# Two Taylor expansions of $\frac{1}{1+\sqrt{2-z}}$ about $z=0$

How do you start expanding this function $$f(z)= \frac{1}{1+\sqrt{2-z}}$$ into two Taylor expansions about $z=0$?

The best I came up is to let $u=\sqrt{2-z}$ and then expand $f(z)$ as a geometric series.

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What do you mean with "two" Taylor expansions? –  Phira Nov 25 '11 at 14:25
Did you consider multiplying numerator and denominator by the conjugate of the denominator? –  Phira Nov 25 '11 at 14:27
Your idea cannot work directly because for the geometric series $u$ has to be small, but if $z$ is small, $u$ isn't. –  Phira Nov 25 '11 at 14:27

Alternatively, with some algebraic manipulations you get at: $$f(z) = \frac{1}{1+\sqrt{2-z}} = \frac{1-\sqrt{2-z}}{\left(1+\sqrt{2-z}\right)\left(1-\sqrt{2-z}\right)} = \frac{1-\sqrt{2-z}}{z-1} = \frac{\sqrt{2-z}-1}{1-z}$$ Suppose, you worked out $\sqrt{2-z}-1 = \sum_{n=0}^\infty c_n z^n$, then $$\frac{\sqrt{2-z}-1}{1-z} = \sum_{n=0}^\infty \left( \sum_{m=0}^n c_m \right) z^n$$
Or, keeping the radical in the denominator, to expand near $z=0$ you should go like this: $$f(z)= \frac{1}{1+\sqrt{2-z}}=\frac{\displaystyle 1}{\displaystyle(1+\sqrt{2})(1+u)}$$ where $$u = \frac{\sqrt{2-z}-\sqrt{2}}{1+\sqrt{2}}$$ is near zero. But this will be more work than Sasha's answer.