Real Derivative and Complex Derivative ¶ A complex function f : C → C f:\mathbb{C}\to \mathbb{C} f : C → C R 2 → R 2 \mathbb{R}^2\to \mathbb{R}^2 R 2 → R 2 f ( z ) = exp ( z ) f(z)=\exp(z) f ( z ) = exp ( z )
f ( x y ) = ( exp ( x ) cos ( y ) exp ( x ) sin ( y ) ) f\begin{pmatrix}
x \\
y
\end{pmatrix} =
\begin{pmatrix}
\exp(x)\cos(y) \\
\exp(x)\sin(y)
\end{pmatrix} f ( x y ) = ( exp ( x ) cos ( y ) exp ( x ) sin ( y ) ) Since the complex version can be rewritten as exp ( z ) = exp ( x + i y ) = exp ( x ) ( cos ( y ) + i sin ( y ) ) \exp(z)=\exp(x+iy)=\exp(x)(\cos(y) + i\sin(y)) exp ( z ) = exp ( x + i y ) = exp ( x ) ( cos ( y ) + i sin ( y ))
However, there does exist crucial differencies. Recall for real functions ( R 2 , R 2 ) (\mathbb{R}^2, \mathbb{R}^2) ( R 2 , R 2 ) A ∈ L ( R 2 , R 2 ) A\in \mathcal L (\mathbb{R}^2, \mathbb{R}^2) A ∈ L ( R 2 , R 2 )
lim x → x 0 ∥ f ( x 0 + h ) − f ( x 0 ) − A h ∥ ∥ h ∥ = 0 \lim_{x\to x_0} \frac{\|f(x_0+h)-f(x_0) - Ah\|}{\|h\|}=0 x → x 0 lim ∥ h ∥ ∥ f ( x 0 + h ) − f ( x 0 ) − A h ∥ = 0 If such A A A Jacobian D f ( x 0 ) Df(x_0) D f ( x 0 )
( ∂ f i ∂ x j ( x 0 ) ) i j ∈ R m × n \left(\frac{\partial f_i}{\partial x_j}(x_0)\right)_{ij} \in \mathbb{R}^{m\times n} ( ∂ x j ∂ f i ( x 0 ) ) ij ∈ R m × n Furthermore, the sufficient condition for this linear transforamtion A A A
Complex derivative, on the other hand, is defined as:
Let f : C → C f:\mathbb{C} \to \mathbb{C} f : C → C derivative of f f f z 0 z_0 z 0
d f d z ( z 0 ) = lim Δ z → 0 f ( z 0 + Δ z ) − f ( z 0 ) Δ z \frac{df}{dz}({z_0})= \lim_{\Delta z\to 0}\frac{f(z_0 + \Delta z) -f(z_0)}{\Delta z} d z df ( z 0 ) = Δ z → 0 lim Δ z f ( z 0 + Δ z ) − f ( z 0 ) Or,
d f d z ( z 0 ) = lim Δ z → 0 u ( z + Δ z ) − u ( z ) Δ z + lim Δ z → 0 v ( z + Δ z ) − v ( z ) Δ z i \frac{df}{dz}({z_0})= \lim_{\Delta z\to 0}\frac{u(z+\Delta z) -u(z)}{\Delta z} + \lim_{\Delta z\to 0}\frac{v(z+\Delta z) -v(z)}{\Delta z} i d z df ( z 0 ) = Δ z → 0 lim Δ z u ( z + Δ z ) − u ( z ) + Δ z → 0 lim Δ z v ( z + Δ z ) − v ( z ) i where u : C → R u : \mathbb{C} \to \mathbb{R} u : C → R v v v
This is very different from f : R 2 → R 2 f:\mathbb{R}^2 \to \mathbb{R}^2 f : R 2 → R 2
If we insist write the same definition for the real analog(i.e. denote a complex number a + b i a+bi a + bi ( a b ) a \choose b ( b a ) f 1 ↔ u , f 2 ↔ v f_1\leftrightarrow u, f_2\leftrightarrow v f 1 ↔ u , f 2 ↔ v
