Compact sets

• Let \((X, d)\) be a metric space and \(\Omega\subseteq X\). An open covering of \(\Omega\) is a family of open sets \((U_{\alpha})_{\alpha\in A}\) (\(A\) is not necessarily countable) such that \[\Omega \subseteq \bigcup_{\alpha\in A} U_{\alpha}.\]

  Definition.

  A set \(\Omega\) in a metric space is said to be compact if and only if every open covering of \(\Omega\) has a finite subcovering.

  Theorem.

  A set \(\Omega\) in a metric space is compact if and only if every sequence \((z_{n})_{n\in\mathbb N} \subseteq \Omega\) has a subsequence that converges to a point in \(\Omega\).

  Theorem.

  A set \(\Omega\subseteq \mathbb C\) is compact if and only if it is closed and bounded.

Compactification of \(\mathbb C\)

• By adjoining a point \(\infty\), which we call the point at \(\infty\), to \(\mathbb{C}\) we obtain the extended complex plane \[\mathbb{C}_{\infty}=\mathbb{C} \cup\{\infty\} .\]

• Next, we introduce the Riemann sphere \[\mathbb S=\left\{\left(x_{1}, x_{2}, x_{3}\right) \in \mathbb{R}^{3} : x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\right\}\] and \[f: \mathbb S \rightarrow \mathbb{C}_{\infty}\] given by \[f\left(\left(x_{1}, x_{2}, x_{3}\right)\right)=\frac{x_{1}+i x_{2}}{1-x_{3}} \quad \text { if } \quad \left(x_{1}, x_{2}, x_{3}\right) \neq(0,0,1),\] and \[f((0,0,1))=\infty.\]

• We write \(N=(0,0,1)\) and we call \(N\) the north pole of \(\mathbb S\). Further the function \(f\) is called stereographic projection.

Stereographic projection is one-to-one

• First, we show that \(f\) is one-one. Let \((x_{1}, x_{2}, x_{3})\) and \((y_{1}, y_{2}, y_{3})\) be distinct elements of \(\mathbb S\) such that \(f\left(\left(x_{1}, x_{2}, x_{3}\right)\right)=f\left((y_{1}, y_{2}, y_{3})\right)\).

• Then \((x_{1}, x_{2}, x_{3})\) and \((y_{1}, y_{2}, y_{3})\) are different from \(N\), and therefore \[\frac{x_{1}+i x_{2}}{1-x_{3}}=\frac{y_{1}+i y_{2}}{1-y_{3}}=z \in \mathbb{C}.\]

• Thus \[|z|^{2}=\frac{x_{1}^{2}+x_{2}^{2}}{\left(1-x_{3}\right)^{2}}=\frac{y_{1}^2+y_{2}^2}{\left(1-y_{3}\right)^{2}} .\]

• Since \(\left(x_{1}, x_{2}, x_{3}\right)\) and \((y_{1}, y_{2}, y_{3})\) are in \(\mathbb S\), we have \[|z|^{2}=\frac{1-x_{3}^{2}}{\left(1-x_{3}\right)^{2}}=\frac{1-y_{3}^{2}}{\left(1-y_{3}\right)^{2}}\quad \iff\quad |z|^{2}=\frac{1+x_{3}}{1-x_{3}}=\frac{1+y_{3}}{1-y_{3}}.\]

• Therefore \[\frac{|z|^{2}-1}{|z|^{2}+1}=x_{3}, \quad \frac{|z|^{2}-1}{|z|^{2}+1}=y_{3}.\]

• Thus \(x_{3}=y_{3}\) and hence \(x_{1}=y_{1},\) and \(x_{2}=y_{2}\).

Stereographic projection is onto

• Next, we show that \(f\) is onto. If \(z=\infty\), we take \(N \in \mathbb S\) so that \(f(N)=\infty\).

