Let Ω⊆C\Omega \subseteq \mathcal{C} be a open set and f:Ω→Cf : \Omega \rightarrow \mathcal{C} be a complex function. ff is holomorphic at z∈Ωz \in \Omega, if the following limit exits:

lim⁡z+h∈Ω,h→0f(z+h)−f(z)h.(1) \begin{array}{lr} \lim_{z + h \in \Omega, h \rightarrow 0} \frac{ f(z + h) - f(z) }{ h }. & (1) \end{array}

The limit is denoted as f′(z)f'(z) and is called the derivative of ff at zz. Note that hh can approach 00 from any direction.

Define

∂f∂x(z)≐lim⁡x∈R,z+x∈Ω,x→0f(z+x)−f(z)x, { \frac{\partial f}{ \partial x} (z) \doteq \lim_{x \in \mathbb{R}, z + x \in \Omega, x \rightarrow 0} \frac{ f(z + x) - f(z) }{ x }, }

which is the limit of (1)(1) when hh approaches to 00 from the real axis.

Similarly, define when limit when hh is purely imaginary:

∂f∂y(z)≐lim⁡y∈R,z+y∈Ω,y→0f(z+i⋅y)−f(z)i⋅y. \small \frac{\partial f}{ \partial y} (z) \doteq \lim_{y \in \mathbb{R}, z + y \in \Omega, y \rightarrow 0} \frac{ f(z + i \cdot y) - f(z) }{ i \cdot y }.

If the derivate of ff exists at zz, we have shown that

∂f∂x=∂f∂y.(2) \begin{aligned} \frac{\partial f}{ \partial x} = \frac{\partial f}{ \partial y}. &\qquad (2) \end{aligned}

Writing f=u+ivf = u + iv, we get

∂f∂x=∂u∂x+i⋅∂v∂x,∂f∂y=1i⋅(∂u∂y+i⋅∂v∂y)=−i⋅∂u∂y+∂v∂y. \begin{aligned} \frac{\partial f}{ \partial x} &= \frac{\partial u}{ \partial x} + i \cdot \frac{\partial v}{ \partial x}, \\ \frac{\partial f}{ \partial y} &= \frac{1}{i} \cdot \left( \frac{\partial u}{ \partial y} + i \cdot \frac{\partial v}{ \partial y} \right) = -i \cdot \frac{\partial u}{ \partial y} + \frac{\partial v}{ \partial y}. \end{aligned}

Equation (2) implies that

∂u∂x=∂v∂y,∂v∂x=−∂u∂y. \frac{\partial u}{ \partial x} = \frac{\partial v}{ \partial y}, \qquad \frac{\partial v}{ \partial x} = - \frac{\partial u}{ \partial y}.

These are the Cauchy-Riemann equations. The next theorem states that the converse is also true, i.e., when the equations are satisfies at some point zz, then the derivative of f=u+ivf = u + iv exists.

Theorem. Let u,v:Ω→Ru, v : \Omega \rightarrow \mathbb{R} and f=u+ivf = u + iv. If uu and vv are continuously differentiable at z∈Ωz \in \Omega and satisfy Cauchy-Riemann equations, then the derivative of ff exits.

Proof. Since uu and vv are continuously differentiable at z=(x,y)z = (x, y), for small Δz=(Δx,Δy)∈R2\Delta z = (\Delta x, \Delta y) \in \mathbb{R}^2,

u(x+Δx,y+Δy)−u(x,y)=∂u∂x(z)⋅Δx+∂u∂y(z)⋅Δy+∥Δz∥1⋅Φu(Δz),v(x+Δx,y+Δy)−u(x,y)=∂v∂x(z)⋅Δx+∂v∂y(z)⋅Δy+∥Δz∥1⋅Φv(Δz) \begin{aligned} u( x + \Delta x , y + \Delta y ) - u(x, y) &= \frac{\partial u}{ \partial x}(z) \cdot \Delta x + \frac{\partial u}{ \partial y}(z) \cdot \Delta y + \Vert \Delta z \Vert_1 \cdot \Phi_u( \Delta z), \\ v( x + \Delta x , y + \Delta y ) - u(x, y) &= \frac{\partial v}{ \partial x}(z) \cdot \Delta x + \frac{\partial v}{ \partial y}(z) \cdot \Delta y + \Vert \Delta z \Vert_1 \cdot \Phi_v( \Delta z) \end{aligned}

where Φu(Δz)→0\Phi_u( \Delta z) \rightarrow 0 and Φv(Δz)→0\Phi_v( \Delta z) \rightarrow 0 as ∥Δz∥1→0\Vert \Delta z \Vert_1 \rightarrow 0.

Therefore,

f(z+Δz)−f(z)=u(x+Δx,y+Δy)−u(x,y)+i(v(x+Δx,y+Δy)−u(x,y))=∂u∂x(z)⋅(Δx+iΔy)+∂u∂y(z)⋅(Δy−iΔx)+∥Δz∥1⋅(Φu(Δz)+Φv(Δz))=(∂u∂x(z)−i∂u∂y(z))⋅(Δx+iΔy)+∥Δz∥1⋅(Φu(Δz)+Φv(Δz)), \small \begin{aligned} f( z + \Delta z ) - f( z ) &= u( x + \Delta x , y + \Delta y ) - u(x, y) + i \Big( v( x + \Delta x , y + \Delta y ) - u(x, y) \Big) \\ &= \frac{\partial u}{ \partial x}(z) \cdot \big( \Delta x + i \Delta y \big) + \frac{\partial u}{ \partial y}(z) \cdot \big( \Delta y - i \Delta x \big) + \Vert \Delta z \Vert_1 \cdot \Big( \Phi_u( \Delta z) + \Phi_v( \Delta z) \Big) \\ &= \Big( \frac{\partial u}{ \partial x}(z) - i \frac{\partial u}{ \partial y}(z) \Big) \cdot \big( \Delta x + i \Delta y \big) + \Vert \Delta z \Vert_1 \cdot \Big( \Phi_u( \Delta z) + \Phi_v( \Delta z) \Big), \end{aligned}

which finishes the proof.

□\square

Reference

[1] E. M. Stein and R. Shakarchi, Complex analysis. Princeton, N.J: Princeton University Press, 2003.