Let f(z) = |Z| and set its domain to be D = {z an element of C: |z| = 1} the unit circle.  Where in D is f complex differntiable

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).

Let f(z) = |Z| and set its domain to be D = {z an element of C: |z| = 1} the unit circle. Where in D is f complex differentiable?

Explore the differentiability of the function f(z) = |z| within the unit circle domain. This complex analysis problem uses the Cauchy-Riemann equations to determine where the function is differentiable. A detailed step-by-step solution is provided, suitable for undergraduate students studying complex analysis.

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