(1)
Along the imaginary axis (h = iΔy):
f'(z₀) = lim[Δy→0] [f(x₀, y₀ + Δy) – f(x₀, y₀)] / (iΔy)
= -∂u/∂y + i∂v/∂y . . .(2)
For the complex differentiability of the function, f(z), the above two expressions, that is equations (1) and (2) must be equal:
∂u/∂x + i∂v/∂x = -∂u/∂y + i∂v/∂y
Now equating the real and imaginary parts:
∂u/∂x = ∂v/∂y . . .(3)
∂v/∂x = -∂u/∂y . . .(4)
Equations (3) and (4) are the Cauchy-Riemann equations, which give the necessary conditions for a complex function to be differentiable at a point. In case they are fulfilled and partial derivatives are continuous in a region, then a function will be said to be complex differentiable or analytic within that region.
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Solved Examples on Cauchy-Riemann Equations
Example 1: Explain the basic verification of Cauchy-Riemann Equations.
Solution:
Consider the complex function f(z) = x² – y² + i(2xy), where z = x + iy.
Step 1: Identify the real and imaginary parts of the function.
u(x, y) = x² – y² (real part)
v(x, y) = 2xy (imaginary part)
Step 2: Calculate the partial derivatives.
∂u/∂x = 2x
∂u/∂y = -2y
∂v/∂x = 2y
∂v/∂y = 2x
Step 3: Check if the Cauchy-Riemann equations are satisfied.
∂u/∂x = ∂v/∂y (2x = 2x)
∂u/∂y = -∂v/∂x (-2y = -2y)
Therefore, the function f(z) = x² – y² + i(2xy) satisfies the Cauchy-Riemann equations and is analytic everywhere.
Example 2: Verify Cauchy-Riemann Equations in polar form.
f(z) = r²(cos 2θ + i sin 2θ), where z = reiθ
Solution:
Consider the function f(z) = r²(cos 2θ + i sin 2θ), where z = reiθ.
Step 1: Express the function in terms of x and y.
x = r cos θ
y = r sin θ
r² = x² + y²
cos 2θ = cos²θ – sin²θ = (x² – y²) / (x² + y²)
sin 2θ = 2 sin θ cos θ = 2xy / (x² + y²)
Therefore, f(z) = (x² – y²) + i(2xy)
Step 2: Identify the real and imaginary parts.
u(x, y) = x² – y²
v(x, y) = 2xy
Step 3: Calculate the partial derivatives.
∂u/∂x = 2x
∂u/∂y = -2y
∂v/∂x = 2y
∂v/∂y = 2x
Step 4: Verify the Cauchy-Riemann equations.
∂u/∂x = ∂v/∂y (2x = 2x)
∂u/∂y = -∂v/∂x (-2y = -2y)
The function satisfies the Cauchy-Riemann equations and is analytic everywhere.
Example 3: Finding the analytic function given partial derivatives
Solution:
Given ∂u/∂x = 2x and ∂v/∂x = -2y, find the analytic function f(z).
Step 1: Use the Cauchy-Riemann equations to find ∂u/∂y and ∂v/∂y.
∂u/∂x = ∂v/∂y, so ∂v/∂y = 2x
∂u/∂y = -∂v/∂x, so ∂u/∂y = 2y
Step 2: Integrate ∂u/∂x with respect to x to find u(x, y).
u(x, y) = ∫ (2x) dx = x² + C(y)
Step 3: Differentiate u(x, y) with respect to y and compare with ∂u/∂y.
∂u/∂y = C'(y) = 2y
Integrate to find C(y): C(y) = y² + C
Step 4: Substitute back into u(x, y).
u(x, y) = x² + y² + C
Step 5: Integrate ∂v/∂y with respect to y to find v(x, y).
v(x, y) = ∫ (2x) dy = 2xy + D(x)
Step 6: Differentiate v(x, y) with respect to x and compare with ∂v/∂x.
∂v/∂x = 2y + D'(x) = -2y
D'(x) = -4y, which is impossible since D is a function of x only.
Therefore, D(x) = 0
Step 7: Write the analytic function f(z).
f(z) = u(x, y) + iv(x, y) = (x² + y² + C) + i(2xy)
f(z) = z̄z + C + i Im(z²), where C is a real constant.
Example 4: Prove that the following function is nowhere analytic or Show that f(z) = |z|² = x² + y² is nowhere analytic.
Solution:
Step 1: Identify the real and imaginary parts.
u(x, y) = x² + y²
v(x, y) = 0
Step 2: Calculate the partial derivatives.
∂u/∂x = 2x
∂u/∂y = 2y
∂v/∂x = 0
∂v/∂y = 0
Step 3: Check the Cauchy-Riemann equations.
∂u/∂x ≠ ∂v/∂y (2x ≠ 0) for all x ≠ 0
∂u/∂y ≠ -∂v/∂x (2y ≠ 0) for all y ≠ 0
The Cauchy-Riemann equations are not satisfied for any point (x, y) except (0, 0). Therefore, f(z) = |z|² is nowhere analytic.
