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Differential coefficients and derivatives

For $w = f(z)$ defined in the region $D$ and a point $z_0$ in $D$,

$\displaystyle \lim_{z \to z_{0}}\frac{f(z) - f(z_{0})}{z - z_{0}} = \lim_{\Delta z \to 0}\frac{f(z_{0} + \Delta z) - f(z_{0})}{\Delta z} $

exists and the absolute value of the limit is finite, $f(z)$ is said to be differnetiable at $z_{0}$ and this limit is denoted by $f'(z_{0})$. Furthermore, this is called coefficient of derivative at $z_0$.

When all $z$ in $D$ are differentiable, the derivative $f'(z)$ is a function of $z$ in $D$. This is called a derivative of $w = f(z)$ and denoted by $\frac{dw}{dz}, w', f'(z)$.

Theorem 3..3  

(1) If $f(z),g(z)$ are differentiable in $D$, then the followings hold.

(i) $\{f(z) \pm g(z)\}' = f'(z) \pm g'(z)\}$

(ii) $\{f(z)\cdot g(z)\}' = f'(z)g(z) + f(z)g'(z)$

(iii) $\{\frac{f(z)}{g(z)}\}' = \frac{f'(z)g(z) - f(z)g'(z)}{\{g(z)\}^2}$

(2) If $F(w)$ is differentiable in $w$, $w = f(z)$ is differentiable in $z$, then the composite function $F(f(z))$ is differentiable in $z$ and $\frac{dF(w)}{dz} = \frac{dF(w)}{dw}\cdot \frac{df(z)}{dz}$.

(3) If $f(z)$ is differentiable, then $f(z)$ is continuous.

A function $f(z) = u(x,y) + iv(x,y)$ defined in $D$ is differentiable at $z_{0} = x_{0} + iy_{0}$ in $D$. Then keeping $y$ constant and bringing $\Delta z$ colser to 0, we have

$\displaystyle f'(z) = \lim_{\Delta x \to 0}\frac{f(z_{0} + \Delta x) - f(z_{0})... ...f}{\partial x} = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$

Similarly, keeping $x$ constant and bringing $\Delta z$ closer to 0, we hafe

$\displaystyle f'(z) = \lim_{\Delta y \to 0}\frac{f(z_{0} + i\Delta y) - f(z_{0}... ...}{\partial y} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}$

Thus,

$\displaystyle f'(z) = \frac{\partial f}{\partial x} = -i\frac{\partial f}{\partial y}$

Compare the real part to the imaginary part, we obtain

$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} $

This equation is called a Cauchy-Riemann differential equation.

Theorem 3..4   A function $f(z) = u(x,y) + iv(x,y)$ defined in $D$ is differentiable at $z_{0} = x_{0} + iy_{0}$ if and only if $u(x,y),v(x,y)$ is totally differentiable at $(x_{0},y_{0}$ and satisfies

$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial v}{\partial x} = \frac{\partial u}{\partial y} $

Then

$\displaystyle f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} $

When $f(z)$ is differentiable at any point in the domain $D$, $f(z)$ is said to be analytic in $D$.

An analytic function in $Z$ entire plane $(\vert z\vert < \infty)$ is called entire function.

Note1. When a function is analytic at $z_{0}$, it means that it is analytic including not only $z_{0}$ but also its neighborhood.

Note2. If $f(z)$ is analytic, it is continuous (becouse it is differentiable).

Theorem 3..5   If $f(z) = u + iv$ is analytic in $D$ and if the second partial derivatives of $u,v$ are continuous, then $u,v$ is harmonic function. That is it satisfies Laplace's differential equation.

$\displaystyle \Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u... ...a v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$

Note We assumed that $u$ and $v$ have continuous second-order partial derivatives, but We don't need this assumption because the holomorphic function is separately proved to be differentiable many times.

Theorem 3..6   If $u(x,y),v(x,y)$ satisfies the Cauchy-Riemann's equation in the region $D$ and it has a continuous partial derivatives, then $f(z) = u + iv$ is analytic in $D$.

Note This theorem is a strong enough condition and effective for determining anayticity.

Example 3..1   For $w = z^2 = x^2 - y^2 + 2ixy$, $u = x^2 - y^2, v = 2xy$ are polynomials and its partial derivatives are continuous. Furthermore, $u_{x} = 2x = v_{y}, u_{y} = -2y = -v_{x}$. Thus by the theorem 3.6, $w = z^2$ is analytic on whole plane.

Example 3..2   For $w = \bar{z} = x - iy$, $u = x, v = -y$. Then $u_{x} = 1$ and $v_{y} = -1$ are not equal. Then the condition for anlytic in Cauchy-Riemann's equation is not satisfied. Thus, $w = \bar{z}$ is not analytic.

Example 3..2   For $f(z)$ is analytic and $f'(z) \equiv 0$, prove that $f(z)$ is constant.

