Continuity and differentiability in the complex case

We provide the defintions of continuous and differentiable function in $\mathbb {C}$$\mathbb {C}$. We will see that it is not possible to simply extend the properties of real functions.

Definition

We consider ${\mathcal {D}}\subset \mathbb {C}$$\mathcal{D}\subset \mathbb {C}$ and $z_{0}=x_{0}+iy_{0}\in {\mathcal {D}}$$z_{0}=x_{0}+iy_{0}\in \mathcal{D}$. We say that $f:{\mathcal {D}}\rightarrow \mathbb {C}$$f:\mathcal{D}\rightarrow \mathbb {C}$ is continuous at the point $z_{0}$$z_{0}$ if:

$$\forall \,\epsilon >0\;\exists \,\delta >0\,\colon \,\forall \,z\in {\mathcal {D}},\,\mid z-z_{0}\mid <\delta \,\Rightarrow \mid f(z)-f(z_{0})\mid <\epsilon .$$

$$\forall \, \epsilon >0\; \exists \, \delta >0\, \colon \, \forall \, z\in \mathcal{D},\, \mid z-z_{0}\mid <\delta \, \Rightarrow \mid f(z)-f(z_{0})\mid <\epsilon .$$

 

By recalling that $\forall \,z\in \mathbb {C}$$\forall \, z\in \mathbb {C}$ we have that $z=x+iy$$z=x+iy$ and so that $f(z)=f(x+iy)=u(x,y)+iv(x,y)$$f(z)=f(x+iy)=u(x,y)+iv(x,y)$, we state and prove the following

Theorem

We consider ${\mathcal {D}}\subset \mathbb {C} {\mbox{ and }}{\mathcal {\tilde {D}}}\subset \mathbb {R} ^{2}.$$\mathcal{D}\subset \mathbb {C}\mbox{ and }\mathcal{\tilde{D}}\subset \mathbb {R}^{2}.$ Also $z_{0}=x_{0}+iy_{0}\in {\mathcal {D}}$$z_{0}=x_{0}+iy_{0}\in \mathcal{D}$. The function $f:{\mathcal {D}}\rightarrow \mathbb {C}$$f:\mathcal{D}\rightarrow \mathbb {C}$ is continuous at the point $z_{0}$$z_{0}$ if and only if the functions $u(x,y)\;\&\;v(x,y)$$u(x,y)\; \& \; v(x,y)$ as defined in (1.1) are continuous, in the notion of the continuity of functions defined on $\mathbb {R} ^{n}$$\mathbb {R}^{n}$ made during analysis II, in point $(x_{0},y_{0})\in {\mathcal {\tilde {D}}}$$(x_{0},y_{0})\in \mathcal{\tilde{D}}$.

 

Proof

Let $u,v$$u,v$ be continuous at the point $(x_{0},y_{0})\in {\mathcal {\tilde {D}}}$$(x_{0},y_{0})\in \mathcal{\tilde{D}}$. We have that:

$$\mid f(z)-f(z_{0})\mid =\mid u(x,y)+iv(x,y)-u(x_{0},y_{0})-iv(x_{0},y_{0})=$$

$$\mid f(z)-f(z_{0})\mid =\mid u(x,y)+iv(x,y)-u(x_{0},y_{0})-iv(x_{0},y_{0})=$$

$$=\mid u(x,y)-u(x_{0},y_{0})+i(v(x,y)-v(x_{0},y_{0}))\mid =$$

$$=\mid u(x,y)-u(x_{0},y_{0})+i(v(x,y)-v(x_{0},y_{0}))\mid =$$

$$={\sqrt {(u(x,y)-u(x_{0},y_{0}))^{2}+(v(x,y)-v(x_{0},y_{0}))^{2}}}$$

$$=\sqrt{(u(x,y)-u(x_{0},y_{0}))^{2}+(v(x,y)-v(x_{0},y_{0}))^{2}}$$

$$\leq \mid u(x,y)-u(x_{0},y_{0})\mid +\mid v(x,y)-v(x_{0},y_{0})\mid .$$

$$\leq \mid u(x,y)-u(x_{0},y_{0})\mid +\mid v(x,y)-v(x_{0},y_{0})\mid .$$

We used the definition of $f(z)$$f(z)$ and the elementar inequality:

${\sqrt {\sum x_{i}^{2}}}\leq \sum \mid x_{i}\mid .$$\sqrt{\sum x_{i}^{2}}\leq \sum \mid x_{i}\mid .$

In an analogous way:

$$\mid z-z_{0}\mid \leq \mid x-x_{0}\mid +\mid y-y_{0}\mid .$$

$$\mid z-z_{0}\mid \leq \mid x-x_{0}\mid +\mid y-y_{0}\mid .$$

We know, by hypothesis, that $u,v$$u,v$ are continuous at $(x_{0},y_{0})$$(x_{0},y_{0})$, that is:

$$\forall \,\epsilon _{1}\;\exists \,\delta _{1}\,\colon \,\forall \,(x,y)\in {\mathcal {\tilde {D}}}\;\parallel (x,y)-(x_{0},y_{0})\parallel \leq \delta _{1}\Rightarrow \mid u(x,y)-u(x_{0},y_{0})\mid \leq \epsilon _{1},$$

$$\forall \, \epsilon _{1}\; \exists \, \delta _{1}\, \colon \, \forall \, (x,y)\in \mathcal{\tilde{D}}\; \parallel (x,y)-(x_{0},y_{0})\parallel \le \delta _{1}\Rightarrow \mid u(x,y)-u(x_{0},y_{0})\mid \le \epsilon _{1},$$

$$\forall \,\epsilon _{2}\;\exists \,\delta _{2}\,\colon \,\forall \,(x,y)\in {\mathcal {\tilde {D}}}\;\parallel (x,y)-(x_{0},y_{0})\parallel \leq \delta _{2}\Rightarrow \mid v(x,y)-v(x_{0},y_{0})\mid \leq \epsilon _{2}.$$

$$\forall \, \epsilon _{2}\; \exists \, \delta _{2}\, \colon \, \forall \, (x,y)\in \mathcal{\tilde{D}}\; \parallel (x,y)-(x_{0},y_{0})\parallel \le \delta _{2}\Rightarrow \mid v(x,y)-v(x_{0},y_{0})\mid \le \epsilon _{2}.$$

Therefore we have that, called $\epsilon =\epsilon _{1}+\epsilon _{2}\;\&\;\delta =\delta _{1}+\delta _{2}$$\epsilon =\epsilon _{1}+\epsilon _{2}\; \& \; \delta =\delta _{1}+\delta _{2}$:

$$\forall \,z\in {\mathcal {D}},\,\mid z-z_{0}\mid <\delta \,\Rightarrow \mid f(z)-f(z_{0})\mid <\epsilon .$$

$$\forall \, z\in \mathcal{D},\, \mid z-z_{0}\mid <\delta \, \Rightarrow \mid f(z)-f(z_{0})\mid <\epsilon .$$

Vice versa let $f$$f$ be continuous at $z_{0}=x_{0}+iy_{0}\in {\mathcal {D}}$$z_{0}=x_{0}+iy_{0}\in \mathcal{D}$. From the definition of continuity for complex function we have that:

$$\forall \,\epsilon >0\;\exists \,\delta >0\,\colon \,\forall \,z\in {\mathcal {D}},\,\mid z-z_{0}\mid <\delta \,\Rightarrow \mid f(z)-f(z_{0})\mid <\epsilon .$$

$$\forall \, \epsilon >0\; \exists \, \delta >0\, \colon \, \forall \, z\in \mathcal{D},\, \mid z-z_{0}\mid <\delta \, \Rightarrow \mid f(z)-f(z_{0})\mid <\epsilon .$$

