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I have $f(x+iy)=y^2$. This satisfies the Cauchy-Riemann equations for all $z_0=(x,0)$. Now I want to check whether $f$ is differentiable for all $(x,0)$. So I use the definition, taking the limit as $h$ tends to $0$, and letting $h=a+bi$:

$$\frac{f(z_0+h)-f(z_0)}{h}=\frac{f(x+a+bi)-f(x)}{a+ib}=\frac{b^2}{a+ib}$$

But how am I meant to compute the limit of $\frac{b^2}{a+ib}$ as $h$ tends to $0$, which I suppose means $(a,b)$ tends to $(0,0)$. Can someone help me do this using the definition please? Thanks.

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  • $\begingroup$ Your f(x+iy) = u(x,y) + i v(x,y) where u and v are the real and imaginary parts of f(x+iy). The Cauchy-Riemann equations are that ∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x. Have you really shown that they hold? $\endgroup$
    – Dan Asimov
    Feb 29 at 19:17
  • $\begingroup$ @DanAsimov Well $u=y^2$ and $v=0$. You get $0 = 0$ and $2y = -0$ so they hold whenever $y=0$ and $x$ is any real number. So the cauchy riemann equations hold for all $(x,0)$? $\endgroup$
    – adisnjo
    Feb 29 at 19:18
  • $\begingroup$ While the CR equations hold at every point on the real axis, the do not hold on any open set around such points. And hence $y^2$ is not holomorphic. This means that $y^2$ is nowhere complex differentiable. $\endgroup$
    – Mark Viola
    Feb 29 at 19:29
  • $\begingroup$ @MarkViola I am confused because the other poster says different as the limit tends to $0$ which is finite so it should be differentiable everywhere on the real axis? $\endgroup$
    – adisnjo
    Feb 29 at 19:30
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    $\begingroup$ Original poster, unknown gender so the correct form is 'they'. $\endgroup$ Feb 29 at 19:32

2 Answers 2

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If $x\in\Bbb R$,\begin{align}\lim_{z\to x}\frac{\operatorname{Im}^2z-\operatorname{Im}^2x}{z-x}&=\lim_{z\to x}\frac{\operatorname{Im}^2z}{z-x}\\&=\lim_{z\to x}\frac{\operatorname{Im}(z-x)}{z-x}\operatorname{Im}z\end{align}(since $x\in\Bbb R$ and therefore $\operatorname{Im}x=0$). But $\lim_{z\to x}\operatorname{Im}z=0$ and $\left|\frac{\operatorname{Im}(z-x)}{z-x}\right|\leqslant1$ for each $z\in\Bbb C\setminus\{x\}$. Therefore$$\lim_{z\to x}\frac{\operatorname{Im}(z-x)}{z-x}\operatorname{Im}z=0.$$In other words, your function is differentiable at $x$.

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  • $\begingroup$ It is not Complex Differentiable anywhere. $\endgroup$
    – Mark Viola
    Feb 29 at 19:43
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    $\begingroup$ @MarkViola It is complex differentiable at the points of $\Bbb R$ (and only at those points). $\endgroup$ Feb 29 at 19:54
  • $\begingroup$ That is not complex differentiable. Complex differentiability applies to open sets around a point or on the sphere. It is tantamount to differentiability on $\mathbb{R^2}$. It is synonymous with analyticity. $\endgroup$
    – Mark Viola
    Feb 29 at 20:48
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    $\begingroup$ @MarkViola Where did you get that idea? Take,for instance, Serge Lang's Complex Analysis. It says: “Let $U$ be an open set, and let $z$ be a point of $U$. Let $f$ be a function on $U$. We say that $f$ is complex differentiable at $z$ if the limit$$\lim_{h\to0}\frac{f(z+h)-f(z)}h$$ exists.” Are you claiming that the function $\operatorname{Im}^2$ is not differentiable at the points of $\Bbb R$ according to this definition? $\endgroup$ Feb 29 at 21:37
  • $\begingroup$ Hi Jose my friend. We are using different definitions. I am suggesting that the definition of Complex Differentiability is not as Lang's book suggests. As others define it, it is synonymous with being analytic (or holomorphic). Does that make sense. Having a derivative and being differentiable are two different things $\endgroup$
    – Mark Viola
    Feb 29 at 21:47
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Notice that $$0\leq\left|\dfrac{b^2}{a+ib}\right|^2=\dfrac{b^4}{a^2+b^2}\leq \dfrac{b^4}{b^2}=b^2\to 0,$$ so $$\dfrac{b^2}{a+ib}\to 0$$ and $f$ is differentiable in all points of the form $(x,0)$.

Edit: It could also be used that $f(x,y)=u(x,y)+ iv(x,y)$ is complex differentiable at $x_0+iy_0$ iff $u$ and $v$ are differentiable at $(x_0,y_0)$ and they verify the CR equations; I understood the purpose of the question was to prove that $f$ is holomorphic using exclusively the limit definition.

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  • $\begingroup$ So my working was all correct prior to that? Is there another way of solving this question using the definition without replacing $h$ with $a+ib$. Seems a bit long. $\endgroup$
    – adisnjo
    Feb 29 at 19:20
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    $\begingroup$ Yes, the working was correct. As the function is given in function of the coordinates (that is, in $x$ and $y$) instead of directly in function of the complex number $z$ the most natural way to work this out is using coordinates for $h$ too. $\endgroup$ Feb 29 at 19:23
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    $\begingroup$ I am using the complex norm, not the absolute value; as $a_n\to 0$ iff $|a_n|\to 0$ (per definition) $\endgroup$ Feb 29 at 19:28
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    $\begingroup$ $f$ must be defined in an open set (in this case $\mathbb{C}$), but the conditions may only be satisfied at one point. $\endgroup$ Feb 29 at 21:51
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    $\begingroup$ Well, that is a weaker form of the Theorem. Then, it's better to use the limit definition as we have already done. $\endgroup$ Feb 29 at 22:11

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