Necessary Conditions for Differentiability
Theorem: Cauchy-Riemann (C-R) Equations
Let \( f(z) = u(x,y) + i v(x,y) \) be differentiable at a point \( \alpha = a + ib \).
Then the partial derivatives of \( u \) and \( v \) must satisfy the following equations at \( (a,b) \):
Then the partial derivatives of \( u \) and \( v \) must satisfy the following equations at \( (a,b) \):
\( u_x = v_y \quad \text{and} \quad u_y = -v_x \)
Derivation
Step 1: Definition of Derivative
If \( f \) is differentiable at \( \alpha \), the limit exists independently of the path:
\[ f'(\alpha) = \lim_{\Delta z \to 0} \frac{f(\alpha + \Delta z) – f(\alpha)}{\Delta z} \]
where \( \Delta z = \Delta x + i \Delta y \).
Using \( f(z) = u + iv \): \[ \frac{\Delta f}{\Delta z} = \frac{[u(a+\Delta x, b+\Delta y) – u(a,b)] + i[v(a+\Delta x, b+\Delta y) – v(a,b)]}{\Delta x + i \Delta y} \]
Using \( f(z) = u + iv \): \[ \frac{\Delta f}{\Delta z} = \frac{[u(a+\Delta x, b+\Delta y) – u(a,b)] + i[v(a+\Delta x, b+\Delta y) – v(a,b)]}{\Delta x + i \Delta y} \]
Path I: Approach along Real Axis (x-axis)
Set \( \Delta y = 0 \), so \( \Delta z = \Delta x \to 0 \).
\[ \begin{align} f'(\alpha) &= \lim_{\Delta x \to 0} \frac{u(a+\Delta x, b) – u(a,b)}{\Delta x} + i \lim_{\Delta x \to 0} \frac{v(a+\Delta x, b) – v(a,b)}{\Delta x} \\ &= u_x(a,b) + i v_x(a,b) \quad \dots(1) \end{align} \]
Set \( \Delta y = 0 \), so \( \Delta z = \Delta x \to 0 \).
\[ \begin{align} f'(\alpha) &= \lim_{\Delta x \to 0} \frac{u(a+\Delta x, b) – u(a,b)}{\Delta x} + i \lim_{\Delta x \to 0} \frac{v(a+\Delta x, b) – v(a,b)}{\Delta x} \\ &= u_x(a,b) + i v_x(a,b) \quad \dots(1) \end{align} \]
Path II: Approach along Imaginary Axis (y-axis)
Set \( \Delta x = 0 \), so \( \Delta z = i \Delta y \to 0 \).
\[ \begin{align} f'(\alpha) &= \lim_{\Delta y \to 0} \frac{u(a, b+\Delta y) – u(a,b)}{i \Delta y} + i \lim_{\Delta y \to 0} \frac{v(a, b+\Delta y) – v(a,b)}{i \Delta y} \\ &= \frac{1}{i} u_y(a,b) + \frac{i}{i} v_y(a,b) \\ &= -i u_y(a,b) + v_y(a,b) \\ &= v_y(a,b) – i u_y(a,b) \quad \dots(2) \end{align} \] (Note: \( 1/i = -i \))
Set \( \Delta x = 0 \), so \( \Delta z = i \Delta y \to 0 \).
\[ \begin{align} f'(\alpha) &= \lim_{\Delta y \to 0} \frac{u(a, b+\Delta y) – u(a,b)}{i \Delta y} + i \lim_{\Delta y \to 0} \frac{v(a, b+\Delta y) – v(a,b)}{i \Delta y} \\ &= \frac{1}{i} u_y(a,b) + \frac{i}{i} v_y(a,b) \\ &= -i u_y(a,b) + v_y(a,b) \\ &= v_y(a,b) – i u_y(a,b) \quad \dots(2) \end{align} \] (Note: \( 1/i = -i \))
Step 3: Equating Real and Imaginary Parts
Since the limit must be unique, equate (1) and (2):
\[ u_x + i v_x = v_y – i u_y \]
Real Parts: \( u_x = v_y \)
Imaginary Parts: \( v_x = -u_y \)
Real Parts: \( u_x = v_y \)
Imaginary Parts: \( v_x = -u_y \)
The Cauchy-Riemann Equations:
\( \boxed{ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} } \)
\( \boxed{ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} } \)
Examples & Applications
Ex 1: Conjugate Function \( f(z) = \bar{z} \)
\( f(z) = x – iy \implies u = x, \ v = -y \).