We see that:

$$\mid u(x,y)-u(x_{0},y_{0})\mid \leq \mid f(z)-f(z_{0})\mid \;\&\;\mid v(x,y)-v(x_{0},y_{0})\mid \leq \mid f(z)-f(z-z_{0})\mid ,$$

$$\mid u(x,y)-u(x_{0},y_{0})\mid \leq \mid f(z)-f(z_{0})\mid \; \& \; \mid v(x,y)-v(x_{0},y_{0})\mid \le \mid f(z)-f(z-z_{0})\mid ,$$

$$\mid x-x_{0}\mid \leq \mid z-z_{0}\mid \;\&\;\mid y-y_{0}\mid \leq \mid z-z_{0}\mid .$$

$$\mid x-x_{0}\mid \le \mid z-z_{0}\mid \; \& \; \mid y-y_{0}\mid \le \mid z-z_{0}\mid .$$

So, given $\epsilon _{1}$$\epsilon _{1}$ and $\epsilon _{2}$$\epsilon _{2}$, we obtain:

$$\exists \,\delta _{1}\,\colon \,\forall \,(x,y)\in {\mathcal {\tilde {D}}}\;\parallel (x,y)-(x_{0},y_{0})\parallel \leq \delta _{1}\Rightarrow \mid u(x,y)-u(x_{0},y_{0})\mid \leq \epsilon _{1},$$

$$\exists \, \delta _{1}\, \colon \, \forall \, (x,y)\in \mathcal{\tilde{D}}\; \parallel (x,y)-(x_{0},y_{0})\parallel \le \delta _{1}\Rightarrow \mid u(x,y)-u(x_{0},y_{0})\mid \le \epsilon _{1},$$

$$\exists \,\delta _{2}\,\colon \,\forall \,(x,y)\in {\mathcal {\tilde {D}}}\;\parallel (x,y)-(x_{0},y_{0})\parallel \leq \delta _{2}\Rightarrow \mid v(x,y)-v(x_{0},y_{0})\mid \leq \epsilon _{2}.$$

$$\exists \, \delta _{2}\, \colon \, \forall \, (x,y)\in \mathcal{\tilde{D}}\; \parallel (x,y)-(x_{0},y_{0})\parallel \le \delta _{2}\Rightarrow \mid v(x,y)-v(x_{0},y_{0})\mid \le \epsilon _{2}.$$

That is, $u,v$$u,v$ are continuous at $(x_{0},y_{0})\in {\mathcal {\tilde {D}}}.$$(x_{0},y_{0})\in \mathcal{\tilde{D}}.$

 

Now, we introduce the definition of differentiable functions in $\mathbb {C}$$\mathbb {C}$; it will be clear that, while the definition of continuity is simply an extension of the known one for the functions in $\mathbb {R}$$\mathbb {R}$, the notion of differentiability is more complicated that the pure extension of a known fact.

Definition

Consider ${\mathcal {D}}\subset \mathbb {C}$$\mathcal{D}\subset \mathbb {C}$ and $z_{0}=x_{0}+iy_{0}\in {\mathcal {D}}$$z_{0}=x_{0}+iy_{0}\in \mathcal{D}$. We say that $f:{\mathcal {D}}\rightarrow \mathbb {C}$$f:\mathcal{D}\rightarrow \mathbb {C}$ is differentiable at the point $z_{0}$$z_{0}$ if the following limit exists and is unique:

$$\lim _{z\rightarrow z_{0}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}\equiv f^{\prime }(z_{0}).$$

$$\lim _{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}\equiv f^{\prime }(z_{0}).$$

The value of this limit is the derivative of the function $f$$f$ at the point $z_{0}$$z_0$.

 

Remark 1: Stating that the limit is unique is equivalent to state that it must exists and be the same for every path followed to reach $z_{0}$$z_{0}$.