Example 5: Determine where the function f(z) = z² + z̄ is analytic.
Solution:
Step 1: Express the function in terms of x and y.
z = x + iy
z̄ = x – iy
f(z) = (x + iy)² + (x – iy) = (x² – y² + 2ixy) + (x – iy)
f(z) = (x² – y² + x) + i(2xy – y)
Step 2: Identify the real and imaginary parts.
u(x, y) = x² – y² + x
v(x, y) = 2xy – y
Step 3: Calculate the partial derivatives.
∂u/∂x = 2x + 1
∂u/∂y = -2y
∂v/∂x = 2y
∂v/∂y = 2x – 1
Step 4: Check the Cauchy-Riemann equations.
∂u/∂x = ∂v/∂y: 2x + 1 = 2x – 1
∂u/∂y = -∂v/∂x: -2y = -2y
The first equation is satisfied when 2x + 1 = 2x – 1, which simplifies to 2 = 0. This is impossible for any real x.
Therefore, f(z) = z² + z̄ is nowhere analytic.
Example 6: Determine if f(z) = r² sin 2θ + ir² cos 2θ is analytic, where z = reiθ.
Solution:
Step 1: Express the function in Cartesian coordinates.
x = r cos θ, y = r sin θ
r² = x² + y²
sin 2θ = 2 sin θ cos θ = 2xy / (x² + y²)
cos 2θ = cos²θ – sin²θ = (x² – y²) / (x² + y²)
f(z) = 2xy + i(x² – y²)
Step 2: Identify the real and imaginary parts.
u(x, y) = 2xy
v(x, y) = x² – y²
Step 3: Calculate the partial derivatives.
∂u/∂x = 2y
∂u/∂y = 2x
∂v/∂x = 2x
∂v/∂y = -2y
Step 4: Verify the Cauchy-Riemann equations.
∂u/∂x = ∂v/∂y (2y = -2y)
∂u/∂y = -∂v/∂x (2x = -2x)
The Cauchy-Riemann equations are not satisfied for any (x, y), except (0, 0).
Therefore, f(z) = r² sin 2θ + ir² cos 2θ is not analytic anywhere except possibly at z = 0.
Example 7: Find an analytic function f(z) = u + iv such that u = x³ – 3xy².
Solution:
Step 1: Given u(x, y) = x³ – 3xy², calculate its partial derivatives.
∂u/∂x = 3x² – 3y²
∂u/∂y = -6xy
Step 2: Use the Cauchy-Riemann equations to find ∂v/∂x and ∂v/∂y.
∂v/∂y = ∂u/∂x = 3x² – 3y²
∂v/∂x = -∂u/∂y = 6xy
Step 3: Integrate ∂v/∂y with respect to y to find v(x, y).
v(x, y) = ∫ (3x² – 3y²) dy = 3x²y – y³ + C(x)
Step 4: Differentiate v(x, y) with respect to x and compare with ∂v/∂x.
∂v/∂x = 6xy + C'(x) = 6xy
Therefore, C'(x) = 0, so C(x) = C (a constant)
Step 5: Write the analytic function f(z).
f(z) = u + iv = (x³ – 3xy²) + i(3x²y – y³ + C)
f(z) = z³ – z̄³ + iC, where C is a real constant.
Example 8: Find the harmonic conjugate of u(x, y) = ex cos y.
Solution:
Step 1: Calculate the partial derivatives of u(x, y).
∂u/∂x = ex cos y
∂u/∂y = -ex sin y
Step 2: Use the Cauchy-Riemann equations to find ∂v/∂x and ∂v/∂y.
∂v/∂y = ∂u/∂x = ex cos y
∂v/∂x = -∂u/∂y = ex sin y
Step 3: Integrate ∂v/∂x with respect to x to find v(x, y).
v(x, y) = ∫ (ex sin y) dx = ex sin y + C(y)
Step 4: Differentiate v(x, y) with respect to y and compare with ∂v/∂y.
∂v/∂y = ex cos y + C'(y) = ex cos y
Therefore, C'(y) = 0, so C(y) = C (a constant)
Step 5: Write the harmonic conjugate v(x, y).
v(x, y) = ex sin y + C, where C is a real constant.
The analytic function f(z) = u + iv = ex cos y + i(ex sin y + C) = ez + iC.
Example 9: Prove that if f(z) = u + iv is an entire function and u² + v² is constant, then f(z) is constant.
Solution:
Step 1: Assume u² + v² = k, where k is a constant.
Step 2: Differentiate u² + v² with respect to x and y.
∂(u² + v²)/∂x = 2u(∂u/∂x) + 2v(∂v/∂x) = 0
∂(u² + v²)/∂y = 2u(∂u/∂y) + 2v(∂v/∂y) = 0
Step 3: Use the Cauchy-Riemann equations.
∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x
Step 4: Substitute the Cauchy-Riemann equations into the equations from step 2.
