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Has $\frac{z}{\bar{z}}$ complex derivative at $z=0$? I got zero from contour integral when contour is $z=\gamma(t)= e^{it}$, $t=[0,2\pi]$ so it should be analytic on unit circle.

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How do you define this function at $z=0$? – Siminore Aug 28 '12 at 16:59
It doesn't make sense to ask for the derivative of a function at a point unless the function is defined at that point. – m_t_ Aug 28 '12 at 19:29
Well, how would you define function at $z=0$? General form of function is $\frac{1}{|z|^2}z^2$. I guess it is not uniquely defined at $z=0$? – alvoutila Aug 29 '12 at 13:47
up vote 9 down vote accepted

Have you noticed that $$ \lim_{\substack{z \to 0 \\ z \in \mathbb{R}}} \frac{z}{\overline{z}} =1 $$ and $$ \lim_{\substack{z \to 0 \\ z \in i\mathbb{R}}} \frac{z}{\overline{z}} =-1 ? $$ The fact that the integral along a single, particular loop is zero has no relevance. There is Morera's theorem, but it deals with a continuous function defined in an open domain, and by assumption the integral along any loop must be zero.

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None of the statements I've seen of Morera's theorem make any assumption of a simply connected domain. It does, however, require the function to be continuous in that domain. – Robert Israel Aug 28 '12 at 17:38
There is also the fact that if an analytic function is bounded in a neighbourhood of an isolated singularity, the singularity is removable. Combined with Morera's theorem, we get the following: suppose $f$ is continuous and bounded in a punctured disk $U = \{z: 0 < |z - a| < r\}$, and $\oint_\Gamma f(z)\ dz = 0$ for all triangles $\Gamma$ in $U$. Then $f(a)$ can be defined so that $f$ is analytic in $U \cup \{a\}$. – Robert Israel Aug 28 '12 at 17:45
Yes, you are probably right. The domain need not be simply connected, since Morera's theorem is a local statement about the existence of a primitive. I'll remove this comment. – Siminore Aug 29 '12 at 7:50
Is Morera's theorem converse of Cauchy's integral theorem? is following statement true: " I got zero from contour integral when contour is z=γ(t)=eit, t=[0,2π] so it should be analytic on unit circle." – alvoutila Sep 5 '12 at 11:05
@alvoutila It is not sufficient. You must compute the integral along any contour, and find zero. A single, specific, contour has no relevance. – Siminore Sep 5 '12 at 17:06

The function $f(z):=z/\bar z$ is a priori undefined at $z:=0$; but maybe it can be defined there in a reasonable way, i.e., such that the extended $f$ is continuous there. So let's check this. Writing $z=re^{i\phi}$ with $r\geq0$ and $\phi\in{\mathbb R}$ we see that $$f(z)={re^{i\phi}\over re^{-i\phi}}=e^{2i\phi}=\cos(2\phi)+i\sin(2\phi)\ ,$$ independently of $r\geq0$. This implies that $f$ assumes arbitrary values of absolute value $1$ in points $z$ arbitrarily close to $0$. It follows that the limit $\lim_{z\to0} f(z)$ does not exist; so there is no "reasonable way" to define $f(0)$.

Since it is impossible to make $f$ continuous at $0$, a fortiori $f$ cannot be made differentiable there.

So let's forget about the origin. Maybe our $f$ is analytic in some region $\Omega\subset{\mathbb C}$ which does not contain the origin. If this were the case then the function $g(z):=z/ f(z)$, being the quotient of two analytic functions (with nonzero denominator), would also be analytic in $\Omega$.

Now $g(z)=\bar z$. So let's test whether this has a complex derivative at some point $z_0\in\dot{\mathbb C}$. To this end we have to consider the limit $$\lim_{h\to 0}{g(z_0+h)-g(z_0)\over h}=\lim_{h\to 0}{\overline{z_0+h}- \bar z_0\over h} =\lim_{h\to 0}{\bar h\over h}=\overline{\lim_{h\to 0}{ h\over \bar h}}\ .$$ As shown in the first part of the answer the limit on the right hand side does not exist. This proves that $g$, and whence $f$, does not have a complex derivative at any point $z_0\in\dot{\mathbb C}$.

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what set is $S^1$? One can follow that if $f=z/\bar{z}$, limit $\lim_{z \rightarrow 0} f(z)$ does not exist, but how one deduces that when $f=e^{2i\phi}$? – alvoutila Aug 28 '12 at 20:55
@alvoutila: $S^1$ is the one-dimensional unit sphere, i.e. the set of all points $\zeta\in{\mathbb C}$ with $|\zeta|=1$. See my edit. – Christian Blatter Aug 29 '12 at 8:07
But how would you show that if $f=e^{i2t}$ then limit $\lim_{z \rightarrow 0} f(z)$ does not exist? – alvoutila Aug 29 '12 at 13:15
@Blatter: If $f=\frac{z}{\bar{z}}$ and $g(z)=zf(z)$, how then $g(z)=\bar{z}$? Should it be $\frac{z^2}{\bar{z}}$? – alvoutila Aug 29 '12 at 14:31
@Blatter: If $f=\frac{z}{\bar{z}}$ and $g(z)=zf(z)$, how then $g(z)=\bar{z}$? Should it be $\frac{z^2}{\bar{z}}$? How do you get from the first part of the answer that the limit on the right hand side does not exist? Is this $\lim_{h \rightarrow 0} \frac{\bar{h}}{h}$ possible to show that it does not exist? – alvoutila Aug 29 '12 at 15:00

Differentiability implies continuity. But this functions is not continuous at $0$, even if you assign it a value there. Its limit along the real axis is $1$, and its limit along the imaginary axis is $-1$.

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