My question comes from section 1.6 of Complex Analysis (4th edition) by Serge Lang:
Let $U$ be an open subset of $\mathbb{C}$ and let \begin{align*} f(x + iy) = u(x,y) + i v(x,y), \qquad x + iy \in U \end{align*}
be a complex-valued function on $U$. Lang argues that if $f$ is holomorphic at some point $z = x + iy$, then the real vector field associated with $f$ is differentiable at $(x,y)$. However, I don't quite follow the argument. The relevant passage is below:
At a fixed $z$, let $f'(z) = a + ib$. Let $w = h + ik$, with $h,k$ real. Suppose that $$ f(z + w) - f(z) = f'(z) w + \sigma(w) w, $$
where $$ \lim_{w \to 0} \sigma(w) = 0. $$
Then $$ f'(z) w = (a + ib)(h + ik) = ah - bk + i(ak + bh). $$
On the other hand, let $$ F: U \to \mathbb{R}^2 $$ be the map such that $$ F(x,y) = \big( u(x,y), v(x,y) \big). $$
We call $F$ the (real) vector field associated with $f$. Then $$ F(x + h, y + k) - F(x,y) = (ah - bk, bh + ak) + \sigma_1(h,k) h + \sigma_2(h,k) k, \hspace{1cm} (1) $$
where $\sigma_1(h,k), \, \sigma_2(h,k)$ are functions tending to $0$ as $(h,k)$ tends to $0$. Hence if we assume that $f$ is holomorphic, we conclude that $F$ is differentiable in the sense of real variables, and that its derivative is represented by the (Jacobian) matrix $$ J_F(x,y) = \begin{pmatrix} a & -b \\ b & \phantom{-} a \end{pmatrix} = \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\[10pt] \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix}. $$
My questions: How was equation (1) obtained? Moreover, if it appears that $\sigma = \sigma_1 + i \sigma_2$, in which case $\sigma_1$ and $\sigma_2$ are real-valued functions. But then in the right-hand side of (1) we are adding the scalar quantity $\sigma_1(h,k) k + \sigma_2(h,k) k$ to the vector $(ah - bk,bh + ak)$...how does that make sense?
Here is my attempt at deducing the differentiability of $F$: We have \begin{align*} F(x,y) &= \Big( \text{Re}\big(f(x + iy) \big), \text{Im} \big(f(x + iy) \big) \Big) \\[5pt] &= \big( \text{Re}\big(f(z) \big), \text{Im} \big(f(z) \big) \big) \end{align*}
and
\begin{align*} F(x + h, y + k) &= \Big( \text{Re}\big(f(x + h + i(y+k)) \big), \text{Im} \big(f(x + h + i(y+k)) \big) \Big) \\[5pt] &= \Big( \text{Re} \big(f(z + w) \big), \text{Im}\big(f(z+w) \big) \Big) \end{align*}
Now using the fact that $\text{Re}(z_1 + z_2) = \text{Re}(z_1) + \text{Re}(z_2)$ and $\text{Im}(z_1 + z_2) = \text{Im}(z_1) + \text{Im}(z_2)$, it follows that
\begin{align*} F(x + h, y + k) - F(x,y) &= \big( \text{Re} \big(f(z + w) \big), \text{Im}\big(f(z+w) \big) \big) - \big( \text{Re}\big(f(z) \big), \text{Im} \big(f(z) \big) \big) \\[5pt] &= \big( \text{Re} \big( f(z + w) - f(z) \big), \text{Im} \big( f(z + w) - f(z) \big) \big). \end{align*}
Now writing $\sigma(w) = \sigma(h + ik) = \sigma_1(h,k) + i \sigma_2(h,k)$, we have \begin{align*} f(z + w) - f(z) &= f'(z) w + \sigma(w) w \\[3pt] &= (a + ib)(h + ik) + \big[\sigma_1(h,k) + i \sigma_2(h,k) \big] (h + ik) \\[3pt] &= (ah - bk) + i(ak + bh) + h \sigma_1(h,k) - k \sigma_2(h,k) + i \left[k \sigma_1(h,k) + h \sigma_2(h,k) \right] \\[3pt] &= \big[ ah - bk + h \sigma_1(h,k) - k \sigma_2(h,k) \big] + i \big[ak + bh + k \sigma_1(h,k) + h \sigma_2(h,k) \big]. \end{align*}
Hence, I get \begin{align*} & F(x + h, y + k) - F(x,y) \\[3pt] = \; &\big( ah - bk + h \sigma_1(h,k) - k \sigma_2(h,k), ak + bh + k \sigma_1(h,k) + h \sigma_2(h,k) \big) \\[5pt] = \; & \big(ah - bk, ak + bh \big) + \big( h \sigma_1(h,k) - k \sigma_2(h,k), k \sigma_1(h,k) + h \sigma_2(h,k) \big), \end{align*}
which is different than what the book has...So does the book have a mistake, or did I make a mistake in my calculation?
$\sigma_1, \sigma_2$ are not the functions as you defined them in your proof. Let us write $$\sigma(w) = \sigma(h,k) = \bar \sigma_1(h,k) + i \bar \sigma_2(h,k)$$ with real-valued functions $\bar \sigma_1, \bar \sigma_2$. Then we get \begin{align*} & F(x + h, y + k) - F(x,y) \\[3pt] = \; & \big(ah - bk, ak + bh \big) + \big( h \bar \sigma_1(h,k) - k \bar \sigma_2(h,k), k \bar \sigma_1(h,k) + h \bar \sigma_2(h,k) \big) \\ = \; & \big(ah - bk, ak + bh \big) + \big(\bar \sigma_1(h,k),\bar \sigma_2(h,k)\big)h + \big(-\bar \sigma_2(h,k), \bar \sigma_1(h,k)\big)k . \end{align*}
Now take $\sigma_1(h,k) = \big(\bar \sigma_1(h,k),\bar \sigma_2(h,k)$ and $\sigma_2(h,k) = \big(-\bar \sigma_2(h,k), \bar \sigma_1(h,k)\big)$.
These are functions with values in $\mathbb R^2$ tending to $0$ as $(h,k)$ tends to $0$.