lim ( x , y ) T → ( x 0 , y 0 ) T ( ∂ f 1 ∂ ( x , y ) ∂ f 2 ∂ ( x , y ) ) \lim_{(x,y)^T \to (x_0,y_0)^T}
\begin{pmatrix}
\frac{\partial f_1}{\partial (x, y)} \\
\frac{\partial f_2}{\partial (x, y)}
\end{pmatrix} ( x , y ) T → ( x 0 , y 0 ) T lim ( ∂ ( x , y ) ∂ f 1 ∂ ( x , y ) ∂ f 2 ) Which doesn’t make sense at all(Although one could argue ∂ f i ∂ ( x , y ) \frac{\partial f_i}{\partial (x,y)} ∂ ( x , y ) ∂ f i ( ∇ f i ) T (\nabla f_i)^T ( ∇ f i ) T
To make it worse, even if we pretend such notation is sensible, they are still different. The limit of the real part lim Δ z → 0 u ( z + Δ z ) − u ( z ) Δ z \lim_{\Delta z\to 0}\frac{u(z+\Delta z)-u(z)}{\Delta z} lim Δ z → 0 Δ z u ( z + Δ z ) − u ( z ) ∂ f 1 ∂ ( x , y ) \frac{\partial f_1}{\partial (x,y)} ∂ ( x , y ) ∂ f 1
So the real difference is that complex number should be viewed as a whole where division is well-defined. For me, if I ever found it hard to convince myself R 2 → R 2 \mathbb{R}^2 \to \mathbb{R}^2 R 2 → R 2 C → C \mathbb{C} \to \mathbb{C} C → C ( a , b ) (a,b) ( a , b )
The next example illustrate this key difference:
Consider f ( x + i y ) = x 2 + i y f(x+iy)=x^2 + iy f ( x + i y ) = x 2 + i y Interactive Plot Of The Real Analog
f ( x y ) = ( x 2 y ) f\begin{pmatrix}
x \\
y
\end{pmatrix}=
\begin{pmatrix}
x^2 \\
y
\end{pmatrix} f ( x y ) = ( x 2 y ) This function is clearly differentiable since the jacobian is:
( 2 x 0 0 1 ) \begin{pmatrix}
2x & 0 \\
0 & 1
\end{pmatrix} ( 2 x 0 0 1 ) which clearly continuous(component-wise) for all x , y x,y x , y
However, this is not true for the complex analog. We now show the complex function is not differentiable at 0 ∈ C 0\in \mathbb{C} 0 ∈ C
Suppose it is, then the two limits below should be equal at 0 ∈ C 0\in \mathbb{C} 0 ∈ C
lim Δ x → 0 f ( x 0 + Δ x + y 0 i ) − f ( x 0 + y 0 i ) Δ x = lim Δ y → 0 f ( x 0 + ( y 0 + Δ y ) ) − f ( x 0 + y 0 i ) i Δ y \lim_{\Delta x\to 0}\frac{f(x_0 + \Delta x + y_0i) -f(x_0 + y_0 i)}{\Delta x} = \lim_{\Delta y\to 0}\frac{f(x_0 + (y_0 +\Delta y)) -f(x_0 + y_0 i)}{i\Delta y } Δ x → 0 lim Δ x f ( x 0 + Δ x + y 0 i ) − f ( x 0 + y 0 i ) = Δ y → 0 lim i Δ y f ( x 0 + ( y 0 + Δ y )) − f ( x 0 + y 0 i ) But computing the limit tells us that:
Unless x 0 = 1 2 x_0 = \frac{1}{2} x 0 = 2 1 x 0 = 1 2 x_0=\frac12 x 0 = 2 1
One might ask, doesn’t the seem problem exists for the real analog? It’s quite obvious the answer is no, since we never compute a single limit of a finite difference.