• Thus, we may suppose that \(z \in \mathbb{C}\). We take \[x_{1}=\frac{z+\overline{z}}{|z|^{2}+1}, \quad x_{2}=\frac{z-\overline{z}}{i\left(|z|^{2}+1\right)}, \quad x_{3}=\frac{|z|^{2}-1}{|z|^{2}+1}.\]

• Then \[\begin{aligned} x_{1}^{2}+x_{2}^{2}+x_{3}^{2}&=\frac{(z+\overline{z})^{2}-(z-\overline{z})^{2}+\left(|z|^{2}-1\right)^{2}}{\left(|z|^{2}+1\right)^{2}}\\ &=\frac{4|z|^{2}+\left(|z|^{2}-1\right)^{2}}{\left(|z|^{2}+1\right)^{2}}=1 \end{aligned}\]

  implying \(\left(x_{1}, x_{2}, x_{3}\right) \in \mathbb S\). Further, we check that \(f\left(\left(x_{1}, x_{2}, x_{3}\right)\right)=z\).

Geometric interpretation of the stereographic projection

• Let \(z=x+i y\) be a point in the complex plane and we identify it by \((x, y, 0)\) in \(\mathbb{R}^{3}\). We consider the line joining \(z\) to \(N\) in \(\mathbb{R}^{3}\).

• Its parametric representation is \[t N+(1-t) z,\quad \text{ for } \quad -\infty<t<\infty .\]

• The points on this line are \[((1-t) x,(1-t) y, t),\quad \text{ for } \quad -\infty<t<\infty .\]

• Thus, the line intersects \(\mathbb S\) if and only if \[(1-t)^{2} x^{2}+(1-t)^{2} y^{2}+t^{2}=1,\] which we re-write as \[(1-t)|z|^{2}=(1-t) x^{2}+(1-t) y^{2}=1+t .\]

• Hence \[t=\frac{|z|^{2}-1}{|z|^{2}+1} .\]

• Thus, the line intersects \(\mathbb S\) at \(Z\) given by \[\begin{aligned} Z=&\Bigg(\left(1-\frac{|z|^{2}-1}{|z|^{2}+1}\right) x,\left(1-\frac{|z|^{2}-1}{|z|^{2}+1}\right) y, \frac{|z|^{2}-1}{|z|^{2}+1}\Bigg)\\ & =\left(\frac{2 x}{|z|^{2}+1}, \frac{2 y}{|z|^{2}+1}, \frac{|z|^{2}-1}{|z|^{2}+1}\right) \\ & =\left(\frac{z+\bar{z}}{|z|^{2}+1}, \frac{z-\bar{z}}{i\left(|z|^{2}+1\right)}, \frac{|z|^{2}-1}{|z|^{2}+1}\right) . \end{aligned}\]

• We summarize the above procedure as follows:

  • Let \(z=(x, y)\) be a point in the complex plane.

  • The line passing through \((x, y, 0)\) and \((0,0,1)\) intersects \(\mathbb S\) exactly at one point \(Z=\left(x_{1}, x_{2}, x_{3}\right)\) given above and \(f\left(\left(x_{1}, x_{2}, x_{3}\right)\right)=z\).

  • We call \(\left(x_{1}, x_{2}, x_{3}\right)\) the spherical coordinates of \(z\) and \((0,0,1)\) the spherical coordinates of \(\infty\).

Distance between points in the extended complex plane

• Let \(z \in \mathbb{C}_{\infty}\) and \(z^{\prime} \in \mathbb{C}_{\infty}\). Let \((x_{1}, x_{2}, x_{3})\) and \((x_{1}^{\prime}, x_{2}^{\prime}, x_{3}^{\prime})\) be spherical coordinates of \(z\) and \(z^{\prime}\), respectively. Namely, \[x_{1}=\frac{z+\overline{z}}{|z|^{2}+1}, \quad x_{2}=\frac{z-\overline{z}}{i\left(|z|^{2}+1\right)}, \quad x_{3}=\frac{|z|^{2}-1}{|z|^{2}+1},\] and \[x_{1}'=\frac{z'+\overline{z}'}{|z'|^{2}+1}, \quad x_{2}'=\frac{z'-\overline{z}'}{i\left(|z'|^{2}+1\right)}, \quad x_{3}'=\frac{|z'|^{2}-1}{|z'|^{2}+1}.\]

