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We know that for any complex function to be complex differentiable i.e. holomorphic, 2 properties must be true - the cauchy riemann conditions hold and the limit at that point exists. I understand that entirely.

but how to prove that the complex conjugate of z is nowhere differentiable - clearly the cauchy riemann do not hold, but is this enough to show it is nowhere differentiable? or do we have to find the limit too?

from the above stated theorem, can the independent conditions be taken individually to imply 'simple' ie NOT COMPLEX differentiabilty, and if both hold then the function is holomorphic? or is it more like the condition for differentiability/continuity of real functions where differentiability implies continuity but not vice versa?

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  • $\begingroup$ Existence of the limit is not enough. The function must be differentiable (as a function of 2 real variables) at the point. $\endgroup$ – Julián Aguirre Dec 12 '17 at 19:11
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Let $f(z)=z^{5}/|z|^{4}$ for $z\ne 0$, $f(0)=0$, $f$ satisfies Cauchy-Riemann equations at $z=0$, but $f$ is not complex-differentiable at $z=0$.

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The second part of the question asks if the Cauchy Riemann equations are enough to get real-differentiability. This is also not true, as there are examples of functions which have all real partials $0$ and are still not differentiable at $0$. You can express one of these in complex notation using real and imaginary parts (although it may be an ugly expression), and it will trivially satisfy the Cauchy-Riemann equations, since everything is $0$. Take for example a 'butterfly' type function, $\frac{xy}{x^2 + y^2}$

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You're missing an elementary point of logic. You seem to know that if $f'(a)$ exists, then both of the following must hold: i) The Cauchy-Riemann equations for $f$ hold at $a;$ ii) $\lim_{z\to a} f(z)$ exists.

Now you have a function, namely $f(z)=\bar z,$ for which i) fails to hold at every $a.$ Could $f'(a)$ exist at some $a?$ No!!! If it did, then the CR equations for $f$ would hold at $a,$ contradiction.

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