Question #223831

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

Expert's answer

"f(z)=|z|^{2}= z \\bar{z}\\\\"

If f is differentiable at "z=z_0"

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

Now

"\\frac{\\partial f}{\\partial \\bar{z}}=z\\\\\nfor \\space z(\\not=0) \\in \\nsubseteq \\frac{\\partial f}{\\partial \\bar{z}} \\not=0\\\\\n\\implies f(z)=|z|^2" is not differential at "z(\\not=)=0 \\in \\nsubseteq"

To check at z=0

"f'(0)= \\lim\\limits_{z \\to 0 }\\frac{f(z)-f(0)}{z-0}\\\\\nf'(0)= \\lim\\limits_{z \\to 0 }\\frac{z \\bar{z}-0}{z}\\\\\nf'(0)=0"

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

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