Question 3.10: Prove the following assuming that f(z) and g(z) are analytic...
Prove the following assuming that f(z) and g(z) are analytic in a region R.
(a) \frac{d}{d z}\{f(z)+g(z)\}=\frac{d}{d z} f(z)+\frac{d}{d z} g(z)
(b) \frac{d}{d z}\{f(z) g(z)\}=f(z) \frac{d}{d z} g(z)+g(z) \frac{d}{d z} f(z)
(c) \frac{d}{d z}\left\{\frac{f(z)}{g(z)}\right\}=\frac{g(z) \frac{d}{d z} f(z)-f(z) \frac{d}{d z} g(z)}{[g(z)]^2} if g(z) \neq 0
(a)
\begin{aligned} \frac{d}{d z}\{f(z)+g(z)\} &=\lim _{\Delta z \rightarrow 0}\frac{f(z+\Delta z)+g(z+\Delta z)-\{f(z)+g(z)\}}{\Delta z} \\ &=\lim _{\Delta z \rightarrow 0} \frac{f(z+\Delta z)-f(z)}{\Delta z}+\lim _{\Delta z \rightarrow 0} \frac{g(z+\Delta z)-g(z)}{\Delta z} \\ &=\frac{d}{d z} f(z)+\frac{d}{d z} g(z) \end{aligned}(b) \begin{aligned} \frac{d}{d z}\{f(z) g(z)\} &=\lim _{\Delta z \rightarrow 0} \frac{f(z+\Delta z) g(z+\Delta z)-f(z) g(z)}{\Delta z} \\ &=\lim _{\Delta z \rightarrow 0} \frac{f(z+\Delta z)\{g(z+\Delta z)-g(z)\}+g(z)\{f(z+\Delta z)-f(z)\}}{\Delta z} \\ &=\lim _{\Delta z \rightarrow 0} f(z+\Delta z)\left\{\frac{g(z+\Delta z)-g(z)}{\Delta z}\right\}+\lim _{\Delta z \rightarrow 0} g(z)\left\{\frac{f(z+\Delta z)-f(z)}{\Delta z}\right\} \\ &=f(z) \frac{d}{d z} g(z)+g(z) \frac{d}{d z} f(z)\end{aligned}
Note that we have used the fact that \lim _{\Delta z \rightarrow 0} f(z+\Delta z)=f(z) which follows since f(z) is analytic and thus continuous (see Problem 3.4).
Another Method
Let U = f(z), V = g(z). Then ΔU = f(z + Δz) – f(z) and ΔV = g(z + Δz) – g(z), i.e., f(z + Δz) = U + ΔU, g(z + Δz) = V + ΔV. Thus
\begin{aligned} \frac{d}{d z} U V &=\lim _{\Delta z \rightarrow 0} \frac{(U+\Delta U)(V+\Delta V)-U V}{\Delta z}=\lim _{\Delta z \rightarrow 0} \frac{U \Delta V+V \Delta U+\Delta U \Delta V}{\Delta z} \\ &=\lim _{\Delta z \rightarrow 0}\left(U \frac{\Delta V}{\Delta z}+V \frac{\Delta U}{\Delta z}+\frac{\Delta U}{\Delta z} \Delta V\right)=U \frac{d V}{d z}+V \frac{d U}{d z} \end{aligned}where it is noted that ΔV → 0 as Δz→ 0, since V is supposed analytic and thus continuous. A similar procedure can be used to prove (a)
c) We use the second method in (b). Then
\begin{aligned} \frac{d}{d z}\left(\frac{U}{V}\right) &=\lim _{\Delta z \rightarrow 0} \frac{1}{\Delta z}\left\{\frac{U+\Delta U}{V+\Delta V}-\frac{U}{V}\right\}=\lim _{\Delta z \rightarrow 0} \frac{V \Delta U-U \Delta V}{\Delta z(V+\Delta V) V} \\ &=\lim _{\Delta z \rightarrow 0} \frac{1}{(V+\Delta V) V}\left\{V \frac{\Delta U}{\Delta z}-U \frac{\Delta V}{\Delta z}\right\}=\frac{V(d U / d z)-U(d V / d z)}{V^2} \end{aligned}The first method of (b) can also be used.