# Complex Differentiability Limit Definition

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.

• 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? Feb 29 at 19:17
• @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 $(x,0)$? Feb 29 at 19:18
• 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. Feb 29 at 19:29
• @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? Feb 29 at 19:30
• Original poster, unknown gender so the correct form is 'they'. Feb 29 at 19:32

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 $$x$$.

• It is not Complex Differentiable anywhere. Feb 29 at 19:43
• @MarkViola It is complex differentiable at the points of $\Bbb R$ (and only at those points). Feb 29 at 19:54
• 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. Feb 29 at 20:48
• @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? Feb 29 at 21:37
• 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 Feb 29 at 21:47

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.

• 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. Feb 29 at 19:20
• 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. Feb 29 at 19:23
• I am using the complex norm, not the absolute value; as $a_n\to 0$ iff $|a_n|\to 0$ (per definition) Feb 29 at 19:28
• $f$ must be defined in an open set (in this case $\mathbb{C}$), but the conditions may only be satisfied at one point. Feb 29 at 21:51
• Well, that is a weaker form of the Theorem. Then, it's better to use the limit definition as we have already done. Feb 29 at 22:11