Now, we provide some necessary conditions for differentiability of a complex function. We will see that, in some cases, these conditions can become sufficient. We repeat that the limit $\lim _{z\rightarrow z_{0}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}\equiv f^{\prime }(z_{0}).$$\lim _{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}\equiv f^{\prime }(z_{0}).$ must exist for every path used to reach $z_{0}$$z_{0}$; we study two simple paths: those for which we arrive at $z_{0}$$z_{0}$ along parallel axis to those of $Re(z)\;\&\;Im(z)$$Re(z)\; \& \; Im(z)$ in the complex plane:

• Along the axis parallel to $Re(z)$$Re(z)$: A generic $z\in \mathbb {C}$$z\in \mathbb {C}$ who belongs to this axis has the form $z=x+iy_{0}$$z=x+iy_{0}$. Therefore we have that:

$${\frac {f(z)-f(z_{0})}{z-z_{0}}}={\frac {u(x,y_{0})+iv(x,y_{0})-u(x_{0},y_{0})-iv(x_{0},y_{0})}{x-x_{0}}}\ .$$

$$\frac{f(z)-f(z_{0})}{z-z_{0}}=\frac{u(x,y_{0})+iv(x,y_{0})-u(x_{0},y_{0})-iv(x_{0},y_{0})}{x-x_{0}} \ .$$

By taking the limit for $z\rightarrow z_{0}$$z\rightarrow z_{0}$ we have that:

$$\lim _{z\rightarrow z_{0}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}=\lim _{x\rightarrow x_{0}}{\frac {u(x,y_{0})+iv(x,y_{0})-u(x_{0},y_{0})-iv(x_{0},y_{0})}{x-x_{0}}}=$$

$$\lim _{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}=\lim _{x\rightarrow x_{0}}\frac{u(x,y_{0})+iv(x,y_{0})-u(x_{0},y_{0})-iv(x_{0},y_{0})}{x-x_{0}}=$$

$$=\lim _{x\rightarrow x_{0}}\left[{\frac {u(x,y_{0})-u(x_{0},y_{0})}{x-x_{0}}}+i{\frac {v(x,y_{0})-v(x_{0},y_{0})}{x-x_{0}}}\right]=$$

$$=\lim _{x\rightarrow x_{0}}\left[\frac{u(x,y_{0})-u(x_{0},y_{0})}{x-x_{0}}+i\frac{v(x,y_{0})-v(x_{0},y_{0})}{x-x_{0}}\right]=$$

$$={\frac {\partial u}{\partial x}}(x_{0},y_{0})+i{\frac {\partial v}{\partial x}}(x_{0},y_{0}).$$

$$=\frac{\partial u}{\partial x}(x_{0},y_{0})+i\frac{\partial v}{\partial x}(x_{0},y_{0}).$$

• Along the axis parallel to $Im(z)$$Im(z)$: A generic $z\in \mathbb {C}$$z\in \mathbb {C}$ who belongs to this axis has the form $z=x_{0}+iy$$z=x_{0}+iy$. Therefore, as before, we have that:

$$\lim _{z\rightarrow z_{0}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}={\frac {\partial v}{\partial y}}(x_{0},y_{0})-i{\frac {\partial u}{\partial y}}(x_{0},y_{0}).$$

$$\lim _{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}=\frac{\partial v}{\partial y}(x_{0},y_{0})-i\frac{\partial u}{\partial y}(x_{0},y_{0}).$$

We remind that if $f(z)$$f(z)$ is differentiable at $z_{0}$$z_{0}$ the limit at $z_{0}$$z_0$ must exist and be unique. The two found limits, if $f(z)$$f(z)$ is differentiable at $z_{0}$$z_{0}$, must be the same. We give the following

Definition

The condition:

$${\begin{cases}{\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}}\\{\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}\end{cases}}$$

$$\begin{cases} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \end{cases}$$

are called the Cauchy-Riemann equations.

