The book in question is An Introduction to Complex Function Thoery by Bruce Palka.
Palka defines differentiability in the real sense as that a continuous function $f:A\to \mathbb{C}$ is differentiable at $z = x + iy$ in the real sense at $z_0 \in A$ if there exists complex numbers $c, d \in \mathbb{C}$ such that $f(z) = f(z_0) + c(z – z_0) + d(\overline{z} – \overline{z_0}) + E(z)$ where $E(z) \to 0$ as $z \to z_0$.
Palka's definition of complex differentiability is that (assume the same function $f$) $f$ has a complex derivative at $z_0 \in A$ if the limit $\lim_{z\to z_0}\frac{f(z) – f(z_0)}{z – z_0}$ exists.
Palka discusses the general philosophy between these two notions in pages 97-98 by noting that the derivative of $f$ is in the real sense the matrix $\begin{bmatrix}\alpha & \beta\\ \gamma & \delta\end{bmatrix}$ with $\alpha = u_x, \beta = u_y, \gamma = v_x, \delta = v_y$ when $f = u + iv$ giving the $f$ the linear approximation form at $z_0$ as
$$f(z) = f(z_0) + \begin{bmatrix}\alpha & \beta\\\ \gamma & \delta\end{bmatrix}\begin{bmatrix}x – x_0\\\ y – y_0\end{bmatrix} + E(z)$$
when a complex number $z$ is viewed as a two dimensional vector.
Then Palka states (without any computation) that the $c$ and $d$ w.r.t. the elements of the said matrix are $c = \frac{1}{2}\left(\alpha + \delta + i\left(\gamma – \beta\right)\right)$ and $d = \frac{1}{2}\left(\alpha – \delta + i\left(\gamma + \beta\right)\right)$ when $x = \frac{z + \overline{z}}{2} , y = \frac{z – \overline{z}}{2i}$.
Unfortunately I cannot seem to be able to derive these same equalities. Namely after applying the matrix to the vector $\begin{bmatrix}x – x_0\\\ y – y_0\end{bmatrix}$ and substituting the form of $x, x_0, y, y_0$, we get
$$\frac{1}{2}\begin{bmatrix}(z – z_0)(\alpha – i\beta) + (\overline{z} – \overline{z}_0)(\alpha + i\beta)\\\ (z – z_0)(\gamma – i\delta) + (\overline{z} – \overline{z}_0)(\gamma + i\delta)\end{bmatrix}$$
and I can't see any clear connection between $c$ and $d$ from this. What should I do? Thanks!
Best Answer
$f(z) = f(z_0) + (\alpha+i\gamma)(x - x_0) + (\beta+i\delta)(y - y_0) + E(z), z=x+iy$ and note that $x-x_0=\frac{(z-z_0)+\overline {(z-z_0)}}{2}$ and $y-y_0=\frac{(z-z_0)-\overline {(z-z_0)}}{2i}$, substitute in the above and separate the terms containing $z-z_0$ and $\overline {z-z_0}=\bar z-\bar z_0$ respectively, identifying the coefficients, so $c=\frac{1}{2}(\alpha+i\gamma-i\beta+\delta)$ and $d=\frac{1}{2}(\alpha+i\gamma+i\beta-\delta)$ and we are done!