Question

Prove that f(z)=z² is analytic

Answer 1

The answer to your question slightly depends on the definitions that you use. In the context of complex analysis one usually say that f is analytic at z0 if it is differentiable in some neighbourhood of z0. For differentiability we have to consider the limit

limz→0f(z0+z)−f(z0)z

In our case f(z)=|z|2=x2+y2 where we use z=x+iy. When z0=0 we have to consider the limit of |z|2/z as z→0. This limit is equal to 0 since the modulus of our expression is equal to |z| which goes to zero. If modulus goes to zero, then expression goes to zero as well. This proves that f is differentiable at 0. On the other hand, if z0=x0+iy0≠0, then we could first consider the limit along the real line, that is when z=x. In this case

limz→0f(z0+z)−f(z0)z=limx→02x0x+x2x=2x0

If we consider the limit along vertical direction, i.e. z=iy then we get

limz→0f(z0+z)−f(z0)z=limy→02y0y+y2iy=−2iy0

As we can see, one limit is real and one is purely imaginary, this means that they can not be equal unless x0=y0=0. This proves that f(z)=|z|2 can not be differentiable at any point except z0=0 (where it is differentiable by previous argument). Since the function is not differentiable in a any neighbourhood of zero, it is not analytic at zero.

In fact, it is easy to see from Cauchy-Riemann equations that a real-valued non-constant function can not be analytic. It still could be differentiable (essentially by accident) at some points, but can not be analytic