Check for analyticity of a complex function Prove that $f(z)=|z|^4$ is differentiable but not analytic at $z=0$
My Attempt : $|z|=\sqrt{x^2+y^2} so |z|^4=(x^2+y^2)^2$
Now, we see that at $z=0$ all the four partial derivatives are equal to zero so CR equations are satisfied at z=0. also as $f(z)$ is differentiable at z=0 it must be continueous there.
So why is the given function not analytic at z=0 ? am i missing out on something ?
 A: The crucial point is that the CR equations
$$ 4x(x^2+y^2) = 0 \text{ and } 4y(x^2+y^2)=0 $$
are satisfied iff $x=y=0$. Therefore, elsewhere $f(z)$ fails the necessary (but not sufficient) condition for being differentiable, i.e., CR equations fail. Any open neighbourhood will contains points other than the origin for which the equations do not hold.
This is the same reasoning used to show that $|z|^2$ is differentiable only at the origin, and hence nowhere analytic.
A: Analytic (holomorphic) means that the function is differentiable on an open set in $\mathbb{C}$. A single point is not an open set, thus it is not analytic (holomorphic). The reason we care about about this property is that it allows us to write the function in a Taylor series and obtain interesting and important information about the function that way. Many results that work for holomorphic functions don't work for functions that are not differentiable on open sets solely because we cannot write such functions in a Taylor series. Cauchy's theorem and anything that results from it does not hold, for instance.