• Then we define \[\begin{aligned} d\left(z, z^{\prime}\right)=\sqrt{\left(x_{1}-x_{1}^{\prime}\right)^{2}+\left(x_{2}-x_{2}^{\prime}\right)^{2} +\left(x_{3}-x_{3}^{\prime}\right)^{2}} . \end{aligned}\]

• Thus

  \[\begin{aligned} d\left(z, z^{\prime}\right)^{2}&=\left(x_{1}-x_{1}^{\prime}\right)^{2}+\left(x_{2}-x_{2}^{\prime}\right)^{2}+\left(x_{3}-x_{3}^{\prime}\right)^{2}\\ &=2-2\left(x_{1} x_{1}^{\prime}+x_{2} x_{2}^{\prime}+x_{3} x_{3}^{\prime}\right). \end{aligned}\]

• Now we note \[\begin{aligned} &x_{1} x_{1}^{\prime}+x_{2} x_{2}^{\prime}+x_{3} x_{3}^{\prime} \\ &=\frac{(z+\overline{z})\left(z^{\prime}+\overline{z}^{\prime}\right)}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)}-\frac{(z-\overline{z})\left(z^{\prime}-\overline{z}^{\prime}\right)}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)} +\frac{\left(|z|^{2}-1\right)\left(\left|z^{\prime}\right|^{2}-1\right)}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)} \\ & =\frac{\left(|z|^{2}-1\right)\left(\left|z^{\prime}\right|^{2}-1\right)+2 z \overline{z}^{\prime}+2 \overline{z} z^{\prime}}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)} \\ & =\frac{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)-2\left(|z|^{2}+1\right)-2\left(\left|z^{\prime}\right|^{2}+1\right)+4+2 z \overline{z}^{\prime}+2 \overline{z} z^{\prime}}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)} \end{aligned}\]

• Further \[\begin{aligned} &2\left(|z|^{2}+1\right)+2\left(\left|z^{\prime}\right|^{2}+1\right)-4-2 z \overline{z}^{\prime}-2 \overline{z} z^{\prime}\\ &=2\left(|z|^{2}+\left|z^{\prime}\right|^{2}-z \overline{z}^{\prime}-\overline{z} z^{\prime}\right)=2\left|z-z^{\prime}\right|^{2} . \end{aligned}\]

• Hence \[2\left(x_{1} x_{1}^{\prime}+x_{2} x_{2}^{\prime}+x_{3} x_{3}^{\prime}\right)=2-\frac{4\left|z-z^{\prime}\right|^{2}}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)}.\]

• Since \(d\left(z, z^{\prime}\right)^{2}=2-2\left(x_{1} x_{1}^{\prime}+x_{2} x_{2}^{\prime}+x_{3} x_{3}^{\prime}\right)\), we obtain \[d\left(z, z^{\prime}\right)^2=\frac{4\left|z-z^{\prime}\right|^{2}}{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)}\]

• Hence \[d\left(z, z^{\prime}\right)=\frac{2\left|z-z^{\prime}\right|}{\sqrt{\left(|z|^{2}+1\right)\left(\left|z^{\prime}\right|^{2}+1\right)}}\]

• If \(z^{\prime}=\infty\). Then \(x_{1}^{\prime}=0, x_{2}^{\prime}=0\) and \(x_{3}^{\prime}=1\) and \[d(z, \infty)^{2}=2-2 x_{3}=2-2\left(\frac{|z|^{2}-1}{|z|^{2}+1}\right)=\frac{4}{|z|^{2}+1} .\]

• Hence \[d(z, \infty)=\frac{2}{\sqrt{|z|^{2}+1}}.\]

• Thus \(\mathbb{C}_{\infty}\) is a metric space.

• The above metric \(d\) induces the following topology on \(\mathbb{C}_{\infty}\):

  • Let \(E \subseteq \mathbb{C}_{\infty}\).

  • If \(\infty \notin E\), then E is open in \(\mathbb{C}_{\infty}\) if and only if it is open in \(\mathbb{C}\).

  • If \(\infty \in E\), then \(E\) is open in \(\mathbb{C}_{\infty}\) if and only its complement in \(\mathbb{C}_{\infty}\) is a closed and bounded (i.e. compact) subset of \(\mathbb{C}\).

• Finally, we check that \(f\) and \(f^{-1}\) are continuous to conclude the following result.

