Analytic Functions on Domains
Theorem #C1
Let \( f: D \subseteq \mathbb{C} \to \mathbb{C} \) where \( D \) is a domain (i.e., an open and connected set).
Let \( f(z) = u(x,y) + i v(x,y) \) be analytic in \( D \).
Statement: If \( f'(z) = 0, \forall z \in D \implies f \) is constant.
Statement: If \( f'(z) = 0, \forall z \in D \implies f \) is constant.
Proof:
Given that \( f'(z) = 0 \) for all \( z \in D \). Recall the formula for the complex derivative in terms of partial derivatives:
\[ f'(z) = u_x + i v_x \]
Using the Cauchy-Riemann (C-R) equations, we know \( v_x = -u_y \). Substituting this back, we get:
\[ f'(z) = u_x – i u_y \]
Since \( f'(z) = 0 \), both the real and imaginary parts must be zero:
If \( u_y = 0 \) everywhere in \( D \), then \( u \) is independent of \( y \).
\( u_x – i u_y = 0 \implies \begin{cases} u_x = 0 \\ u_y = 0 \end{cases} \)
If \( u_x = 0 \) everywhere in \( D \), then \( u \) is independent of \( x \).If \( u_y = 0 \) everywhere in \( D \), then \( u \) is independent of \( y \).
Since the gradient \( \nabla u = (0,0) \) on a connected set \( D \), it follows that:
\[ u(x,y) = \text{constant} \]
Similarly, applying the C-R equations again (\( u_x = v_y \) and \( u_y = -v_x \)):
\[ v_x = 0 \quad \text{and} \quad v_y = 0 \implies v(x,y) = \text{constant} \]
Conclusion: Since both real part \( u \) and imaginary part \( v \) are constants:
\[ f(z) = u + iv = \text{constant on } D. \quad \blacksquare \]
Important Note: The Importance of Connectedness
Why must \( D \) be a domain (connected)? Consider the following counter-example.
Let \( D = A \cup B \) where \( A \) and \( B \) are disjoint open sets (e.g., two separated disks).
\( A = \{ z : |z| < 1 \} \)
\( B = \{ z : |z – 4| < 1 \} \)
Define a function \( f: D \to \mathbb{C} \) as:
\[
f(z) = \begin{cases}
1 & \text{if } z \in A \\
2 & \text{if } z \in B
\end{cases}
\]
Here, \( f'(z) = 0 \) for all \( z \in D \) (derivative of a constant is 0 locally).
However, the Range of \( f = \{1, 2\} \).
Result: \( f \) is NOT constant on \( D \) because \( D \) is not connected.
Let \( D = A \cup B \) where \( A \) and \( B \) are disjoint open sets (e.g., two separated disks).
\( A = \{ z : |z| < 1 \} \)
\( B = \{ z : |z – 4| < 1 \} \)
A
f(z)=1
B
f(z)=2
However, the Range of \( f = \{1, 2\} \).
Result: \( f \) is NOT constant on \( D \) because \( D \) is not connected.
Mutual Harmonic Conjugates Theorem
Theorem
Let \( f: D \subseteq \mathbb{C} \to \mathbb{C} \) such that \( w = f(z) = u + iv \) and \( D \) is a connected domain.
Statement: If \( u \) and \( v \) are harmonic conjugates of each other, then both \( u \) and \( v \) are constant.
Consequently, \( f(z) \) is constant.
Statement: If \( u \) and \( v \) are harmonic conjugates of each other, then both \( u \) and \( v \) are constant.
Consequently, \( f(z) \) is constant.
Proof:
The condition that “u and v are harmonic conjugates of each other” implies two separate analyticity conditions:
1. \( v \) is the harmonic conjugate of \( u \):
By definition, the function \( f = u + iv \) is analytic.
Therefore, the Cauchy-Riemann equations (Set A) hold: \[ \begin{cases} u_x = v_y \\ u_y = -v_x \end{cases} \quad \dots \text{(A)} \]
By definition, the function \( f = u + iv \) is analytic.
Therefore, the Cauchy-Riemann equations (Set A) hold: \[ \begin{cases} u_x = v_y \\ u_y = -v_x \end{cases} \quad \dots \text{(A)} \]
2. \( u \) is the harmonic conjugate of \( v \):
By definition, the function \( g = v + iu \) is analytic (Real part is \(v\), Imaginary part is \(u\)).