In a way, C → C \mathbb{C} \to \mathbb{C} C → C R → R \mathbb{R}\to \mathbb{R} R → R
lim Δ z → 0 u ( z + Δ z ) − u ( z ) Δ z , lim Δ z → 0 v ( z + Δ z ) − v ( z ) Δ z \lim_{\Delta z\to 0}\frac{u(z+\Delta z) -u(z)}{\Delta z}, \lim_{\Delta z\to 0}\frac{v(z+\Delta z) -v(z)}{\Delta z} Δ z → 0 lim Δ z u ( z + Δ z ) − u ( z ) , Δ z → 0 lim Δ z v ( z + Δ z ) − v ( z ) resembles the R 2 → R \mathbb{R}^2 \to \mathbb{R} R 2 → R
import matplotlib.pyplot as plt
import numpy as np
import ipywidgets as wdgt
slider = wdgt.FloatSlider(value=1, min=1, max=3, step=0.5)
def plot(r):
fig, ax = plt.subplots()
X, Y = np.meshgrid(np.linspace(-r, r, 10), np.linspace(-r, r, 10))
ax.quiver(X, Y, X**2, Y)
plt.show()
wdgt.interactive(plot, r=slider)
Cauchy-Riemann Condition ¶ Expanding (9)
lim Δ x → 0 f ( x 0 + Δ x + y 0 i ) − f ( x 0 + y 0 i ) Δ x = lim Δ y → 0 f ( x 0 + ( y 0 + Δ y i ) ) − f ( x 0 + y 0 i ) Δ y lim Δ x → 0 u ( x 0 + Δ x , y 0 ) − u ( x 0 , y 0 ) Δ x + i v ( x 0 + Δ x , y 0 ) − v ( x 0 , y 0 ) Δ x = lim Δ y → 0 u ( x 0 , y 0 + Δ y ) − u ( x 0 , y 0 ) i Δ y + i v ( x 0 , y 0 + Δ y ) − v ( x 0 , y 0 ) i Δ y ∂ u ∂ x + i ∂ v ∂ x = − ∂ u ∂ y i + ∂ v ∂ y \begin{align*}
\lim_{\Delta x\to 0} \frac{f(x_0+\Delta x + y_0i) - f(x_0+y_0i)}{\Delta x} &= \lim_{\Delta y\to 0} \frac{f(x_0 + (y_0+\Delta yi)) - f(x_0+y_0i)}{\Delta y} \\
\lim_{\Delta x\to 0}\frac{u(x_0+\Delta x, y_0)-u(x_0,y_0)}{\Delta x} + i \frac{v(x_0+\Delta x, y_0)-v(x_0,y_0)}{\Delta x} &=
\lim_{\Delta y\to 0}\frac{u(x_0, y_0+\Delta y)-u(x_0,y_0)}{i\Delta y} + i \frac{v(x_0, y_0+\Delta y)-v(x_0,y_0)}{i\Delta y}\\
\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} &= -\frac{\partial u}{\partial y}i + \frac{\partial v}{\partial y}
\end{align*} Δ x → 0 lim Δ x f ( x 0 + Δ x + y 0 i ) − f ( x 0 + y 0 i ) Δ x → 0 lim Δ x u ( x 0 + Δ x , y 0 ) − u ( x 0 , y 0 ) + i Δ x v ( x 0 + Δ x , y 0 ) − v ( x 0 , y 0 ) ∂ x ∂ u + i ∂ x ∂ v = Δ y → 0 lim Δ y f ( x 0 + ( y 0 + Δ y i )) − f ( x 0 + y 0 i ) = Δ y → 0 lim i Δ y u ( x 0 , y 0 + Δ y ) − u ( x 0 , y 0 ) + i i Δ y v ( x 0 , y 0 + Δ y ) − v ( x 0 , y 0 ) = − ∂ y ∂ u i + ∂ y ∂ v Note that this equation essentially unwind the “mixing of components” we mentioned in Difference 2 : Mixing Of Components
∂ u ∂ x = ∂ v ∂ y ∂ v ∂ x = − ∂ u ∂ y \boxed{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad
\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}} ∂ x ∂ u = ∂ y ∂ v ∂ x ∂ v = − ∂ y ∂ u This is known as the Cauchy-Riemann Condition(C-R condition).
