Let $f(z) = |z|$ where $z \in \mathbb{C}$, and the domain $D$ is defined as the unit circle $D = \{ z \in \mathbb{C} : |z| = 1 \}$.

Step 1: Analyze $f(z) = |z|$

The function $f(z) = |z|$ gives the modulus of the complex number $z$. Explicitly, for $z = x + iy$ (where $x$ and $y$ are real numbers), the modulus $|z|$ is defined as:

$|z| = \sqrt{x^2 + y^2}$

For $z \in D$, we know that $|z| = 1$, meaning that the modulus is constant on the unit circle.

Step 2: Conditions for Complex Differentiability

A function is complex differentiable at a point $z_0$ if it satisfies the Cauchy-Riemann equations at that point. The Cauchy-Riemann equations for $f(z) = u(x, y) + iv(x, y)$, where $u(x, y)$ is the real part and $v(x, y)$ is the imaginary part, are:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

In the case of $f(z) = |z|$, we have:

• $u(x, y) = \sqrt{x^2 + y^2}$ (the modulus of $z$),
• $v(x, y) = 0$ (since there is no imaginary part in the function).

Step 3: Applying Cauchy-Riemann Equations

To check for complex differentiability, we need to compute the partial derivatives of $u(x, y)$ and $v(x, y)$:

• $\frac{\partial u}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}$,
• $\frac{\partial u}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}$,
• $\frac{\partial v}{\partial x} = 0$,
• $\frac{\partial v}{\partial y} = 0$.

For the Cauchy-Riemann equations to hold:

1. $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \Rightarrow \frac{x}{\sqrt{x^2 + y^2}} = 0$,
2. $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \Rightarrow \frac{y}{\sqrt{x^2 + y^2}} = 0$.

Both equations imply that $x = 0$ and $y = 0$, which only happens at the origin. However, the domain $D$ excludes the origin because it is the unit circle, where $|z| = 1$. Therefore, $f(z) = |z|$ does not satisfy the Cauchy-Riemann equations at any point on the unit circle.

Step 4: Conclusion

Since the Cauchy-Riemann equations are not satisfied anywhere on the unit circle $D$, the function $f(z) = |z|$ is nowhere complex differentiable within the domain $D$. Thus, $f(z)$ is not complex differentiable at any point on the unit circle.

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Related Questions:

1. How do the Cauchy-Riemann equations apply to functions defined on other domains?
2. Can a function be real-differentiable but not complex-differentiable?
3. What are the differences between holomorphic and complex-differentiable functions?
4. How is complex differentiability different from real differentiability in multi-variable calculus?
5. Can a function with a non-differentiable modulus be differentiable in the complex plane?

Tip:

For complex differentiability, always check the Cauchy-Riemann equations as they are essential in verifying whether a function is holomorphic (complex-differentiable).