 

For what said about the differentiability of a function at $z_{0}$$z_0$, these equations are necessary conditions for the differentiability of a function at $z_{0}$$z_{0}$. But, they are not sufficient: if the function is differentiable the two limits computed by getting closer along the real and the imaginary axis must be the same because the uniqueness of the limit along each path is valid. If the function is not differentiable, the two limits can coincide even if the limit is not unique along each path. It can exist a path that denies the existence of a limit and so making the function not differentiable, even if it satisfies the Cauchy-Riemann equations.

Theorem

We consider ${\mathcal {D}}\subset \mathbb {C}$$\mathcal{D}\subset \mathbb {C}$ and $z_{0}=x_{0}+iy_{0}\in {\mathcal {D}}$$z_{0}=x_{0}+iy_{0}\in \mathcal{D}$. If $f:{\mathcal {D}}\rightarrow \mathbb {C}$$f:\mathcal{D}\rightarrow \mathbb {C}$ and it is derivable at the point $z_{0}$$z_{0}$, then it satisfies the Cauchy-Riemann condition

 

Remark 2: If Cauchy-Riemann equations are satisfied, then:

$${\frac {\partial f}{\partial {\bar {z}}}}(x_{0},y_{0})=0\ .$$

$$\frac{\partial f}{\partial \bar{z}}(x_{0},y_{0})=0 \ .$$

Proof

We notice that:

$${\frac {\partial f}{\partial x}}={\frac {\partial u}{\partial x}}+i{\frac {\partial v}{\partial x}},{\mbox{ from Cauchy-Riemann equations it follows that:}}$$

$$\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x},\mbox{ from Cauchy-Riemann equations it follows that:}$$

$$={\frac {\partial v}{\partial y}}+i{\frac {\partial v}{\partial x}}={\frac {\partial v}{\partial y}}-i{\frac {\partial u}{\partial y}}=-i{\frac {\partial f}{\partial y}}.$$

$$=\frac{\partial v}{\partial y}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}=-i\frac{\partial f}{\partial y}.$$

So, the Cauchy-Riemann equations are equivalent to:

$${\frac {\partial f}{\partial x}}=-i{\frac {\partial f}{\partial y}}.$$

$$\frac{\partial f}{\partial x}=-i\frac{\partial f}{\partial y}.$$

We compute ${\frac {\partial f}{\partial {\bar {z}}}}$$\frac{\partial f}{\partial \bar{z}}$ as derivative of a composed function, since we can write:

$$x={\frac {z+{\bar {z}}}{2}}\;\&\;y=i\left({\frac {{\bar {z}}-z}{2}}\right),$$

$$x=\frac{z+\bar{z}}{2}\; \& \; y=i\left(\frac{\bar{z}-z}{2}\right),$$

so we have:

$${\frac {\partial f}{\partial {\bar {z}}}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial {\bar {z}}}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial {\bar {z}}}}={\frac {1}{2}}{\frac {\partial f}{\partial x}}+{\frac {i}{2}}{\frac {\partial f}{\partial y}}=0.$$

$$\frac{\partial f}{\partial \bar{z}}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \bar{z}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar{z}}=\frac{1}{2}\frac{\partial f}{\partial x}+\frac{i}{2}\frac{\partial f}{\partial y}=0.$$

 