Let’s recall what had happened so far, we discussed the idea of complex derivative being a limit of finite difference leading to computing limits of the form R 2 → R \mathbb{R}^2\to \mathbb{R} R 2 → R x 0 , y 0 x_0, y_0 x 0 , y 0
We have effectively shown that CR condition is necessary for complex differentiability. It turns out however, that CR condition is also sufficient .
We need to show, ∀ ϵ > 0 \forall \epsilon >0 ∀ ϵ > 0 ∃ δ > 0 \exists \delta >0 ∃ δ > 0 ∣ Δ z ∣ < δ |\Delta z|<\delta ∣Δ z ∣ < δ
∣ f ( z 0 + Δ z ) − f ( z 0 ) Δ z − ( ∂ u ∂ x + i ∂ v ∂ x ) ∣ < ϵ \left| \frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} - \left(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}\right) \right| < \epsilon ∣ ∣ Δ z f ( z 0 + Δ z ) − f ( z 0 ) − ( ∂ x ∂ u + i ∂ x ∂ v ) ∣ ∣ < ϵ We consider the linaer approximation of f ( z 0 + Δ z ) − f ( z 0 ) f(z_0+\Delta z)-f(z_0) f ( z 0 + Δ z ) − f ( z 0 ) z 0 z_0 z 0
f ( z 0 + Δ z ) − f ( z 0 ) = u ( x 0 + Δ x , y 0 + Δ y ) − u ( x 0 , y 0 ) + i [ v ( x 0 + Δ x , y 0 + Δ y ) − v ( x 0 , y 0 ) ] + O ( Δ x 2 ) + i O ( Δ y 2 ) = ∂ u ∂ x Δ x + ∂ u ∂ y Δ y + i ( ∂ v ∂ x Δ x + ∂ v ∂ y Δ y ) + O ( Δ x 2 ) + i O ( Δ y 2 ) = ( ∂ u ∂ x + i ∂ v ∂ x ) Δ x + ( ∂ u ∂ y + i ∂ v ∂ y ) Δ y + O ( Δ x 2 ) + i O ( Δ y 2 ) \begin{align*}
f(z_0+\Delta z)-f(z_0)&=\textcolor{blue}{u(x_0+\Delta x,y_0+\Delta y)-u(x_0,y_0)} + i[\textcolor{red}{v(x_0+\Delta x, y_0+\Delta y)-v(x_0,y_0)}]
\\ &+ O(\Delta x^2) + iO(\Delta y^2) \\
&= \textcolor{blue}{\frac{\partial u}{\partial x}\Delta x +\frac{\partial u}{\partial y} \Delta y} +i\left(\textcolor{red}{\frac{\partial v}{\partial x}\Delta x +\frac{\partial v}{\partial y}\Delta y} \right) \\ &+ O(\Delta x^2)+iO(\Delta y^2) \\
&= \left(\textcolor{blue}{\frac{\partial u}{\partial x}}+i\textcolor{red}{\frac{\partial v}{\partial x}}\right)\Delta x+\left(\textcolor{blue}{\frac{\partial u}{\partial y}} + i\textcolor{red}{\frac{\partial v}{\partial y}}\right)\Delta y\\ &+ O(\Delta x^2)+iO(\Delta y^2) \\
\end{align*} f ( z 0 + Δ z ) − f ( z 0 ) = u ( x 0 + Δ x , y 0 + Δ y ) − u ( x 0 , y 0 ) + i [ v ( x 0 + Δ x , y 0 + Δ y ) − v ( x 0 , y 0 ) ] + O ( Δ x 2 ) + i O ( Δ y 2 ) = ∂ x ∂ u Δ x + ∂ y ∂ u Δ y + i ( ∂ x ∂ v Δ x + ∂ y ∂ v Δ y ) + O ( Δ x 2 ) + i O ( Δ y 2 ) = ( ∂ x ∂ u + i ∂ x ∂ v ) Δ x + ( ∂ y ∂ u + i ∂ y ∂ v ) Δ y + O ( Δ x 2 ) + i O ( Δ y 2 ) We need to expose i Δ y i\Delta y i Δ y