Functions on the complex plane

• Let \(E\subseteq\mathbb C\) be a set of complex numbers. A function \(f\) defined on \(E\) is a rule that assigns to each \(z\) in \(E\) a unique complex number \(w\).

• The number \(w\) is called the value of \(f\) at \(z\) and is denoted by \(f(z)\). The set \(E\) is called the domain of definition of \(f\).

• Suppose \(w = u + iv\) is the value of a function \(f\) at \(z = x + iy\), meaning \[u + iv = f(x + iy).\]

• The real numbers \(u\) and \(v\) depend on the real variables \(x\) and \(y\).

• Consequently, \(f(z)\) can be expressed as a pair of real-valued functions of the real variables \(x\) and \(y\) as follows: \[f(z) = u(x,y) + iv(x,y).\]

• The functions \(u\) and \(v\) are usually denoted by \(\operatorname{Re} f(z)\) and \(\operatorname{Im} f(z)\) respectively.

Examples

• Consider the function \(f(z) = z^2\). Then \(f(x+iy) = x^2 - y^2 + i2xy\). Therefore, we have \(\operatorname{Re} f(z) = x^2 - y^2\) and \(\operatorname{Im} f(z) = 2xy\). The domain of \(f\) is the entire complex plane.

• Let us consider \(f(z) = z^{1/n}\), where we assign to each complex number its \(n\)-th root. Then \(f\) is not a function in the sense of the definition above, since for each \(z \neq 0\), there are \(n\) roots of \(z\).

• Let us express \(z\) as \(z = |z|(\cos\theta + i \sin\theta)\), where \(\theta \in (-\pi, \pi]\) is the argument of \(z\). Then the roots of \(z\) are given by \[z_k = \sqrt[n]{|z|} \left( \cos\frac{\theta + 2k\pi}{n} + i \sin\frac{\theta + 2k\pi}{n} \right), \quad k = 0, 1,\ldots, n-1.\]

• Clearly, for each \(z \in \mathbb{C}\), we have \(n\) values of \(z^{1/n}\) determined by \(k\). In this case, we say that the \(n\)-th root has \(n\) branches. We can make \(z^{1/n}\) a single-valued function by choosing a specific \(k\).

• A branch corresponding to \(k = 0\) is called the principal branch. Each branch is a single-valued function.

Continuity of complex functions

• The function \(f\) is said to be continuous on \(E\) if it is continuous at every point of \(E\).

• Sums and products of continuous functions are also continuous.

• The function \(f\) is continuous if and only if its real and imaginary parts are continuous.

Properties of continuous functions

• Since the convergence notions for the complex numbers and points in \(\mathbb{R}^2\) are the same, the function \(f\) of the complex argument \(z = x + i y\) is continuous if and only if it is continuous when viewed as a function of the two real variables \(x\) and \(y\).

• By the triangle inequality, it is immediate that if \(f\) is continuous, then the real-valued function defined by \(z \mapsto |f(z)|\) is continuous.

• We say that \(f\) attains a maximum at the point \(z_0 \in E\) if \[|f(z)| \leq \left| f(z_0) \right| \quad \text{for all } \quad z \in E,\] with the inequality reversed for the definition of a minimum.

  Theorem.

  A continuous complex function on a compact set \(E\) is bounded and attains a maximum and minimum on \(E\).

Examples

• The function \(f(z) = z^2\) is continuous on \(\mathbb{C}\), which can be easily shown using the definition of continuity.

• The principal branch of the square root function \(z^{1/2}\) is continuous on \(\mathbb{C} \setminus (-\infty, 0]\).

• Recall that \[z_k = \sqrt{|z|} \left( \cos\frac{\theta + 2k\pi}{2} + i \sin\frac{\theta + 2k\pi}{2} \right), \quad k = 0, 1.\]

• This exclusion is necessary because for \(z = x + iy\) where \(x \leq 0\) and \(y \to 0\) we may observe the following:

  • If \(y \to 0^+\), then \(\arg(z) \to \pi\), leading to \(z_0 \to i \sqrt{|x|}\).

