Let f f f be a function whose domain contains a neighborhood ∣ z − z 0 ∣ < ϵ |z-z_0| < \epsilon ∣ z − z 0 ∣ < ϵ of z 0 z_0 z 0 . The derivative of f f f at z 0 z_0 z 0 is
f ′ ( z 0 ) = lim z → z 0 f ( z ) − f ( z 0 ) z − z 0 f'(z_0) = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0} f ′ ( z 0 ) = z → z 0 lim z − z 0 f ( z ) − f ( z 0 )
provided this limit exists. f f f is differentiable at z 0 z_0 z 0 when f ′ ( z 0 ) f'(z_0) f ′ ( z 0 ) exists.
If f f f is differentiable at z 0 z_0 z 0 , then f f f is continuous at z 0 z_0 z 0 .
Let c ∈ C c \in \mathbb{C} c ∈ C , and let f f f , g g g be functions differentiable at z ∈ C z \in \mathbb{C} z ∈ C . Then:
d d z c = 0 \dfrac{\text{d}}{\text{d}z}\,c = 0 d z d c = 0
d d z ( c f ( z ) ) = c f ′ ( z ) \dfrac{\text{d}}{\text{d}z}\,(cf(z)) = cf'(z) d z d ( c f ( z )) = c f ′ ( z )
d d z z n = n z n − 1 \dfrac{\text{d}}{\text{d}z}\,z^n = nz^{n-1} d z d z n = n z n − 1 , for positive integer n n n
d d z ( f ( z ) + g ( z ) ) = f ′ ( z ) + g ′ ( z ) \dfrac{\text{d}}{\text{d}z}\,(f(z)+g(z)) = f'(z)+g'(z) d z d ( f ( z ) + g ( z )) = f ′ ( z ) + g ′ ( z )
d d z ( f ( z ) g ( z ) ) = f ( z ) g ′ ( z ) + f ′ ( z ) g ( z ) \dfrac{\text{d}}{\text{d}z}\,(f(z)g(z)) = f(z)g'(z)+f'(z)g(z) d z d ( f ( z ) g ( z )) = f ( z ) g ′ ( z ) + f ′ ( z ) g ( z )
d d z f ( z ) g ( z ) = g ( z ) f ′ ( z ) − f ( z ) g ′ ( z ) ( g ( z ) ) 2 \dfrac{\text{d}}{\text{d}z}\,\dfrac{f(z)}{g(z)} = \dfrac{g(z)f'(z)-f(z)g'(z)}{(g(z))^2} d z d g ( z ) f ( z ) = ( g ( z ) ) 2 g ( z ) f ′ ( z ) − f ( z ) g ′ ( z ) , provided g ( z ) ≠ 0 g(z) \ne 0 g ( z ) = 0
Cauchy-Riemann Equations
Let f ( z ) = u ( x , y ) + i v ( x , y ) f(z) = u(x,y)+iv(x,y) f ( z ) = u ( x , y ) + i v ( x , y ) , and suppose f ′ ( z 0 ) f'(z_0) f ′ ( z 0 ) exists at z 0 = x 0 + i y 0 z_0 = x_0+iy_0 z 0 = x 0 + i y 0 . Then the first-order partial derivatives of u u u and v v v exist at ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) and satisfy
u x = v y , u y = − v x u_x = v_y, \qquad u_y = -v_x u x = v y , u y = − v x
at ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) , and
f ′ ( z 0 ) = u x ( x 0 , y 0 ) + i v x ( x 0 , y 0 ) f'(z_0) = u_x(x_0,y_0)+iv_x(x_0,y_0) f ′ ( z 0 ) = u x ( x 0 , y 0 ) + i v x ( x 0 , y 0 )
The equations follow from forcing the derivative limit to agree along the real and imaginary directions of approach.
