Table of contents | |
Cauchy-Riemann Equations | |
Cauchy-Riemann Equations in Polar Form | |
Cauchy-Riemann Equations and Their Applications | |
Solved Numericals |
In these equations, the variables represent components of a complex-valued function, with both real and imaginary parts. The partial derivatives in these equations must satisfy the Cauchy-Riemann conditions. Holomorphic complex differentiable functions are characterized by these equations, making them valuable for solving partial differential equations. Due to their significance, the Cauchy-Riemann equations are frequently encountered in various competitive exams. The property of a complex function being holomorphic means it is differentiable and related to a specific subset of complex functions.
Summary: The Cauchy-Riemann equations in polar form provide a framework for examining complex functions using polar coordinates, allowing for the analysis of partial differential equations in a more versatile manner. This approach is essential for working with complex numbers and functions in various mathematical contexts.
The polar form represents complex numbers in a rectangular shape when plotted on a graph and is also a conjugate function of harmony.
Q1. The value of k that makes the complex-valued function analytic, where z = x + iy, is _________.
(Answer in integer)
Ans: 1.999 - 2.001
For a function f(z) = u + iv to be analytic, then u and v should obey Cauchy-Riemann equations.
C-R Equations:
Given,
f(Z) = u(x, y) + iv(x, y)
f(Z) = e-kx cos 2y - ie-kx sin 2y
Since f(Z) is analytic, it will satisfy Cauchy-Riemann equation,
Q2. If is analytic, then a, b, c, d, e are ______.
Solution: Given:
Now,
Consider ux = vy & uy = -vx (C-R equations)
⇒ a = 1, 2b = -3e, 2d = 4
& 2b = -12, 4c = e
∴ a = 1, b = - 6, c = 1, d = 2 & e = 4.
Q3. If f(z) = u + iv is an analytic function of z = x + iy and u – v = ex (cosy - siny), then f(z) in terms of z is
Solution:
f(z) = u + iv
⇒ i f(z) = - v + i u
⇒ (1 + i) f(z) = (u - v) + i(u + v)
⇒ F(z) = U + iv, where F(z) = (1 + i) f(z)
U = u – v, V = u + v
Now,
Let F(z) be an analytic function
dV = ex (sin y + cos y) dx + ez(cosy – siny) dy
∴ dV = d[ex(siny + cosy)]
Now,
On integrating
V = ex (siny + cosy) + c1
F(z) = U + iV = ex(cosy - siny) + i ex (siny + cosy) + ic1
F = ex(cosy + isiny) + iex (cosy + isiny) + ic1
F(z) = (1 + i) ex + iy + ic1 = (1 + i)ez + ic1
⇒ (1 + i) F(z) = (1 + i) ez + ic1
∴ f(z) = ez + (1 + i) c
Q4. An analytic function of a complex variable is defined as f(z) = x2- y2 + iψ (x, y), where ψ(x, y) is a real function. The value of the imaginary part of f(Z) at z = (1 + i) is _______ (round off to 2 decimal places).
Ans: 1.99 - 2.01
Solution: f(z) = ϕ + iψ
ϕ = Real part, ψ = Imaginary part
If f(z) is analytic function
Calculation:
Given, f(z) = x2 – y2 + i ψ (x, y)
ψ = 2 xy
Given, z = 1 + i
Comparing it with z = x + iy, we get:
∴ x = 1, y = 1
44 videos|101 docs|58 tests
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1. What are the Cauchy-Riemann equations? |
2. How are the Cauchy-Riemann equations represented in polar form? |
3. What are some applications of the Cauchy-Riemann equations? |
4. How are the Cauchy-Riemann equations relevant to mechanical engineering? |
5. Are there any practical examples where the Cauchy-Riemann equations are used in mechanical engineering? |
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