📂Complex Anaylsis

The Conditions for the Inverses of Cauchy-Riemann Equations to Hold

Theorem

In the complex domain □ref.□ defined by complex function □ref.□ that takes real values for function □ref.□, $$f(z) = f(x+iy) = u(x,y) + iv(x,y)$$ can be represented, and if □ref.□ has a continuous first-order partial derivative with respect to □ref.□ and satisfies the system of differential equations, $$\begin{cases} u_{x} (x,y) = v_{y} (x,y) \\ u_{y} (x,y) = -v_{x} (x,y) \end{cases}$$ then □ref.□ is analytic at □ref.□.

Explanation

Analysis always has the problem of being too wordy and unappealing to read. In short, for the inverse of Cauchy-Riemann equations to hold, the partial derivatives need to be continuous. Naturally, most functions we deal with meet this condition easily.

Proof

Part 1. Form of □ref.□

Without loss of generality, for two real numbers □ref.□, if we set □ref.□, then for □ref.□, $$f(z+h) - f(z) = [ u(x+\alpha,y+\beta) - u(x,y) ] + i [ v(x+\alpha,y+\beta) - v(x,y) ]$$ can be obtained. Similarly, without loss of generality, if we only consider □ref.□, then by the Mean Value Theorem, $$\begin{align*} & u(x+\alpha,y+\beta) - u(x,y) \\ =& u(x+\alpha,y+\beta) - u(x,y+\beta) + u(x,y+\beta) - u(x,y) \\ =& [u(x+\alpha,y+\beta) - u(x,y+\beta)] + [u(x,y+\beta) - u(x,y)] \\ =& \alpha u_{x} (x+\theta \alpha,y+\beta) + \beta u_{y} (x,y+\phi \beta) \end{align*}$$ there exist □ref.□ and □ref.□ that satisfy.

Mean Value Theorem: If a function □ref.□ is continuous on □ref.□ and differentiable on □ref.□, then there exists at least one □ref.□ in □ref.□ that satisfies □ref.□.

Here, setting □ref.□ is because when using the Mean Value Theorem on □ref.□ at □ref.□, $$\begin{align*} {{ \partial u } \over { \partial x }} \left( c , y + \beta \right) =& {{ u(x+\alpha,y+\beta) - u(x,y+\beta) } \over { \left( x + \alpha \right) - x }} \\ \implies \alpha u_{x} \left( c , y + \beta \right) =& u(x+\alpha,y+\beta) - u(x,y+\beta) \end{align*}$$ □ref.□ can be represented as a number □ref.□ between □ref.□ and □ref.□.

---

Part 2. Form of □ref.□

Now, for some □ref.□ and □ref.□, $$\begin{align*} u_{x} (x+\theta \alpha,y+\beta) =& u_{x} + \varepsilon_{1} \\ u_{y} (x,y+\phi \beta) =& u_{y} + \varepsilon_{2} \end{align*}$$ if set, since it was stated in the assumption that □ref.□ and □ref.□ are continuous, then when □ref.□, it will be □ref.□. (This is why continuity is necessary.) Therefore, $$u(x+\alpha,y+\beta) - u(x,y) = \alpha u_{x} + \beta u_{y} + \alpha \varepsilon_{1} + \beta \varepsilon_{2}$$ and in the same way, for some □ref.□ and □ref.□, $$v(x+\alpha,y+\beta) - v(x,y) = \alpha v_{x} + \beta v_{y} + \alpha \eta_{1} + \beta \eta_{2}$$ can also be set. Returning to □ref.□, since □ref.□ satisfies the Cauchy-Riemann equations, $$\begin{align*} & f(z+h) - f(z) \\ =& [ u(x+\alpha,y+\beta) - u(x,y) ] + i [ v(x+\alpha,y+\beta) - v(x,y) ] \\ =& [\alpha u_{x} + \beta u_{y} + \alpha \varepsilon_{1} + \beta \varepsilon_{2}] + i [\alpha v_{x} + \beta v_{y} + \alpha \eta_{1} + \beta \eta_{2}] \\ =& h(u_{x} + i v_{x}) + \alpha \xi_{1} + \beta \xi_{2} \end{align*}$$ where □ref.□ and □ref.□. Now, $$f ' (z) = \lim_{h \to 0} {{f(z+h) - f(z)} \over h} = \lim_{h \to 0} \left( u_{x} + i v_{x} + {{\alpha \xi_{1} + \beta \xi_{2}} \over h} \right)$$ thus, proving □ref.□ concludes the proof.

---

Part 3. □ref.□

From the inequality $$\left| {{\alpha \xi_{1} + \beta \xi_{2}} \over {h}} \right| \le { {\max (|\alpha|,|\beta|)} \over {\sqrt{\alpha^2 + \beta^2}} } |\xi_{1} + \xi_{2}| \le |\xi_{1} + \xi_{2}| \le |\xi_{1}| + |\xi_{2}|$$ since $$\lim_{h \to 0} \xi_{1} = 0 \\ \lim_{h \to 0} \xi_{2} = 0$$ the following holds: $$\lim_{h \to 0} {{\alpha \xi_{1} + \beta \xi_{2}} \over h} = 0$$

■

Renewal

• August 19, 2023, by Daesik Ryu, Focused reinforcement on the Mean Value Theorem content