Proof Hint
Take Δ z = Δ x \Delta z = \Delta x Δ z = Δ x in the limit, get f ′ ( z 0 ) = u x + i v x f'(z_0) = u_x + iv_x f ′ ( z 0 ) = u x + i v x
Take Δ z = i Δ y \Delta z = i\,\Delta y Δ z = i Δ y in the limit, get f ′ ( z 0 ) = v y − i u y f'(z_0) = v_y - iu_y f ′ ( z 0 ) = v y − i u y
Equate real parts and imaginary parts
The equations are necessary but not sufficient. They can hold at z 0 z_0 z 0 while f ′ ( z 0 ) f'(z_0) f ′ ( z 0 ) does not exist. See Sufficient Conditions for Differentiability .
The Jacobian of ( x , y ) ↦ ( u , v ) (x,y) \mapsto (u,v) ( x , y ) ↦ ( u , v ) is then
( u x − v x v x u x ) \begin{pmatrix} u_x & -v_x \\ v_x & u_x \end{pmatrix} ( u x v x − v x u x )
which acts as multiplication by u x + i v x u_x + iv_x u x + i v x . A complex-differentiable map is locally a rotation plus a uniform scaling.
Examples:
f ( z ) = z 2 f(z) = z^2 f ( z ) = z 2 : u x = v y = 2 x u_x = v_y = 2x u x = v y = 2 x , u y = − v x = − 2 y u_y = -v_x = -2y u y = − v x = − 2 y , and f ′ ( z ) = 2 x + i 2 y = 2 z f'(z) = 2x + i2y = 2z f ′ ( z ) = 2 x + i 2 y = 2 z .
f ( z ) = z ˉ f(z) = \bar{z} f ( z ) = z ˉ : u x = 1 u_x = 1 u x = 1 , v y = − 1 v_y = -1 v y = − 1 , so the equations fail everywhere and z ˉ \bar{z} z ˉ is nowhere differentiable.
Sufficient Conditions for Differentiability
Let f ( z ) = u ( x , y ) + i v ( x , y ) f(z) = u(x,y)+iv(x,y) f ( z ) = u ( x , y ) + i v ( x , y ) be defined throughout a neighborhood U U U of z 0 = x 0 + i y 0 z_0 = x_0+iy_0 z 0 = x 0 + i y 0 . f ′ ( z 0 ) f'(z_0) f ′ ( z 0 ) exists iff :
The first-order partial derivatives of u u u and v v v with respect to x x x and y y y exist everywhere in U U U .
Those partial derivatives are continuous at ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) and satisfy the Cauchy-Riemann equations at ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) .
When these hold,
f ′ ( z 0 ) = u x ( x 0 , y 0 ) + i v x ( x 0 , y 0 ) f'(z_0) = u_x(x_0,y_0)+iv_x(x_0,y_0) f ′ ( z 0 ) = u x ( x 0 , y 0 ) + i v x ( x 0 , y 0 )
Let f ( z ) = u ( r , θ ) + i v ( r , θ ) f(z) = u(r,\theta)+iv(r,\theta) f ( z ) = u ( r , θ ) + i v ( r , θ ) be defined throughout a neighborhood of a non-zero point z 0 = r 0 e i θ 0 z_0 = r_0e^{i\theta_0} z 0 = r 0 e i θ 0 . f ′ ( z 0 ) f'(z_0) f ′ ( z 0 ) exists iff :
The first-order partial derivatives of u u u and v v v with respect to r r r and θ \theta θ exist everywhere in the corresponding polar-coordinate neighborhood.
Those partial derivatives are continuous at ( r 0 , θ 0 ) (r_0,\theta_0) ( r 0 , θ 0 ) and satisfy
r u r = v θ , u θ = − r v r ru_r = v_\theta, \qquad u_\theta = -rv_r r u r = v θ , u θ = − r v r
at ( r 0 , θ 0 ) (r_0,\theta_0) ( r 0 , θ 0 ) .
When these hold,
f ′ ( z 0 ) = e − i θ 0 ( u r ( r 0 , θ 0 ) + i v r ( r 0 , θ 0 ) ) f'(z_0) = e^{-i\theta_0}(u_r(r_0,\theta_0)+iv_r(r_0,\theta_0)) f ′ ( z 0 ) = e − i θ 0 ( u r ( r 0 , θ 0 ) + i v r ( r 0 , θ 0 ))