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In David Wunsch's Complex Variables with Applications, Example 3 on page 266 asks the reader to find a Maclaurin expansion of $f(x)=(z+1)^{1/2}$ where the principal branch is used. The principal branch is identified as $$e^{(1/2)\log(z+1)}$$ and its derivative is given as $$e^{(1/2)\log(z+1)}\frac{1}{2(z+1)}=\frac{(z+1)^{1/2}}{2(z+1)}.$$

First question: Why can't I just differentiate $(z+1)^{1/2}$ as I do in regular calculus, as $\frac12(z+1)^{-1/2}$. This is the same answer as above, is it not? If not, can someone show me a good counterexample demonstrating what can go wrong with this thinking?

Later, David states that $(z+1)^{1/2}/(2(z+1))$ must be interpreted as $$\frac{(z+1)^{1/2}}{2(z+1)}=\frac{e^{(1/2)\log(z+1)}}{2(z+1)},$$ then states that at $z=0$, this last expression equals 1/2. However, if I use $\frac12(z+1)^{-1/2}$, then substitute $z=0$, I also get 1/2.

Second Question: Again, can someone provide a good counterexample of how I can get in trouble with this kind of thinking?

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