Complex Differentiability
Definition
Let \( f: D \subset \mathbb{C} \longrightarrow \mathbb{C} \) and let \( \alpha \in D^\circ \) (i.e., \( \alpha \) is an interior point).
We say \( f \) is differentiable at \( \alpha \) if and only if the limit of the difference quotient exists finitely: \[ \lim_{z \to \alpha} \phi(z) = \lim_{z \to \alpha} \frac{f(z) – f(\alpha)}{z – \alpha} \] If this limit exists, it is denoted by \( f'(\alpha) \).
We say \( f \) is differentiable at \( \alpha \) if and only if the limit of the difference quotient exists finitely: \[ \lim_{z \to \alpha} \phi(z) = \lim_{z \to \alpha} \frac{f(z) – f(\alpha)}{z – \alpha} \] If this limit exists, it is denoted by \( f'(\alpha) \).
The Case of the Conjugate Function
Example: Show that \( f(z) = \bar{z} \) is nowhere differentiable in \( \mathbb{C} \).
Proof:
Proof:
Let \( \alpha = a + ib \) be any point in \( \mathbb{C} \).
Form the difference quotient:
\[
\begin{align}
\phi(z) &= \frac{\bar{z} – \bar{\alpha}}{z – \alpha} \\
&= \frac{(x-iy) – (a-ib)}{(x+iy) – (a+ib)} \\
&= \frac{(x-a) – i(y-b)}{(x-a) + i(y-b)}
\end{align}
\]
Path Dependence Test:
Let \( z \to \alpha \) along the line \( y – b = m(x – a) \) (a line with slope \( m \)).
Substitute \( (y-b) \) with \( m(x-a) \): \[ \begin{align} \lim_{z \to \alpha} \phi(z) &= \lim_{x \to a} \frac{(x-a) – i \cdot m(x-a)}{(x-a) + i \cdot m(x-a)} \\ &= \lim_{x \to a} \frac{(x-a)(1 – im)}{(x-a)(1 + im)} \\ &= \frac{1 – im}{1 + im} \end{align} \]
Substitute \( (y-b) \) with \( m(x-a) \): \[ \begin{align} \lim_{z \to \alpha} \phi(z) &= \lim_{x \to a} \frac{(x-a) – i \cdot m(x-a)}{(x-a) + i \cdot m(x-a)} \\ &= \lim_{x \to a} \frac{(x-a)(1 – im)}{(x-a)(1 + im)} \\ &= \frac{1 – im}{1 + im} \end{align} \]
Conclusion:
The limit depends on \( m \) (the slope of the path).
Since the limit is not unique, \( f(z) = \bar{z} \) is not differentiable at \( \alpha \).
Since \( \alpha \) was arbitrary, \( f \) is nowhere differentiable.
Since the limit is not unique, \( f(z) = \bar{z} \) is not differentiable at \( \alpha \).
Since \( \alpha \) was arbitrary, \( f \) is nowhere differentiable.
Crucial Conceptual Gap: Real vs. Complex
For \( f(z) = \bar{z} = x – iy \):
- \( u(x,y) = x \) is differentiable everywhere in \( \mathbb{R}^2 \).
- \( v(x,y) = -y \) is differentiable everywhere in \( \mathbb{R}^2 \).
- \( f \) is continuous everywhere.
HOWEVER: \( f \) is differentiable nowhere in the complex sense.
Lesson: Smoothness of real components \( u, v \) is necessary but NOT sufficient for complex differentiability.
Specific Examples
Ex 1: Modulus Squared \( f(z) = |z|^2 \)
Check differentiability at \( \alpha = 0 \).
Proof:
Check differentiability at \( \alpha = 0 \).
Proof:
Form the difference quotient for \( z \neq 0 \):
\[ \phi(z) = \frac{f(z) – f(0)}{z – 0} = \frac{|z|^2 – 0}{z} \]
Since \( |z|^2 = z \bar{z} \), we substitute:
\[ \phi(z) = \frac{z \bar{z}}{z} \]
Since \( z \neq 0 \) in the limit, we can cancel \( z \):
\[ \phi(z) = \bar{z} \]
Now, evaluate the limit as \( z \to 0 \):
\[ \lim_{z \to 0} \phi(z) = \lim_{z \to 0} \bar{z} \]
To see if this limit exists, consider the magnitude:
\[ |\bar{z} – 0| = |z| \]
As \( z \to 0 \), \( |z| \to 0 \) regardless of the path (angle).
\[ f'(0) = 0 \]
Conclusion: The limit exists and is unique. Thus, \( f(z) = |z|^2 \) is differentiable at \( z=0 \).