2u(∂u/∂x) + 2v(-∂u/∂y) = 0
2u(∂v/∂x) + 2v(∂v/∂y) = 0
Step 5: Multiply the first equation by ∂u/∂x and the second by ∂v/∂x, then add them.
2u(∂u/∂x)² + 2v(∂v/∂x)² + 2u(∂v/∂x)² + 2v(∂u/∂x)(∂v/∂x) = 0
2(u² + v²)[(∂u/∂x)² + (∂v/∂x)²] = 0
Step 6: Since u² + v² = k (a non-zero constant), we have:
2k[(∂u/∂x)² + (∂v/∂x)²] = 0
This implies that ∂u/∂x = ∂v/∂x = 0. Using the Cauchy-Riemann equations, we can also conclude that ∂u/∂y = ∂v/∂y = 0.
Therefore, u and v are constant, which means f(z) is constant.
Example 10: Find an analytic function f(z) whose real part is u(x, y) = x3 – 3x2y – 3xy2 + y3.
Solution:
Step 1: Calculate the partial derivatives of u(x, y).
∂u/∂x = 3x2 – 6xy – 3y2
∂u/∂y = -3x2– 6xy + 3y2
Step 2: Use the Cauchy-Riemann equations to find ∂v/∂x and ∂v/∂y.
∂v/∂y = ∂u/∂x = 3x2 – 6xy – 3y2
∂v/∂x = -∂u/∂y = 3x2 + 6xy – 3y2
Step 3: Integrate ∂v/∂x with respect to x to find v(x, y).
v(x, y) = ∫ (3x2 + 6xy – 3y2) dx = x3+ 3x2y – 3xy2+ C(y)
Step 4: Differentiate v(x, y) with respect to y and compare with ∂v/∂y.
∂v/∂y = 3x2 + 6xy – 6xy + C'(y) = 3x2 – 6xy – 3y2
Therefore, C'(y) = -3y2, so C(y) = -y3 + C (a constant)
Step 5: Write the analytic function f(z).
f(z) = u + iv = (x3 – 3x2y – 3xy2 + y3) + i(x3+ 3x2y – 3xy2 – y3+ C)
f(z) = (x + iy)3 + iC = z3+ iC, where C is a real constant.
Practice Questions: Cauchy-Riemann Equations
Q1: Verify if the function f(z) = x³ – 3xy² + i(3x²y – y³) satisfies the Cauchy-Riemann equations.
Q2: Determine where the function f(z) = ex cos y + i(ex sin y) is analytic.
Q3: Find an analytic function f(z) = u + iv such that u = ex cos y.
Q4: Show that the function f(z) = |z| is nowhere analytic.
Q5: Given that ∂u/∂x = 2x – y and ∂v/∂x = x + 2y, find the analytic function f(z).
Q6: Verify if the function f(z) = rn (cos nθ + i sin nθ), where n is a positive integer and z = re(iθ), satisfies the Cauchy-Riemann equations in polar form.
Q7: Find the harmonic conjugate of u(x, y) = x² – y² – 2x.
Q8: Prove that if f(z) = u + iv is an entire function and u² – v² is constant, then f(z) is constant.
Q9: Determine if the function f(z) = z + z̄² is analytic anywhere.
Q10: Find an analytic function f(z) whose imaginary part is v(x, y) = x³ – 3xy².
Q11: Verify if the function f(z) = ln(x² + y²) + i arctan(y/x) satisfies the Cauchy-Riemann equations.
Q12: Given that u(x, y) = x³ – 3xy² + ax, where a is a real constant, find the value of a for which f(z) = u + iv is analytic.
Q13: Determine the regions in the complex plane where f(z) = Re(z²) + i Im(z) is analytic.
Q14: Find an analytic function f(z) = u + iv such that u – v = ex sin y.
Q15: Verify if the function f(z) = (x² – y²)/(x² + y²) + i(2xy)/(x² + y²) satisfies the Cauchy-Riemann equations for z ≠ 0.
FAQs on Cauchy-Riemann Equations
What are the Cauchy-Riemann Equations?
They are a set of partial differential equations that provide necessary conditions for a complex function to be complex differentiable (analytic).
How are the Cauchy-Riemann Equations expressed?
For a complex function f(x+iy) = u(x,y) + iv(x,y), the equations are:
∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x
What is the significance of these equations in complex analysis?
They help determine if a function is analytic, which is crucial for various applications in complex analysis, including contour integration and conformal mapping.
Are the Cauchy-Riemann Equations sufficient for analyticity?
No, they are necessary but not sufficient. The function must also have continuous partial derivatives for it to be analytic.
How do these equations relate to harmonic functions?
If a function satisfies the Cauchy-Riemann Equations, its real and imaginary parts are harmonic functions, satisfying Laplace’s equation.
What is the polar form of the Cauchy-Riemann Equations?
In polar coordinates (r,θ), they are:
(1/r)(∂u/∂θ) = ∂v/∂r and ∂u/∂r = -(1/r)(∂v/∂θ)
How are these equations used in physics?
They appear in various physics applications, including fluid dynamics, electromagnetism, and thermodynamics, often relating to potential theory.
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