# MATH3401

Lecture 14

### Differentiability

Example 1: Find the derivative of $$f(z)=4z^2$$ from first principles, i. e. using the definition.

Put $$w = f(z)$$ and take $$z_0\in \C$$, then

$$\ds \lim_{\Delta z\ra 0} \frac{\Delta w}{\Delta z} = \lim_{\Delta z\ra 0} \frac{f(z_0+\Delta z)- f(z_0)}{\Delta z}$$ $$=\ds \lim_{\Delta z\ra 0} \frac{4(z_0+\Delta z )^2-4z^2_0}{\Delta z}$$

$$=\ds \lim_{\Delta z\ra 0} \frac{4z_0^2+8z_0\Delta z + 4(\Delta z)^2- 4z_0^2}{\Delta z}$$ $$=\ds \lim_{\Delta z\ra 0} 8z_0 + 4 \Delta z$$ $$= 8 z_0$$.

Thus $$f'(z)= 8z$$.

### Differentiability

Example 2: Find the derivative of $$f(z)=|z|^2$$.

$$f'$$ does not exist, except at $$z=0$$. Proof: soon.

Note: differentiability $$\Rightarrow$$ continuity

differentiability $$\nLeftarrow$$ continuity

Example for $$\nLeftarrow$$: $$|z|^2$$ (or $$|z|$$, or others).

### Formulae

(cf. $$f:\R \ra \R$$)

• $$\ds \frac{d}{dz}(c) = 0$$; $$c$$ constant
• $$\ds \frac{d}{dz}z^n = nz^{n-1}$$; $$n\in\Z$$
• $$\ds \frac{d}{dz} e^z = e^z$$
• $$\ds \frac{d}{dz} \sin z = \cos z$$
• $$\ds \frac{d}{dz} \cos z = -\sin z$$

### Formulae

For $$f, g$$ differentiable functions:

• $$\ds \big(f\pm g\big)^{\prime} = f^{\prime} \pm g^{\prime}$$
• $$\ds \big(f g\big)^{\prime} = fg^{\prime} + f^{\prime}g$$
• $$\ds \left(\frac{f}{g}\right)^{\prime} = \frac{gf^{\prime} - f g^{\prime}}{g^2}$$, $$g\neq 0$$
• Chain rule: $$f$$ differentiable at $$z_0$$ and $$g$$ differentiable at $$f(z_0)$$,
then $$g\circ f$$ is differentiable at $$z_0$$ and $\big(g \circ f\big)^{\prime} = g^{\prime}\big(f(z_0)\big)f^{\prime}(z_0).$ Write $$\dfrac{dg}{dz} = \dfrac{dg}{dw} \dfrac{dw}{dz}$$ where $$w=f(z).$$

### Cauchy-Riemann

$$f : z \mapsto w = u(x,y) + iv(x,y)$$ . Suppose $$f$$ is differentiable at $$z_0 = x_0+iy_0$$. Set $$\Delta z = \Delta x + i \Delta y$$ ... (♣️) $f^{\prime} (z_0) = \lim_{\Delta z\ra 0}\frac{ \Delta w }{ \Delta z} \quad (1)$

Key point: Derivative is independent of how $$\Delta z \ra 0$$ (⚛️)

### Cauchy-Riemann

Note $$\Delta w = f(z_0+\Delta z)-f(z_0)$$

$$\qquad = u(x_0 +\Delta x, y_0 +\Delta y) + i v(x_0 +\Delta x, y_0 +\Delta y)$$

$$\qquad \;\;\,- u(x_0 , y_0 ) - i v(x_0 , y_0 ) \quad (2)$$

Now, from (1) we have

$$f^{\prime}(z_0) = \ds \lim_{\left(\Delta x, \Delta y\right)\ra (0,0)} \Re \left(\frac{\Delta w}{\Delta z}\right) + i \lim_{\left(\Delta x, \Delta y\right)\ra (0,0)} \Im \left(\frac{\Delta w}{\Delta z}\right) \quad (3)$$

As per (⚛️), the value of the limit is independent of how $$\left(\Delta x, \Delta y\right)\ra (0,0)$$.

### Cauchy-Riemann

To start, let $$\left(\Delta x, \Delta y\right)\ra (0,0)$$ along the $$x$$-axis, i. e. consider $$\Delta z$$ of the form $$\left(\Delta x, 0\right)$$ with $$\Delta x\neq 0$$. So $\frac{\Delta w}{\Delta z} = \frac{u(x_0 +\Delta x, y_0) - u(x_0,y_0)}{\Delta x} + i \frac{v(x_0 +\Delta x, y_0) - v(x_0,y_0)}{\Delta x}$

Hence

$$\ds \lim_{\left(\Delta x, \Delta y\right)\ra (0,0)} \Re \left(\frac{\Delta w}{\Delta z}\right)$$ $$= u_x(x_0,y_0)$$ $$= \ds \frac{\partial u}{\partial x}(x_0,y_0) \quad (4)$$

$$\ds \lim_{\left(\Delta x, \Delta y\right)\ra (0,0)} \Im \left(\frac{\Delta w}{\Delta z}\right)$$ $$= v_x(x_0,y_0)$$ $$= \ds \frac{\partial v}{\partial x}(x_0,y_0)\quad (5)$$

### Cauchy-Riemann

Repeat for $$\Delta z$$ of the form $$\Delta z = 0 + i \Delta y$$ with $$\Delta y\neq 0$$, i. e. $$\Delta z \ra 0$$ on
the $$y$$-axis. For this $$\Delta z$$: $\frac{\Delta w}{\Delta z} = \frac{u(x_0, y_0+\Delta y) - u(x_0,y_0)}{i \Delta y} + i \frac{v(x_0, y_0+\Delta y) - v(x_0,y_0)}{i \Delta y}$

So

$$\ds \lim_{\left(\Delta x, \Delta y\right)\ra (0,0)} \Re \left(\frac{\Delta w}{\Delta z}\right)$$ $$= v_y(x_0,y_0) \quad (6)$$

$$\ds \lim_{\left(\Delta x, \Delta y\right)\ra (0,0)} \Im \left(\frac{\Delta w}{\Delta z}\right)$$ $$= -u_y(x_0,y_0) \quad (7)$$

### Cauchy-Riemann

Keeping in mind (⚛️), (4) & (6) and (5) & (7) imply
the Cauchy-Riemann equations:

If $$f=u+iv$$ is differentiable at $$z_0=x_0+iy_0$$, then \begin{align} u_x & = v_y \\ -v_x & = u_y \end{align} \quad \text{ at }\quad(x_0, y_0).

#### Necessary & sufficient conditions

Note: We have shown that Cauchy-Riemann (C/R) are necessary for complex differentiability. They are not sufficient (later).

Sufficient conditions for $$f^{\prime}(z_0)$$ to exist:

• $$f$$ is defined in a neibourhood of $$z_0$$,
• $$u_x, u_y, v_x, v_y$$ are defined and continuous in a neighbourhood at $$(x_0,y_0)$$,
• C/R hold at $$(x_0,y_0)$$.

Then $$f^{\prime}(z_0)$$ exists.