  • If \(y \to 0^-\), then \(\arg(z) \to -\pi\), leading to \(z_0 \to -i \sqrt{|x|}\).

• We see that the principal branch \(z^{1/2}\) exhibits discontinuity depending on the direction of approach along the imaginary axis.

Complex Differentiation

Properties of holomorphic functions

• If \(f \in H(\Omega)\) and \(g \in H(\Omega)\), then \(f+g \in H(\Omega)\) and \(fg \in H(\Omega)\), so that \(H(\Omega)\) is a ring.

• Moreover, if \(f \in H(\Omega)\) and \(g \in H(\Omega)\) and additionally \(g\) is never \(0\) on \(\Omega\), then \(f/g \in H(\Omega)\).

• The usual differentiation rules apply: for any \(f \in H(\Omega)\) and \(g \in H(\Omega)\) and \(\alpha, \beta\in\mathbb C\) we have \[\begin{aligned} (\alpha f+\beta g)'&=\alpha f'+\beta g',\\ (fg)'&=f'g+fg', \end{aligned}\]

• If \(f \in H(\Omega)\) and \(g \in H(\Omega)\) and additionally \(g\) is never \(0\) on \(\Omega\), then \[\begin{aligned} \bigg(\frac{f}{g}\bigg)'=\frac{f'g-fg'}{g^2}. \end{aligned}\]

• More interestingly, superpositions of holomorphic functions are also holomorphic.

Proof: To prove this, fix \(z_{0} \in \Omega\), and put \(w_{0}=f\left(z_{0}\right)\). Then \[\begin{gathered} f(z)-f\left(z_{0}\right)=\left[f^{\prime}\left(z_{0}\right)+\varepsilon(z)\right]\left(z-z_{0}\right),\\ g(w)-g\left(w_{0}\right)=\left[g^{\prime}\left(w_{0}\right)+\eta(w)\right]\left(w-w_{0}\right), \end{gathered}\] where \(\lim_{z \to z_{0}}\varepsilon(z) = 0\) and \(\lim_{w \to w_{0}}\eta(w) = 0\). Put \(w=f(z)\), and substitute it to the last equation. If \(z \neq z_{0}\), we obtain \[\frac{h(z)-h\left(z_{0}\right)}{z-z_{0}}=\left[g^{\prime}\left(f\left(z_{0}\right)\right)+\eta(f(z))\right]\left[f^{\prime}\left(z_{0}\right)+\varepsilon(z)\right].\] The differentiability of \(f\) forces \(f\) to be continuous at \(z_{0}\). Hence the conclusion follows from the last equation.$$\tag*{$\blacksquare$}$$

Proof: Let \(z_{0} \in \Omega\), and let \((z_{n})_{n\in\mathbb N}\subseteq \Omega \backslash\left\{z_{0}\right\}\) with \(\lim_{n\to \infty}z_{n} =z_{0}\). Then \[\frac{f\left(z_{n}\right)-f\left(z_{0}\right)}{z_{n}-z_{0}}=\frac{f\left(z_{n}\right)-f\left(z_{0}\right)}{g\left(f\left(z_{n}\right)\right)-g\left(f\left(z_{0}\right)\right)}=\left[\frac{g\left(f\left(z_{n}\right)\right)-g\left(f\left(z_{0}\right)\right)}{f\left(z_{n}\right)-f\left(z_{0}\right)}\right]^{-1}\] (Note that \(f\left(z_{n}\right) \neq f\left(z_{0}\right)\) since \(f\) is 1-1 and \(z_{n} \neq z_{0}\).) By continuity of \(f\) at \(z_{0}\), the expression in brackets approaches \(g^{\prime}\left(f\left(z_{0}\right)\right)\) as \(n \rightarrow \infty\). Since \(g^{\prime}\left(f\left(z_{0}\right)\right) \neq 0\), the result follows. $$\tag*{$\blacksquare$}$$

Examples

• The function \(f(z) = z\) is holomorphic on any open set in \(\mathbb{C}\), and \(f^{\prime}(z) = 1\). In fact, any polynomial \[p(z) = a_{0} + a_{1} z + \cdots + a_{n} z^{n}\] is holomorphic in the entire complex plane and \[p^{\prime}(z) = a_{1} + 2a_{2} z + \cdots + n a_{n} z^{n-1}.\]

• The function \(\frac{1}{z}\) is holomorphic on any open set in \(\mathbb{C}\) that does not contain the origin, and \(f^{\prime}(z) = -\frac{1}{z^{2}}\).

