Question #223831

The function f(z)=|z|^{2} is differentiable


1
Expert's answer
2021-08-17T14:19:01-0400

f(z)=z2=zzˉf(z)=|z|^{2}= z \bar{z}\\

If f is differentiable at z=z0z=z_0

Then fzˉz=zˉ=0\frac{\partial f}{\partial \bar{z}}|_{z=\bar{z}}=0\\

Now

fzˉ=zfor z(0)fzˉ0    f(z)=z2\frac{\partial f}{\partial \bar{z}}=z\\ for \space z(\not=0) \in \nsubseteq \frac{\partial f}{\partial \bar{z}} \not=0\\ \implies f(z)=|z|^2 is not differential at z()=0z(\not=)=0 \in \nsubseteq

To check at z=0

f(0)=limz0f(z)f(0)z0f(0)=limz0zzˉ0zf(0)=0f'(0)= \lim\limits_{z \to 0 }\frac{f(z)-f(0)}{z-0}\\ f'(0)= \lim\limits_{z \to 0 }\frac{z \bar{z}-0}{z}\\ f'(0)=0

Hence f(z) is differentiable at z=0 only.


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