Therefore, the Cauchy-Riemann equations (Set B) hold: \[ \begin{cases} v_x = u_y \\ v_y = -u_x \end{cases} \quad \dots \text{(B)} \]
By definition, the function \( g = v + iu \) is analytic (Real part is \(v\), Imaginary part is \(u\)).
Therefore, the Cauchy-Riemann equations (Set B) hold: \[ \begin{cases} v_x = u_y \\ v_y = -u_x \end{cases} \quad \dots \text{(B)} \]
Since both conditions are true simultaneously, we solve the systems (A) and (B) together.
Solving for \( u_x \):
From (A): \( u_x = v_y \)
From (B): \( v_y = -u_x \)
Substituting second into first: \[ u_x = -u_x \implies 2u_x = 0 \implies \mathbf{u_x = 0} \] (Thus, \( u \) is independent of \( x \))
From (A): \( u_x = v_y \)
From (B): \( v_y = -u_x \)
Substituting second into first: \[ u_x = -u_x \implies 2u_x = 0 \implies \mathbf{u_x = 0} \] (Thus, \( u \) is independent of \( x \))
Solving for \( u_y \):
From (A): \( u_y = -v_x \)
From (B): \( v_x = u_y \)
Substituting second into first: \[ u_y = -(u_y) \implies 2u_y = 0 \implies \mathbf{u_y = 0} \] (Thus, \( u \) is independent of \( y \))
From (A): \( u_y = -v_x \)
From (B): \( v_x = u_y \)
Substituting second into first: \[ u_y = -(u_y) \implies 2u_y = 0 \implies \mathbf{u_y = 0} \] (Thus, \( u \) is independent of \( y \))
Conclusion for \( u \):
Since \( \nabla u = (0,0) \) on a connected domain \( D \), \( u \) must be constant.
Since \( \nabla u = (0,0) \) on a connected domain \( D \), \( u \) must be constant.
Conclusion for \( v \):
Similarly, using the equations \( v_x = u_y = 0 \) and \( v_y = -u_x = 0 \), we find that \( \nabla v = (0,0) \).
Thus, \( v \) is also constant.
Similarly, using the equations \( v_x = u_y = 0 \) and \( v_y = -u_x = 0 \), we find that \( \nabla v = (0,0) \).
Thus, \( v \) is also constant.
Final Result:
Since \( u = C_1 \) and \( v = C_2 \), then: \[ f(z) = C_1 + iC_2 = \text{Constant} \quad \blacksquare \]
Since \( u = C_1 \) and \( v = C_2 \), then: \[ f(z) = C_1 + iC_2 = \text{Constant} \quad \blacksquare \]
Theorem: Conditions for Constancy
Theorem: Let \( f: D \subseteq \mathbb{C} \to \mathbb{C} \) be analytic on a domain \( D \) (open and connected). Let \( f(z) = u(x,y) + i v(x,y) \).
If \( f(z) \) satisfies ANY of the following conditions, then \( f \) is constant on \( D \):
If \( f(z) \) satisfies ANY of the following conditions, then \( f \) is constant on \( D \):
- (i) \( \text{Re}(f) = u \) is constant.
- (ii) \( \text{Im}(f) = v \) is constant.
- (iii) \( |f| \) is constant (Modulus is constant).
- (iv) \( \text{Arg}(f) \) is constant.
- (v) \( au + bv = c \) where \( a,b,c \in \mathbb{R} \).
💡 Memory Aid (from notes):
For \( au + bv \): Au = Gold symbol; BV = Constant (Absolute Truth/Parama Satya ⊙).
For \( au + bv \): Au = Gold symbol; BV = Constant (Absolute Truth/Parama Satya ⊙).
Proofs for (i) & (ii): Real or Imaginary Part Constant
(i) If \( u(x,y) = k \) (constant):
Then partial derivatives \( u_x = 0 \) and \( u_y = 0 \).
Using Cauchy-Riemann (C-R) equations (\( u_x = v_y, u_y = -v_x \)): \[ v_y = 0 \quad \text{and} \quad -v_x = 0 \implies v_x = 0 \] Since \( \nabla u = 0 \) and \( \nabla v = 0 \), both \( u \) and \( v \) are constant.
Thus, \( f = u+iv \) is constant.
Then partial derivatives \( u_x = 0 \) and \( u_y = 0 \).