Solution Since $f(z) = u(x,y) + iv(x,y)$ is analytic, by the theorem 3.4, we have

$\displaystyle f'(z) = u_{x} + iv_{x} = v_{y} - iu_{y}$

By assumption, $f'(z) \equiv 0$. Thus, $u_{x} \equiv 0, u_{y} \equiv 0, v_{x} \equiv 0, v_{y} \equiv 0$. Therefore, $u(x,y),v(x,y)$ are constants. Thus, $f(z) = u + iv$ is also constant.

Example 3..3   Show that the following function $u = u (x, y)$ is a harmonic function, and find the holomorphic function $w = u + iv$ that has this in the real part.

$(1) u = xy$ $(2) u = (x-y)(x^2 + 4xy + y^2)$

Solution (1) $u_{x} = y, u_{y} = x, u_{xx} = u_{yy} = 0$ implies $u_{xx} + u_{yy} = 0$. Then $u = xy$ is a harmonic function. Next, $w = u + iv$ is analytic. Then by Cauchy-Riemann's equation

$\displaystyle u_{x} = v_{y} = y, u_{y} = - v_{x} = x$

Thus, integrate the first equation in $y$. Then

$\displaystyle v = \int v_{y} dy = \int y dy = \frac{y^2}{2} + \phi(x) $

Here, $\phi(x)$ is a function of $x$ only. Put this into latter equation, we have $\phi'(x) = -x$. Thus, $\phi(x) = -\frac{x^2}{2} + c_{1}$. This shows that $v = \frac{y^2 - x^2}{2} + c_{1}$. Therefore,

$\displaystyle w = u + iv = xy + i\left(\frac{y^2 - x^2}{2} + c_{1}\right) = \frac{-i}{2}(x^2 - y^2 + 2xyi) + ic_{1} = -\frac{i}{2}z^2 + c (c = c_{1}) $

(2) $u = x^3 + 3x^2y - 3xy^2 - y^3$ implies $u_{x} = 3x^2 + 6xy - 3y^2$, $u_{y} = 3x^2 - 6xy - 3y^2$, $u_{xx} = 6x + 6y$, $u_{yy} = -6x - 6y$. Then, $u_{xx} + u_{yy} = 0$. Thus, $u$ is a harmonic function. Next, $w = u + iv$ is analytic. Then by the Cauchy-Riemann's equation, we have

$\displaystyle u_{x} = v_{y} = 3x^2 + 6xy - 3y^2, u_{y} = - v_{x} = 3x^2 - 6xy - 3y^2$

Now integrate the first equation by $y$. Then

$\displaystyle v = \int v_{y} dy = \int y dy = 3x^2 y + 3xy^2 - y^3 + \phi(x) $

Here, $\phi(x)$ is a function of $x$ only. Now puttin this into the latter equation, we have $\phi'(x) = -3x^2$. Thus, $\phi(x) = -x^3 + c_{1}$. This implies that $v = 3x^2 y + 3xy^2 - y^3 -x^3 + c_{1}$. Therefore,
$\displaystyle w$ $\displaystyle =$ $\displaystyle u + iv = x^3 + 3x^2y - 3xy^2 - y^3 + i(3x^2 y + 3xy^2 - y^3 -x^3 + c_{1})$  
  $\displaystyle =$ $\displaystyle x^3 + 3x^2y - 3xy^2 - y^3 + i(3x^2 y + 3xy^2 - y^3 -x^3) + ic_{1}$  
  $\displaystyle =$ $\displaystyle (1-i)(x+iy)^3 + ic_{1} = (1-i)z^3 + c$  

Exercise3.2
1. Differentiate the following functions.
(a)
$z^3-2z^2 + 3z$
(b)
$(z^2 + i)^3$
(c)
$\frac{z-i}{z+i}$

2. Differentiate the following functions.

(a)
$\tan{z}$
(b)
$\displaystyle{\frac{1}{\cos{z}}}$
(c)
$\sqrt{z^2 + 1}$
(d)
$\sin^{2}{z}$
(e)
$\log(z^2 + 4i)$
(f)
$\displaystyle{i^{\cos{z}}}$
(g)
$\sin^{-1}{(z-i)}$
(h)
$\log(z + \sqrt{z^2 + 1})$
(i)
$\log(\sin^{-1}{z})$
(j)
$\displaystyle{z^z}$

3. When $z = x + iy$, check the analyticity of the following functions and if it is analytic, find the derivative.

(a)
$x - iy$
(b)
$x^2 - y^2 + 2ixy$
(c)
$(x^2 - y^2 - 3x + 2) + i(2xy - 3y)$
(d)
$\displaystyle{\frac{x + iy}{x^2 + y^2}}$

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Next: Complex integration Up: Analytic function,holomorphic function Previous: Analytic function,holomorphic function   Index