Derivatives: \[ u_x = 1, \quad v_y = -1 \] Check C-R: \( u_x \neq v_y \) (since \( 1 \neq -1 \)).
Conclusion: C-R equations are never satisfied. \( f(z) = \bar{z} \) is nowhere differentiable.
\( f(z) = x – iy \implies u = x, \ v = -y \).
Derivatives: \[ u_x = 1, \quad v_y = -1 \] Check C-R: \( u_x \neq v_y \) (since \( 1 \neq -1 \)).
Conclusion: C-R equations are never satisfied. \( f(z) = \bar{z} \) is nowhere differentiable.
Ex 2: Where is \( f(z) = x^2 + y^2 + i(2xy) \) differentiable?
Here \( u = x^2 + y^2 \) and \( v = 2xy \).
Step 1: Calculate Derivatives
\( u_x = 2x, \quad u_y = 2y \)
\( v_x = 2y, \quad v_y = 2x \)
Step 2: Check C-R Conditions
1. \( u_x = v_y \implies 2x = 2x \) (Satisfied for all \( x,y \in \mathbb{R}^2 \))
2. \( u_y = -v_x \implies 2y = -(2y) \implies 4y = 0 \implies y = 0 \)
Conclusion: C-R equations are only satisfied on the line \( y = 0 \) (the Real Axis).
Here \( u = x^2 + y^2 \) and \( v = 2xy \).
Step 1: Calculate Derivatives
\( u_x = 2x, \quad u_y = 2y \)
\( v_x = 2y, \quad v_y = 2x \)
Step 2: Check C-R Conditions
1. \( u_x = v_y \implies 2x = 2x \) (Satisfied for all \( x,y \in \mathbb{R}^2 \))
2. \( u_y = -v_x \implies 2y = -(2y) \implies 4y = 0 \implies y = 0 \)
Conclusion: C-R equations are only satisfied on the line \( y = 0 \) (the Real Axis).
[Visual Representation: The x-axis]
\( D = \{ z = x + iy : y = 0 \} \)
Since this set \( D \) (a line) has no interior points, the function is technically nowhere differentiable in the strict sense of a domain, though C-R holds on the axis.
\( D = \{ z = x + iy : y = 0 \} \)
Ex 3: Where is \( f(z) = z^2 \) differentiable?
\( f(z) = (x^2 – y^2) + i(2xy) \).
\( u = x^2 – y^2, \quad v = 2xy \).
Derivatives:
\( u_x = 2x, \ u_y = -2y \)
\( v_x = 2y, \ v_y = 2x \)
Check:
1. \( u_x = v_y \implies 2x = 2x \) (Always True)
2. \( u_y = -v_x \implies -2y = -(2y) \) (Always True)
Conclusion: Satisfied everywhere. \( D = \mathbb{C} \).
\( f(z) = (x^2 – y^2) + i(2xy) \).
\( u = x^2 – y^2, \quad v = 2xy \).
Derivatives:
\( u_x = 2x, \ u_y = -2y \)
\( v_x = 2y, \ v_y = 2x \)
Check:
1. \( u_x = v_y \implies 2x = 2x \) (Always True)
2. \( u_y = -v_x \implies -2y = -(2y) \) (Always True)
Conclusion: Satisfied everywhere. \( D = \mathbb{C} \).