Using Cauchy-Riemann (C-R) equations (\( u_x = v_y, u_y = -v_x \)): \[ v_y = 0 \quad \text{and} \quad -v_x = 0 \implies v_x = 0 \] Since \( \nabla u = 0 \) and \( \nabla v = 0 \), both \( u \) and \( v \) are constant.
Thus, \( f = u+iv \) is constant.
(ii) If \( v(x,y) = k \) (constant):
Similarly, \( v_x = 0 \) and \( v_y = 0 \). By C-R equations, \( u_y = 0 \) and \( u_x = 0 \).
Thus, \( f \) is constant.
Important Note: If a question states \( f \) is a Real Valued analytic function, then \( f(z) \in \mathbb{R} \implies v(x,y) = 0 \). Since \( v \) is constant (0), by case (ii), \( f \) must be constant.
Similarly, \( v_x = 0 \) and \( v_y = 0 \). By C-R equations, \( u_y = 0 \) and \( u_x = 0 \).
Thus, \( f \) is constant.
Important Note: If a question states \( f \) is a Real Valued analytic function, then \( f(z) \in \mathbb{R} \implies v(x,y) = 0 \). Since \( v \) is constant (0), by case (ii), \( f \) must be constant.
Proof (iii): Modulus \( |f| \) is Constant
Let \( |f(z)| = c \). Then \( |f|^2 = u^2 + v^2 = c^2 \).
If \( c=0 \), then \( u=0, v=0 \implies f=0 \) (constant).
Assume \( c \neq 0 \). Differentiate \( u^2 + v^2 = c^2 \) with respect to \( x \) and \( y \): \[ \begin{align} 2u u_x + 2v v_x &= 0 \implies u u_x + v v_x = 0 \quad \dots(A) \\ 2u u_y + 2v v_y &= 0 \implies u u_y + v v_y = 0 \quad \dots(B) \end{align} \] Using C-R equations (\( v_y = u_x \) and \( v_x = -u_y \)) in (A) and (B): \[ \begin{cases} u u_x – v u_y = 0 \\ v u_x + u u_y = 0 \end{cases} \]
If \( c=0 \), then \( u=0, v=0 \implies f=0 \) (constant).
Assume \( c \neq 0 \). Differentiate \( u^2 + v^2 = c^2 \) with respect to \( x \) and \( y \): \[ \begin{align} 2u u_x + 2v v_x &= 0 \implies u u_x + v v_x = 0 \quad \dots(A) \\ 2u u_y + 2v v_y &= 0 \implies u u_y + v v_y = 0 \quad \dots(B) \end{align} \] Using C-R equations (\( v_y = u_x \) and \( v_x = -u_y \)) in (A) and (B): \[ \begin{cases} u u_x – v u_y = 0 \\ v u_x + u u_y = 0 \end{cases} \]
Writing this system in matrix form \( PX = 0 \):
\[
\begin{pmatrix} u & -v \\ v & u \end{pmatrix}
\begin{pmatrix} u_x \\ u_y \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
Determinant \( |P| = u^2 + v^2 = c^2 \neq 0 \).
Since determinant is non-zero, the only solution is the trivial solution: \[ u_x = 0 \quad \text{and} \quad u_y = 0 \]
Therefore \( u \) is constant \(\implies f\) is constant.
Since determinant is non-zero, the only solution is the trivial solution: \[ u_x = 0 \quad \text{and} \quad u_y = 0 \]
z-plane (Domain)
(x, y coordinates)
Any connected shape D
(x, y coordinates)
Any connected shape D
→
w-plane (Range)
(u, v coordinates)
\( f \) is constant (a single point).
(u, v coordinates)
\(u^2+v^2=c^2\)
If image lies on a circle,\( f \) is constant (a single point).
Proof (iv): Argument \( \text{Arg } f \) is Constant
Let \( \text{Arg } f = c \implies \tan^{-1}\left(\frac{v}{u}\right) = c \).
Then \( \frac{v}{u} = \tan c = \lambda \) (where \( \lambda \in \mathbb{R} \)).
\( v = \lambda u \implies \lambda u – v = 0 \).
Differentiating partially w.r.t \( x \) and \( y \): \[ \lambda u_x – v_x = 0 \quad \text{and} \quad \lambda u_y – v_y = 0 \] Using C-R equations (\( v_x = -u_y, v_y = u_x \)): \[ \begin{cases} \lambda u_x + u_y = 0 \\ -u_x + \lambda u_y = 0 \end{cases} \]
Then \( \frac{v}{u} = \tan c = \lambda \) (where \( \lambda \in \mathbb{R} \)).
