MATH3401

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 \(f'(z)= 8z\).

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

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

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 \(w=f(z).\)

\(f : z \mapsto w = u(x,y) + iv(x,y)\) . 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 \(\Delta z \ra 0\) (⚛️)

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 \(\left(\Delta x, \Delta y\right)\ra (0,0)\).

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

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

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 \[ \begin{align} u_x & = v_y \\ -v_x & = u_y \end{align} \quad \text{ at }\quad(x_0, y_0). \]

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 \((x_0,y_0)\).