If I consider the function $f(z)=z^{2.5}$ $\in \mathbb{R}$, it is continuously differentiable with $f'(z)=2.5z^{1.5}$. But why can I not do the same in $\mathbb{C}$ with $ f'(z)=2.5z^{1.5}$?
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1What is $f(-1)$? Also, who says you can't? – Arthur Feb 14 '19 at 12:33
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$e^{2.5ln(-1)}$ and thats not defined? – tim123 Feb 14 '19 at 12:37
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What's $\ln(-1)$? – Arthur Feb 14 '19 at 12:37
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thats not defined – tim123 Feb 14 '19 at 12:40
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1So there you have the problem with $f(-1)$. Why should the rest of the complex plane be any easier? – Arthur Feb 14 '19 at 12:41
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A more subtle question is why $f(a+bi)=a-bi$ is not holomorphic. – MJD Feb 14 '19 at 12:58
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The function $f(z)=z^{2.5}$ is only well defined on positive real numbers. Everywhere else, you need to provide additional info. For example, it is not clear what $(-1)^{2.5}$ would be equal to.
You could write the function as $z^{2.5} = e^{2.5\cdot \log(z)}$, but then, the function $\log$ is not well defined on the complex numbers. Taking the standard logarithm, the function $\log$ is undefined in $(-\infty, 0]$, and there is no way you can ever continuously define the logarithm function on the entire complex plane.
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