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 Mar12 accepted Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Mar12 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ @Dr.MV, you could have a $C^1$ function that does not satisfy the Cauchy-Riemann equations, and I don't see anything in those notes that would contradict that. Mar12 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ @Dr.MV, no, but that's not true, as Umberto already mentioned in the comments to his answer (and as the book I'm using would also agree). $C^1$ is not the same as holomorphic. Mar12 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ @UmbertoP., thanks for that example, and I don't know why I was thinking that would be the case in the first place. Mar12 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ @Dr.MV, you're absolutely right. I thought the question required me to work without the assumption that $f$ is $C^1$ for that part, and I made that mistake because I also immediately jumped to the conclusion that $C^1$ implies complex differentiable. But assuming $f$ is $C^1$, that follows immediately from the definition of $\frac{\partial f}{\partial \bar{z}}$, since the function will then also satisfy the Cauchy-Riemann equations. Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ @Dr.MV, is there another, easier way to show the above is true without Morera's Theorem? Also, Morera's Theorem doesn't really address that, because here we have an additional factor of $\frac{1}{r^2}$ and the limit, no? And thanks for the upvote :) Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ By the way, if the LHS is zero, that also seems to imply the function is differentiable (this is without assuming $f$ is $C^1$, of course), although I don't quite get why. Any thoughts on this? Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Ah OK, I thought it had something to do with my wording. Mar11 revised Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ edited body Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Wait, so what's wrong with the question that it got downvoted? Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ I understood your notation, but I still can't quite convince myself of the answer to my question. At first I namely thought that equality would hold, but now I'm having doubts. This is unrelated to your argument, but if it held I might have another way of doing this problem. Any thoughts on this? Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Alright, cool. Any input on that additional question in the comments perhaps? Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Also, shouldn't you have $\frac{\partial f}{\partial z}(z_0) (z - z_0)$ in the first line of your Taylor expansion? Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Thanks. By the way, for some reason I thought $\frac{\partial f}{\partial \bar{z}}(z_0) = \lim_{{h} \to 0} \frac{f(z_0 + h) - f(z_0)}{\bar{h}}$, but this is not true, is it? Mar11 comment Show differentiability at a point then find differential where $f(x,y) = (x^2, xy+y^2)$ Have you learned that a function is differentiable at $(a,b)$ if the partial derivatives exist and are continuous at said point? As far as the differential, what definition of it were you given? When you obtain the partial derivatives and look at the latter, it will be straight-forward to find it. Mar11 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Hmm, how did you get that first equality in the comment you just made? I mean, $f(z_0)$ is a constant function, so you could just integrate it directly by finding its antiderivative $f(z_0)z$ and then evaluate it around a closed loop (so that it's indeed $0$), no? Mar10 comment Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Thanks, I'll check your answer in detail in a bit, but I think I was supposed to do this without resorting to Cauchy's Theorem (although I can't see where you've actually used it here, since it seems to be straight-up integration of both sides). In the book that I'm using, the latter namely only appears in the later chapters, so do you know another way to do it? Can I take $f(z_0) = 0$, WLOG? I think I can, and I think that could be useful, but I haven't fully worked it out yet. Mar10 asked Show that $\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0)$ Feb6 accepted Show that Möbius transformations that preserve the unit disk are of the matrix form $\tiny \begin{bmatrix}a & b \\ \bar{b} & \bar{a} \end{bmatrix}$ Feb5 comment Holomorphic branch of the $n^{\mathrm{th}}$ root of $f$ I think there's a typo in the last paragraph, since it should be that $\frac{h}{g}$ is continuous and constant, right?