Ex 2: Complex Polynomial
Let \( f(z) = z^2 + 2z\bar{z} – 2i\bar{z} + 1 \).
Show \( f \) is differentiable at \( \alpha = i \).
Solution:
Let \( f(z) = z^2 + 2z\bar{z} – 2i\bar{z} + 1 \).
Show \( f \) is differentiable at \( \alpha = i \).
Solution:
First, calculate \( f(i) \):
\[ f(i) = i^2 + 2(i)(\bar{i}) – 2i(\bar{i}) + 1 \]
\[ = -1 + 2(i)(-i) – 2i(-i) + 1 \]
\[ = -1 + 2(1) – 2 + 1 = 0 \]
Form the quotient:
\[ \frac{f(z) – f(i)}{z – i} = \frac{z^2 + 2z\bar{z} – 2i\bar{z} + 1}{z – i} \]
Rearrange terms to factor out \( (z-i) \):
\[ f(z) = z^2 + 1 + 2\bar{z}(z – i) \]
Since \( z^2 + 1 = (z-i)(z+i) \):
\[ \text{Numerator} = (z-i)(z+i) + 2\bar{z}(z-i) \]
\[
\begin{align}
\lim_{z \to i} \frac{(z-i)[ (z+i) + 2\bar{z} ]}{z-i} &= \lim_{z \to i} (z+i + 2\bar{z}) \\
&= (i + i + 2(-i)) \\
&= 2i – 2i = 0
\end{align}
\]
Conclusion: The limit exists and is 0. Thus, \( f \) is differentiable at \( z=i \).
Theorem: Differentiability Implies Continuity
If a function \( f: D \to \mathbb{C} \) is differentiable at a point \( \alpha \in D \), then \( f \) is continuous at \( \alpha \).
(Note: The converse is NOT true. Continuity does not imply differentiability, e.g., \( f(z) = \bar{z} \)).
(Note: The converse is NOT true. Continuity does not imply differentiability, e.g., \( f(z) = \bar{z} \)).
Proof:
To prove that \( f \) is continuous at \( \alpha \), we must show that:
\[ \lim_{z \to \alpha} f(z) = f(\alpha) \]
This is equivalent to showing that \( \lim_{z \to \alpha} [f(z) – f(\alpha)] = 0 \).
Step 1: Algebraic Manipulation
For \( z \neq \alpha \), we can write the difference \( f(z) – f(\alpha) \) as:
\[ f(z) – f(\alpha) = \frac{f(z) – f(\alpha)}{z – \alpha} \cdot (z – \alpha) \]
Step 2: Apply Limits
Take the limit as \( z \to \alpha \) on both sides:
\[ \lim_{z \to \alpha} [f(z) – f(\alpha)] = \lim_{z \to \alpha} \left[ \frac{f(z) – f(\alpha)}{z – \alpha} \cdot (z – \alpha) \right] \]
Step 3: Use Differentiability
Since \( f \) is differentiable at \( \alpha \), the first part of the product converges to the derivative \( f'(\alpha) \).
The second part \( (z – \alpha) \) obviously converges to 0.
Applying limit laws (product rule): \[ \begin{align} \lim_{z \to \alpha} [f(z) – f(\alpha)] &= \left[ \lim_{z \to \alpha} \frac{f(z) – f(\alpha)}{z – \alpha} \right] \cdot \left[ \lim_{z \to \alpha} (z – \alpha) \right] \\ &= f'(\alpha) \cdot 0 \\ &= 0 \end{align} \]
Applying limit laws (product rule): \[ \begin{align} \lim_{z \to \alpha} [f(z) – f(\alpha)] &= \left[ \lim_{z \to \alpha} \frac{f(z) – f(\alpha)}{z – \alpha} \right] \cdot \left[ \lim_{z \to \alpha} (z – \alpha) \right] \\ &= f'(\alpha) \cdot 0 \\ &= 0 \end{align} \]
Conclusion
Since \( \lim_{z \to \alpha} [f(z) – f(\alpha)] = 0 \), it follows that:
\[ \lim_{z \to \alpha} f(z) = f(\alpha) \]
Thus, \( f \) is continuous at \( \alpha \). \(\blacksquare\)