We have three necessary and equivalent conditions for the differentiability of a composed function in $z_{0}\in \mathbb {C}$$z_{0}\in \mathbb {C}$:

$${\begin{cases}{\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}}\\{\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}\end{cases}}$$

$$\begin{cases} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \end{cases}$$

$${\frac {\partial f}{\partial x}}=-i{\frac {\partial f}{\partial y}}$$

$$\frac{\partial f}{\partial x}=-i\frac{\partial f}{\partial y}$$

$${\frac {\partial f}{\partial {\bar {z}}}}(x_{0},y_{0})=0$$

$$\frac{\partial f}{\partial \bar{z}}(x_{0},y_{0})=0$$

Definition

We consider ${\mathcal {D}}\subset \mathbb {C}$$\mathcal{D}\subset \mathbb {C}$ and $z_{0}=x_{0}+iy_{0}\in {\mathcal {D}}$$z_{0}=x_{0}+iy_{0}\in \mathcal{D}$. If $f:{\mathcal {D}}\rightarrow \mathbb {C}$$f:\mathcal{D}\rightarrow \mathbb {C}$ is differentiable at the point $z_{0}$$z_{0}$, we say that $f$$f$ is holomorphic in $z_{0}$$z_{0}$. If $\Omega \subset {\mathcal {D}}$$\Omega \subset \mathcal{D}$ wesay that $f$$f$ is holomorphic in $\Omega$$\Omega$ if it has this property at every points $z\in \Omega$$z\in \Omega$.

 

As already said, in some cases the necessary conditions for the differentiability of $f$$f$ can also be sufficient. In fact, more rigorously:

Theorem

Given ${\mathcal {D}}\subset \mathbb {C}$$\mathcal{D}\subset \mathbb {C}$ and $f:{\mathcal {D}}\rightarrow \mathbb {C}$$f:\mathcal{D}\rightarrow \mathbb {C}$, we consider $\Omega \subset {\mathcal {D}}$$\Omega \subset \mathcal{D}$. If partial derivatives of $f$$f$ exist and are continuous in $\Omega$$\Omega$, so Cauchy-Riemann equations are necessary and sufficient conditions for the differentiability of $f$$f$ in $\Omega .$$\Omega .$ Equivalently, we can say that $f$$f$ is holomorphic in $\Omega$$\Omega$ if and only if it satisfies the Cauchy-Riemann equations

 

we do not provide a proof of this fact, but we consider the following examples:

1. $f(z)=f(x,y)=x$$f(z)=f(x,y)=x$, that is the complex function that $z\mapsto Re(z)$$z\mapsto Re(z)$.
   We see that ${\frac {\partial f}{\partial x}}=1\neq 0={\frac {\partial f}{\partial y}}$$\frac{\partial f}{\partial x}=1\neq 0=\frac{\partial f}{\partial y}$. Cauchy-Riemann equations are not satisfied, so we conclude that $f$$f$ is not holomorphic.
2. $f(z)=e^{z}=e^{x}\cos(y)+ie^{x}\sin(y)$$f(z)=e^{z}=e^{x}\cos (y)+ie^{x}\sin (y)$. By using Cauchy-Riemann equations:

$${\begin{cases}e^{x}\cos(y)=e^{x}\cos(y)\\-e^{x}\sin(y)=e^{x}\sin(y)\end{cases}}$$

$$\begin{cases} e^{x}\cos (y)=e^{x}\cos (y)\\ -e^{x}\sin (y)=e^{x}\sin (y) \end{cases}$$

By seeing that it is true $\forall x,y\in \mathbb {R}$$\forall x,y\in \mathbb {R}$ we conclude that $f$$f$ is holomorphic.

1. $f(z)={\sqrt {\mid xy\mid }}$$f(z)=\sqrt{\mid xy\mid }$, $z=x+iy$$z=x+iy$. First of all we see that $v(x,y)\equiv 0$$v(x,y)\equiv 0$, while $u(x,y)={\sqrt {\mid xy\mid }}=Re(f)\equiv f.$$u(x,y)=\sqrt{\mid xy\mid }=Re(f)\equiv f.$ We want to apply the Cauchy-Riemann equations, from which we want to obtain that ${\frac {\partial u}{\partial x}}={\frac {\partial u}{\partial y}}=0$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=0$.