\( v = \lambda u \implies \lambda u – v = 0 \).
Differentiating partially w.r.t \( x \) and \( y \): \[ \lambda u_x – v_x = 0 \quad \text{and} \quad \lambda u_y – v_y = 0 \] Using C-R equations (\( v_x = -u_y, v_y = u_x \)): \[ \begin{cases} \lambda u_x + u_y = 0 \\ -u_x + \lambda u_y = 0 \end{cases} \]
Matrix Form:
\[
\begin{pmatrix} \lambda & 1 \\ -1 & \lambda \end{pmatrix}
\begin{pmatrix} u_x \\ u_y \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
Determinant \( |P| = \lambda^2 – (-1) = \lambda^2 + 1 \).
Since \( \lambda \in \mathbb{R} \), \( \lambda^2 + 1 \geq 1 \neq 0 \).
Thus, \( u_x = 0, u_y = 0 \implies u \) is constant \(\implies f \) is constant.
Since \( \lambda \in \mathbb{R} \), \( \lambda^2 + 1 \geq 1 \neq 0 \).
Thus, \( u_x = 0, u_y = 0 \implies u \) is constant \(\implies f \) is constant.
Proof (v): Linear Combination \( au + bv = c \)
Given \( au + bv = c \).
Differentiate w.r.t \( x \) and \( y \):
\[ a u_x + b v_x = 0 \quad \text{and} \quad a u_y + b v_y = 0 \]
Apply C-R equations (\( v_x = -u_y, v_y = u_x \)):
\[
\begin{cases}
a u_x – b u_y = 0 \\
b u_x + a u_y = 0
\end{cases}
\]
\[
\begin{pmatrix} a & -b \\ b & a \end{pmatrix}
\begin{pmatrix} u_x \\ u_y \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \end{pmatrix}
\]
Determinant \( = a^2 + b^2 \). If \( a, b \) not both zero, Det \( \neq 0 \).
Thus \( u_x = 0, u_y = 0 \implies f \) is constant.
Thus \( u_x = 0, u_y = 0 \implies f \) is constant.
Alternative Elegant Proof for (v) (Multiplier Method)
Let \( f(z) = u + iv \) be analytic and \( au + bv = c \).
Consider the constant \( \lambda = a – ib \).
Define a new function \( g(z) = \lambda f(z) \). Since \( \lambda \) is constant and \( f \) is analytic, \( g \) is analytic.
Expand \( g(z) \): \[ g(z) = (a – ib)(u + iv) = (au + bv) + i(av – bu) \] Let \( g(z) = U + iV \). Here, the real part \( U = au + bv \).
Given \( au + bv = c \), we have \( U = c \) (Constant).
Since the real part of analytic function \( g \) is constant (by Case i), \( g(z) \) must be constant.
\[ g(z) = \mu \implies (a-ib)f(z) = \mu \implies f(z) = \frac{\mu}{a-ib} \] Thus, \( f(z) \) is constant.
Let \( f(z) = u + iv \) be analytic and \( au + bv = c \).
Consider the constant \( \lambda = a – ib \).
Define a new function \( g(z) = \lambda f(z) \). Since \( \lambda \) is constant and \( f \) is analytic, \( g \) is analytic.
Expand \( g(z) \): \[ g(z) = (a – ib)(u + iv) = (au + bv) + i(av – bu) \] Let \( g(z) = U + iV \). Here, the real part \( U = au + bv \).
Given \( au + bv = c \), we have \( U = c \) (Constant).
Since the real part of analytic function \( g \) is constant (by Case i), \( g(z) \) must be constant.
\[ g(z) = \mu \implies (a-ib)f(z) = \mu \implies f(z) = \frac{\mu}{a-ib} \] Thus, \( f(z) \) is constant.
General Theorem: Image on a Curve
Theorem
Let \( f: D \subseteq \mathbb{C} \to \mathbb{C} \) be analytic on a domain \( D \) (open & connected), such that \( w = f(z) = u + iv \).
Statement: If the values \( (u, v) \) lie on any curve \( \phi(u, v) = 0 \) in the w-plane, then \( f \) is constant.