• The function \(f(z) = \overline{z}\) is not holomorphic. Indeed, we have \[\frac{f\left(z_{0} + h\right) - f\left(z_{0}\right)}{h} = \frac{\overline{h}}{h}\] which has no limit as \(h \rightarrow 0\), as one can see by first taking \(h\) real and then \(h\) purely imaginary.

Real differentiability

• The relation (*) can be rewritten in the form \[F(x+h)-F(x)= F'(x)h +r(h), \quad \text{ and } \quad \lim_{h\to 0}\frac{|r(h)|}{|h|}=0. \tag{**}\]

• If \(F=(F_1,\ldots, F_m):E\to\mathbb R^m\) and \(F'(x_0)\) exists at \(x_0\in E\), then \[F'(x_0)= \begin{bmatrix} \frac{\partial F_1}{\partial x_1}(x_0) & \ldots & \frac{\partial F_1}{\partial x_n}(x_0)\\ \vdots & \ddots & \vdots\\ \frac{\partial F_m}{\partial x_1}(x_0) & \ldots & \frac{\partial F_m}{\partial x_n}(x_0) \end{bmatrix}.\]

• We know that if all the partial derivatives \(\frac{\partial F_1}{\partial x_1},\ldots, \frac{\partial F_i}{\partial x_j},\ldots, \frac{\partial F_m}{\partial x_n}\) of \(F\) exist in a neighborhood of \(x_0\in E\) and they are continues at the point \(x_0\), then the function \(F\) is differentiable at \(x_0\).

• The converse is not true. Consider the example \[F(x, y)= \begin{cases} \frac{y^3}{x^2+y^2} & \text{ if } (x, y)\neq(0,0),\\ 0& \text{ if } (x, y)=(0,0). \end{cases}\] The function \(F\) is not differentiable at the point \((0, 0)\), even though the partial derivatives \(\frac{\partial F}{\partial x}\) and \(\frac{\partial F}{\partial y}\) exist at this point. However, these partial derivatives are not continuous at the origin.

• Recall that the function \(f(z) = \overline{z}\) is not holomorphic. Indeed, we have \[\frac{f\left(z_{0} + h\right) - f\left(z_{0}\right)}{h} = \frac{\overline{h}}{h},\] which has no limit as \(h \rightarrow 0\), as one can see by first taking \(h\) real and then \(h\) purely imaginary.

• If the function \(f(z)=\overline{z}\) is considered as a real variable function then it corresponds to the map \(F(x, y) =(x,-y)\), which is indefinitely differentiable in the real sense.

• This example illustrates that the existence of the real derivative need not guarantee that \(f\) is holomorphic.

• This example leads us to associate more generally to each complex valued function \[f=u+i v,\] the mapping from \(F: E\to \mathbb{R}^{2}\), given by \[F(x, y)=(u(x, y), v(x, y)).\]

\(F(x, y)=(u(x, y), v(x, y))\)

• If \(m=n=2\) and \((x_0, y_0)\in E\) is fixed and \(h=(h_1, h_2)\) and \(|h|\to0\), then the relation (**) can be rewritten in the form \[\begin{aligned} u(x_0+h_1, y_0+h_2)-u(x_0, y_0)=& \frac{\partial u}{\partial x}(x_0, y_0)h_1 +\frac{\partial u}{\partial y}(x_0, y_0)h_2\\ &+r_1(h_1, h_2), \end{aligned}\] and \[\begin{aligned} v(x_0+h_1, y_0+h_2)-v(x_0, y_0)=& \frac{\partial v}{\partial x}(x_0, y_0)h_1 +\frac{\partial v}{\partial y}(x_0, y_0)h_2\\ &+r_2(h_1, h_2), \end{aligned}\] where \(r(h_1, h_2)=(r_1(h_1, h_2), r_2(h_1, h_2))\) and \[\lim_{h\to 0}\frac{|r_1(h_1, h_2)|}{|h|}=0, \quad \text{ and } \quad \lim_{h\to 0}\frac{|r_2(h_1, h_2)|}{|h|}=0.\]

Cauchy–Riemann equations

Proof: Fix \(z_0\in\Omega\). Since \(f: \Omega\to \mathbb C\) is holomorphic, we have \[\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=f'(z_0). \tag{C}\]

• To find relation (A), observe that \(z_0=(x_0, y_0)\) can be approached from different directions.