$${\frac {\partial u}{\partial x}}={\frac {1}{2}}{\sqrt {\mid {\frac {y}{x}}\mid }}sgn(x),$$

$$\frac{\partial u}{\partial x}=\frac{1}{2}\sqrt{\mid \frac{y}{x}\mid }sgn(x),$$

$${\frac {\partial u}{\partial y}}={\frac {1}{2}}{\sqrt {\mid {\frac {x}{y}}\mid }}sgn(y).$$

$$\frac{\partial u}{\partial y}=\frac{1}{2}\sqrt{\mid \frac{x}{y}\mid }sgn(y).$$

First of all, we see that both the partial derivatives, if computed respectively at $y=0\;\&\;x=0$$y=0\; \& \; x=0$ are zero.
By computing the incremental limit of $f$$f$ along a line passing through the origin of the axis in the plane $\mathbb {C}$$\mathbb {C}$ we have:

$${\frac {f(z)-f(0)}{z-0}}={\frac {\sqrt {r^{2}\mid \cos(\theta )\sin(\theta )\mid }}{r(\cos(\theta )+i\sin(\theta )}}={\frac {\sqrt {\mid \cos(\theta )\sin(\theta )\mid }}{\cos(\theta )+i\sin(\theta )}}$$

$$\frac{f(z)-f(0)}{z-0}=\frac{\sqrt{r^{2}\mid \cos (\theta )\sin (\theta )\mid}}{r(\cos (\theta )+i\sin (\theta )}=\frac{\sqrt{\mid \cos (\theta )\sin (\theta )\mid}}{\cos (\theta )+i\sin (\theta )}$$

By taking the limit for $z\rightarrow 0$$z\rightarrow 0$ it is evident that it depends on $\theta ,$$\theta ,$ that is the chosen path to reach $z_{0}=0$$z_{0}=0$.
So, $f^{\prime }(0)$$f^{\prime }(0)$ cannot exist.

Remark 3: We will see a relationship between the solutions of the Laplace equation $\nabla ^{2}f=0$$\nabla ^{2}f=0$ and the class of the holomorphic function in $\mathbb {C}$$\mathbb {C}$.

We consider Laplace equation in cartesian coordinates:

$$\nabla ^{2}f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}=0.$$

$$\nabla ^{2}f=\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=0.$$

Let be $g(z)$$g(z)$ a generic holomorphic function on $\Omega$$\Omega$ and be $x=Re(z)\;\&\;y=Im(z)$$x=Re(z)\; \& \; y=Im(z)$. We can state that:

$${\frac {\partial g}{\partial x}}=-i{\frac {\partial g}{\partial y}}.$$

$$\frac{\partial g}{\partial x}=-i\frac{\partial g}{\partial y}.$$

If $g$$g$ is partially differentiable up to the second order, we can derive with respect to $x$$x$ the previous equation, and we have:

$${\frac {\partial ^{2}g}{\partial x^{2}}}=-i{\frac {\partial ^{2}g}{\partial x\partial y}}=-i{\frac {\partial }{\partial y}}\left({\frac {\partial g}{\partial x}}\right)=-{\frac {\partial ^{2}g}{\partial y^{2}}},{\mbox{ that is:}}$$

$$\frac{\partial ^{2}g}{\partial x^{2}}=-i\frac{\partial ^{2}g}{\partial x\partial y}=-i\frac{\partial }{\partial y}\left(\frac{\partial g}{\partial x}\right)=-\frac{\partial ^{2}g}{\partial y^{2}},\mbox{ that is:}$$

$$\nabla ^{2}g=0.$$

$$\nabla ^{2}g=0.$$

The class of the holomorphic function in $\mathbb {C}$$\mathbb {C}$ coincides with the class of the solutions of the Laplace equation in two dimensions.

We do not go beyond in the theory of differentaibility in $\mathbb {C}$$\mathbb {C}$, beacuse the possibility to derive to orders greater than the second (in particular, to derive a function infinite times) will be strictly related to the theory of the integration in $\mathbb {C}$$\mathbb {C}$.