( ∂ u ∂ y + i ∂ v ∂ y ) Δ y ⟶ ( − ∂ u ∂ y i + ∂ v ∂ y ) i Δ y \left(\textcolor{blue}{\frac{\partial u}{\partial y}} + i\textcolor{red}{\frac{\partial v}{\partial y}}\right)\Delta y \longrightarrow \left(\textcolor{blue}{-\frac{\partial u}{\partial y}}i + \textcolor{red}{\frac{\partial v}{\partial y}}\right)i\Delta y ( ∂ y ∂ u + i ∂ y ∂ v ) Δ y ⟶ ( − ∂ y ∂ u i + ∂ y ∂ v ) i Δ y Therefore:
f ( z 0 + Δ z ) − f ( z 0 ) = ( ∂ u ∂ x + i ∂ v ∂ x ) Δ x + ( − ∂ u ∂ y i + ∂ v ∂ y ) i Δ y + O ( Δ x 2 ) + i O ( Δ y 2 ) \begin{align*}
f(z_0+\Delta z)-f(z_0) &= \left(\textcolor{blue}{\frac{\partial u}{\partial x}}+i\textcolor{red}{\frac{\partial v}{\partial x}}\right)\Delta x+\left(\textcolor{blue}{-\frac{\partial u}{\partial y}}i + \textcolor{red}{\frac{\partial v}{\partial y}}\right)i\Delta y\\ &+ O(\Delta x^2)+iO(\Delta y^2)
\end{align*} f ( z 0 + Δ z ) − f ( z 0 ) = ( ∂ x ∂ u + i ∂ x ∂ v ) Δ x + ( − ∂ y ∂ u i + ∂ y ∂ v ) i Δ y + O ( Δ x 2 ) + i O ( Δ y 2 ) Now applying CR condition:
f ( z 0 + Δ z ) − f ( z 0 ) = ( ∂ u ∂ x + i ∂ v ∂ x ) Δ x + ( ∂ v ∂ x i + ∂ u ∂ x ) i Δ y + O ( Δ x 2 ) + i O ( Δ y 2 ) = ∂ u ∂ x ( Δ x + i Δ y ) + i ∂ v ∂ x ( Δ x + i Δ y ) + O ( Δ x 2 ) + i O ( Δ y 2 ) = ( ∂ u ∂ x + i ∂ v ∂ x ) ( Δ x + i Δ y ) ⏟ Δ z + O ( Δ x 2 ) + i O ( Δ y 2 ) ⏟ O ( ∣ Δ z ∣ Δ z ) \begin{align*}
f(z_0+\Delta z)-f(z_0) &= \left(\textcolor{blue}{\frac{\partial u}{\partial x}}+i\textcolor{red}{\frac{\partial v}{\partial x}}\right)\Delta x+\left(\textcolor{red}{\frac{\partial v}{\partial x}}i + \textcolor{blue}{\frac{\partial u}{\partial x}}\right)i\Delta y\\ &+ O(\Delta x^2)+iO(\Delta y^2) \\
&= \textcolor{blue}{\frac{\partial u}{\partial x}}\left(\Delta x + i\Delta y\right) + i\textcolor{red}{\frac{\partial v}{\partial x}}(\Delta x+i\Delta y)
\\ &+ O(\Delta x^2)+iO(\Delta y^2) \\
&= \left(\textcolor{blue}{\frac{\partial u}{\partial x}} + i\textcolor{red}{\frac{\partial v}{\partial x}}\right)\underbrace{(\Delta x +i\Delta y)}_{\Delta z}\\
&+ \underbrace{O(\Delta x^2)+iO(\Delta y^2)}_{O(|\Delta z|\Delta z)} \\
\end{align*} f ( z 0 + Δ z ) − f ( z 0 ) = ( ∂ x ∂ u + i ∂ x ∂ v ) Δ x + ( ∂ x ∂ v i + ∂ x ∂ u ) i Δ y + O ( Δ x 2 ) + i O ( Δ y 2 ) = ∂ x ∂ u ( Δ x + i Δ y ) + i ∂ x ∂ v ( Δ x + i Δ y ) + O ( Δ x 2 ) + i O ( Δ y 2 ) = ( ∂ x ∂ u + i ∂ x ∂ v ) Δ z ( Δ x + i Δ y ) + O ( ∣Δ z ∣Δ z ) O ( Δ x 2 ) + i O ( Δ y 2 ) Therefore the quotient is:
f ( z 0 + Δ z ) − f ( z 0 ) Δ z = ∂ u ∂ x + i ∂ v ∂ x + O ( ∣ Δ z ∣ ) \frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} =\textcolor{blue}{\frac{\partial u}{\partial x}} + i\textcolor{red}{\frac{\partial v}{\partial x} }+O(|\Delta z|) Δ z f ( z 0 + Δ z ) − f ( z 0 ) = ∂ x ∂ u + i ∂ x ∂ v + O ( ∣Δ z ∣ ) and thus the quanity we want to control is:
∣ f ( z 0 + Δ z ) − f ( z 0 ) Δ z − ( ∂ u ∂ x + i ∂ v ∂ x ) ∣ = O ( ∣ Δ z ∣ ) \left| \frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} - \left(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}\right) \right|=O(|\Delta z|) ∣ ∣ Δ z f ( z 0 + Δ z ) − f ( z 0 ) − ( ∂ x ∂ u + i ∂ x ∂ v ) ∣ ∣ = O ( ∣Δ z ∣ ) By definition of big-O, O ( ∣ Δ z ∣ ) < C ∣ Δ z ∣ O(|\Delta z|)<C|\Delta z| O ( ∣Δ z ∣ ) < C ∣Δ z ∣ ∣ Δ z ∣ ∈ R |\Delta z|\in \mathbb{R} ∣Δ z ∣ ∈ R ∀ ϵ > 0 \forall \epsilon >0 ∀ ϵ > 0 δ = ϵ C \delta =\frac{\epsilon}{C} δ = C ϵ O ( ∣ Δ z ∣ ) < C δ = ϵ O(|\Delta z|)<C\delta =\epsilon O ( ∣Δ z ∣ ) < C δ = ϵ
Analytical Function ¶ Now we are ready for the definition of analytic function .
A complex function f : C → C f:\mathbb{C} \to \mathbb{C} f : C → C analytic at z 0 z_0 z 0 if it is differentiable at z z z z z z
A function that is differentiable thourhgout some open set Ω ∈ C \Omega \in \mathbb{C} Ω ∈ C analytic in Ω \Omega Ω
A function that is analytical throughout C \mathbb{C} C entire .
Analytic is a much more restrictive property than merely differentiable(since it requires differentiability in the neighborhood). For instance, we can easily construct complex functions that is differentiable on infinitely many points in C \mathbb{C} C nowhere .
Suppose we want the function to be differentiable on y = x y=x y = x y = − x y=-x y = − x
∂ u ∂ x = 2 x , ∂ v ∂ y = 2 y \frac{\partial u}{\partial x} =2x, \frac{\partial v}{\partial y} = 2y ∂ x ∂ u = 2 x , ∂ y ∂ v = 2 y and
∂ v ∂ x = − 2 x , ∂ u ∂ y = 2 y \frac{\partial v}{\partial x} =-2x, \frac{\partial u}{\partial y} = 2y ∂ x ∂ v = − 2 x , ∂ y ∂ u = 2 y A easy choice would just be:
f = x 2 + y 2 + i ( y 2 − x 2 ) = z z ∗ + i ( z 2 + z ∗ 2 2 ) \begin{align*}
f&=x^2+y^2 +i(y^2-x^2) \\
&= zz^*+i\left(\frac{z^2+{z^*}^2}{2}\right)
\end{align*} f = x 2 + y 2 + i ( y 2 − x 2 ) = z z ∗ + i ( 2 z 2 + z ∗ 2 )
Every open set in C \mathbb{C} C some point not on y = x y=x y = x y = − x y=-x y = − x