• Firstly, with \(z\) of the form \((x_0+h)+iy_0\) in (C), observe that \[f'(z_0)=\lim_{h\to 0}\frac{f(x_0+h, y_0)-f(x_0, y_0)}{h}=\frac{\partial f}{\partial x}.\]

• Secondly, with \(z\) of the form \(x_0+i(y_0+h)\) in (C), observe that \[f'(z_0)=\lim_{h\to 0}\frac{f(x_0, y_0+h)-f(x_0, y_0)}{ih}=\frac{1}{i}\frac{\partial f}{\partial y}.\]

• Thus we obtain equation (A): \[\frac{\partial f}{\partial x}=\frac{1}{i}\frac{\partial f}{\partial y}.\]

• If \(f=u+iv\) for some real valued functions \(u, v:\Omega\to \mathbb C\), then \[\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}.\]

• This implies equations in (B): \[\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \quad \text{ and } \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.\]

• This completes the proof.$$\tag*{$\blacksquare$}$$

Cauchy–Riemann equations and holomorphicity

Proof: Fix \(z_0=x_0+iy_0\in\Omega\) and let \(h=h_1+ih_2\). Recall that \[\begin{aligned} u(x_0+h_1, y_0+h_2)-u(x_0, y_0)= \frac{\partial u}{\partial x}(x_0, y_0)h_1 +\frac{\partial u}{\partial y}(x_0, y_0)h_2+r_1(h_1, h_2), \end{aligned}\] and \[\begin{aligned} v(x_0+h_1, y_0+h_2)-v(x_0, y_0)= \frac{\partial v}{\partial x}(x_0, y_0)h_1 +\frac{\partial v}{\partial y}(x_0, y_0)h_2+r_2(h_1, h_2), \end{aligned}\] where \[\lim_{h\to 0}\frac{|r_1(h_1, h_2)|}{|h|}=0, \quad \text{ and } \quad \lim_{h\to 0}\frac{|r_2(h_1, h_2)|}{|h|}=0.\]

• Thus \[\begin{aligned} f(z_0+h)-f(z_0)=&\frac{\partial u}{\partial x}(x_0, y_0)h_1 +\frac{\partial u}{\partial y}(x_0, y_0)h_2\\ &+i\bigg( \frac{\partial v}{\partial x}(x_0, y_0)h_1 +\frac{\partial v}{\partial y}(x_0, y_0)h_2\bigg)+r(h), \end{aligned}\] where \(r(h)=r_1(h_1, h_2)+ir_2(h_1, h_2)\). Clearly \(\lim_{h\to 0}\frac{|r(h)|}{|h|}=0\).

• By the Cauchy–Riemann equations \(\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\), \(\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\), we find that \[f(z_0+h)-f(z_0)=\bigg(\frac{\partial u}{\partial x}(x_0, y_0)-i\frac{\partial u}{\partial y}(x_0, y_0)\bigg)h+r(h).\]

• Then \[f'(z_0)=\frac{\partial u}{\partial x}(x_0, y_0)-i\frac{\partial u}{\partial y}(x_0, y_0). \quad \tag*{$\blacksquare$}\]

Example

• Note that the hypothesis of real differentiability at the point \(z_0\) is essential and cannot be dispensed with.

• For example, consider the function \[f ( x , y ) = \sqrt{| x y |},\] regarded as a complex function with imaginary part identically zero.

  • Then it has both partial derivatives at \(( x_0 , y_0 ) = ( 0 , 0 )\).

  • It moreover satisfies the Cauchy–Riemann equations at that point.

  • But it is not differentiable in the real sense, and so the first condition, that of real differentiability, is not met.

• Therefore, this function is not complex differentiable.