Let $f(z) = \operatorname{Im}(z)$, where $z = x + iy$ with $x, y$ real. That is,

$f(x + iy) = y.$

Step 1: Compute the Cauchy-Riemann Equations

For $f(z) = u(x, y) + iv(x, y)$, we define

$u(x, y) = y, \quad v(x, y) = 0.$

The Cauchy-Riemann equations state that for $f(z)$ to be holomorphic (complex differentiable), the partial derivatives must satisfy:

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

Computing these derivatives,

$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} y = 0, \quad \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} 0 = 0.$

$\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} y = 1, \quad \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} 0 = 0.$

The first equation $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ holds, but the second equation

$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \Rightarrow 1 = 0$

is false.

Step 2: Conclusion

Since the Cauchy-Riemann equations are not satisfied at any point, particularly at $z = 0$, $f(z) = \operatorname{Im}(z)$ is not differentiable at $z = 0$ or anywhere in $\mathbb{C}$.

Would you like a graphical explanation of this result?

Related Questions

1. What are the Cauchy-Riemann equations, and why are they necessary for differentiability?
2. Can a function be continuous but not differentiable in the complex plane?
3. What are some examples of non-holomorphic functions in complex analysis?
4. Is the real part of $z$ differentiable at $z = 0$?
5. How does the concept of complex differentiability differ from real differentiability?

Tip: A function can be differentiable in the real sense (i.e., partial derivatives exist), but it might still fail to be differentiable in the complex sense if the Cauchy-Riemann equations are not satisfied.