Question 3.39: Suppose w =f(ζ) where ζ =g(z). Assuming f and g are analytic...
Suppose w =f(ζ) \text{ where } ζ =g(z). Assuming f and g are analytic in a region \mathcal{R} , prove that
\frac{d w}{d z}=\frac{d w}{d \zeta} \cdot \frac{d \zeta}{d z}
Let z be given an increment Δz≠0 \text{ so that } z + Δz \text{ is in } \mathcal{R}. Then, as a consequence, ζ and w take on increments Δζ and Δw, respectively, where
\Delta w=f(\zeta+\Delta \zeta)-f(\zeta), \quad \Delta \zeta=g(z+\Delta z)-g(z) (1)
Note that as Δz → 0, we have Δw → 0 and Δζ → 0.
If Δζ ≠0, let us write \epsilon = (Δw/Δζ ) – (dw/dζ ) \text{ so that } \epsilon → 0 \text{ as } Δζ → 0 and
\Delta w=\frac{d w}{d \zeta} \Delta \zeta+\epsilon \Delta \zeta (2)
If Δζ = 0 for values of Δz, then (1) shows that Δw = 0 for these values of Δz. For such cases, we define \epsilon = 0. It follows that in both cases, Δζ ≠0 or Δζ =0, (2) holds. Then dividing (2) by Δz≠0 and taking the limit as Δz → 0, we have
\begin{aligned}\frac{d w}{d z}=\lim _{\Delta z \rightarrow 0} \frac{\Delta w}{\Delta z} &=\lim _{\Delta z \rightarrow 0}\left(\frac{d w}{d \zeta} \frac{\Delta \zeta}{\Delta z}+\epsilon \frac{\Delta w}{\Delta z}\right) \\&=\frac{d w}{d \zeta} \cdot \lim _{\Delta z \rightarrow 0} \frac{\Delta \zeta}{\Delta z}+\lim _{\Delta z \rightarrow 0} \epsilon \cdot \lim _{\Delta z \rightarrow 0} \frac{\Delta w}{\Delta z} \\ &=\frac{d w}{d \zeta} \cdot \frac{d \zeta}{d z}+0 \cdot \frac{d \zeta}{d z}=\frac{d w}{d \zeta} \cdot \frac{d \zeta}{d z}\end{aligned}