Complex Form of C-R Equations
Derivation
Step 1: Coordinate Transformation
We express Cartesian coordinates \((x,y)\) in terms of complex conjugates \((z, \bar{z})\):
\[ z = x + iy, \quad \bar{z} = x – iy \]
Solving for \(x\) and \(y\):
\[ x = \frac{z + \bar{z}}{2}, \quad y = \frac{z – \bar{z}}{2i} \]
Step 2: Chain Rule
Consider \( w = f(z) = u(x,y) + i v(x,y) \). We apply the chain rule to find \( \frac{\partial f}{\partial \bar{z}} \):
\[ \frac{\partial f}{\partial \bar{z}} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \bar{z}} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \bar{z}} \]
Substituting derivatives of \( x \) and \( y \):
\[ \frac{\partial x}{\partial \bar{z}} = \frac{1}{2}, \quad \frac{\partial y}{\partial \bar{z}} = -\frac{1}{2i} = \frac{i}{2} \]
Thus:
\[ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right) \]
Step 3: Expanding f = u + iv
Substitute \( f = u + iv \) into the equation:
\[
\begin{align}
\frac{\partial f}{\partial \bar{z}} &= \frac{1}{2} \left[ (u_x + i v_x) + i (u_y + i v_y) \right] \\
&= \frac{1}{2} \left[ (u_x – v_y) + i (v_x + u_y) \right]
\end{align}
\]
Step 4: Applying C-R Equations
If \( f \) is differentiable, the standard C-R equations hold:
\[ u_x = v_y \quad \text{and} \quad u_y = -v_x \]
Substituting these into our result:
\[ (u_x – v_y) = 0 \quad \text{and} \quad (v_x + u_y) = 0 \]
\[ \implies \frac{\partial f}{\partial \bar{z}} = \frac{1}{2} (0 + i0) = 0 \]
Theorem:
\( f(z) \) satisfies the Cauchy-Riemann equations if and only if: \[ \boxed{ \frac{\partial f}{\partial \bar{z}} = 0 } \]
(Interpretation: An analytic function must be independent of \( \bar{z} \))
\( f(z) \) satisfies the Cauchy-Riemann equations if and only if: \[ \boxed{ \frac{\partial f}{\partial \bar{z}} = 0 } \]
(Interpretation: An analytic function must be independent of \( \bar{z} \))
Examples & Applications
Ex 1: Simple Conjugate
Let \( f(z) = \bar{z} \).
\[ \frac{\partial f}{\partial \bar{z}} = 1 \neq 0 \]
Conclusion: Nowhere differentiable.
Ex 2: Modulus Squared
Let \( f(z) = |z|^2 = z \bar{z} \).
\[ \frac{\partial f}{\partial \bar{z}} = z \]
For C-R to hold, we set derivative to zero:
\[ z = 0 \]
Conclusion: Differentiable only at the origin \( z=0 \).
Ex 3: Polynomial Roots
Let \( f(z) = z^2 \bar{z} – 5|z|^2 + 6\bar{z} \).
Rewrite using \( |z|^2 = z \bar{z} \): \[ f(z) = \bar{z}(z^2 – 5z + 6) \] Calculate derivative w.r.t \( \bar{z} \): \[ \frac{\partial f}{\partial \bar{z}} = z^2 – 5z + 6 \] Set to zero for differentiability points: \[ z^2 – 5z + 6 = 0 \implies (z-2)(z-3) = 0 \] Conclusion: C-R equations hold only at \( z=2 \) and \( z=3 \).
Rewrite using \( |z|^2 = z \bar{z} \): \[ f(z) = \bar{z}(z^2 – 5z + 6) \] Calculate derivative w.r.t \( \bar{z} \): \[ \frac{\partial f}{\partial \bar{z}} = z^2 – 5z + 6 \] Set to zero for differentiability points: \[ z^2 – 5z + 6 = 0 \implies (z-2)(z-3) = 0 \] Conclusion: C-R equations hold only at \( z=2 \) and \( z=3 \).
Correction: The notes mentioned \( z^3 \) but cited roots 2, 3. The polynomial \( z^2-5z+6 \) fits these roots perfectly.
Ex 4: Hidden Conjugate Form
Let \( f(z) = e^x \cos y – i e^x \sin y \).
Factor out \( e^x \): \[ f(z) = e^x (\cos y – i \sin y) \] Using Euler’s Formula \( e^{-iy} = \cos y – i \sin y \): \[ f(z) = e^x \cdot e^{-iy} = e^{x-iy} = e^{\bar{z}} \] Calculate derivative: \[ \frac{\partial f}{\partial \bar{z}} = e^{\bar{z}} \] Since \( e^w \neq 0 \) for any complex number, the derivative is never zero.
Conclusion: Nowhere differentiable.
Factor out \( e^x \): \[ f(z) = e^x (\cos y – i \sin y) \] Using Euler’s Formula \( e^{-iy} = \cos y – i \sin y \): \[ f(z) = e^x \cdot e^{-iy} = e^{x-iy} = e^{\bar{z}} \] Calculate derivative: \[ \frac{\partial f}{\partial \bar{z}} = e^{\bar{z}} \] Since \( e^w \neq 0 \) for any complex number, the derivative is never zero.
Conclusion: Nowhere differentiable.