Statement: If the values \( (u, v) \) lie on any curve \( \phi(u, v) = 0 \) in the w-plane, then \( f \) is constant.
Examples of such curves:
1. Linear: \( au + bv – c = 0 \) (Line)
2. Circular: \( u^2 + v^2 – c = 0 \) (Circle)
3. Parabolic: \( u^2 – 4v = 0 \) (Parabola)
4. Arbitrary: \( u^4 + v^4 + 2u – 3v + 4 = 0 \)
1. Linear: \( au + bv – c = 0 \) (Line)
2. Circular: \( u^2 + v^2 – c = 0 \) (Circle)
3. Parabolic: \( u^2 – 4v = 0 \) (Parabola)
4. Arbitrary: \( u^4 + v^4 + 2u – 3v + 4 = 0 \)
Proof:
Since \( (u,v) \) lies on the curve \( \phi(u,v) = 0 \), we differentiate this equation with respect to \( x \) and \( y \) using the Chain Rule.
1. Differentiate w.r.t \( x \): \[ \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v} \frac{\partial v}{\partial x} = 0 \] Let \( \phi_u \) and \( \phi_v \) be the partial derivatives of \( \phi \). \[ \phi_u u_x + \phi_v v_x = 0 \quad \dots \text{(1)} \]
1. Differentiate w.r.t \( x \): \[ \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v} \frac{\partial v}{\partial x} = 0 \] Let \( \phi_u \) and \( \phi_v \) be the partial derivatives of \( \phi \). \[ \phi_u u_x + \phi_v v_x = 0 \quad \dots \text{(1)} \]
2. Differentiate w.r.t \( y \):
\[ \phi_u u_y + \phi_v v_y = 0 \quad \dots \text{(2)} \]
Now, we use the Cauchy-Riemann (C-R) Equations to express everything in terms of \( u_x \) and \( u_y \):
\[ v_x = -u_y \quad \text{and} \quad v_y = u_x \]
Substitute these into (1) and (2): \[ \begin{align} \text{(1)} \implies & \phi_u u_x – \phi_v u_y = 0 \\ \text{(2)} \implies & \phi_u u_y + \phi_v u_x = 0 \implies \phi_v u_x + \phi_u u_y = 0 \end{align} \]
Substitute these into (1) and (2): \[ \begin{align} \text{(1)} \implies & \phi_u u_x – \phi_v u_y = 0 \\ \text{(2)} \implies & \phi_u u_y + \phi_v u_x = 0 \implies \phi_v u_x + \phi_u u_y = 0 \end{align} \]
We now have a homogeneous system of linear equations for the unknowns \( u_x \) and \( u_y \):
\[
\begin{pmatrix}
\phi_u & -\phi_v \\
\phi_v & \phi_u
\end{pmatrix}
\begin{pmatrix}
u_x \\
u_y
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0
\end{pmatrix}
\]
This is of the form \( PX = 0 \). For non-trivial solutions, the determinant must be zero.
Calculate Determinant \( |P| \): \[ |P| = (\phi_u)(\phi_u) – (-\phi_v)(\phi_v) = (\phi_u)^2 + (\phi_v)^2 \]
This is of the form \( PX = 0 \). For non-trivial solutions, the determinant must be zero.
Calculate Determinant \( |P| \): \[ |P| = (\phi_u)(\phi_u) – (-\phi_v)(\phi_v) = (\phi_u)^2 + (\phi_v)^2 \]
Since \( \phi(u,v) = 0 \) represents a curve, the gradient vector \( (\phi_u, \phi_v) \) is normal to the curve and cannot be zero everywhere (otherwise there is no curve).
Therefore, \( (\phi_u)^2 + (\phi_v)^2 \neq 0 \).
Since the determinant is non-zero, the system has only the trivial solution: \[ u_x = 0 \quad \text{and} \quad u_y = 0 \]
Therefore, \( (\phi_u)^2 + (\phi_v)^2 \neq 0 \).
Since the determinant is non-zero, the system has only the trivial solution: \[ u_x = 0 \quad \text{and} \quad u_y = 0 \]
Since \( \nabla u = (0,0) \) on a connected domain \( D \), \( u \) is constant.
Consequently, by C-R equations, \( v \) is constant.
Thus, \( f(z) = u + iv \) is constant. \(\blacksquare\)
Consequently, by C-R equations, \( v \) is constant.
Thus, \( f(z) = u + iv \) is constant. \(\blacksquare\)