Prove $\frac{1}{(1+z)^2}=1-2z + \mathcal{O}(z^2)$ as $z \to 0$

Question

Prove: $$\frac{1}{(1+z)^2}=1-2z+\mathcal{O}(z^2)$$ as $z\to 0$.

What have you tried so far? Are you familiar with methods for retrieving series expansions? — Scounged, Jul 13 '15 at 14:50
Why did you delete the same question yesterday only to re-ask today? — epimorphic, Jul 13 '15 at 15:59

score 3 · Accepted Answer · answered Jul 13 '15 at 14:55

3

Hint. From the standard identity $$ \frac1{1+z}=1-z+z^2+O(z^3), \quad |z|<1, \tag1 $$ you obtain by differentiating $$ \frac{-1}{(1+z)^2}=-1+2z+O(z^2), \quad |z|<1, \tag2 $$ as desired.

You may prove $(1)$ by the Maclaurin formula.

answered Jul 13 '15 at 14:55

Olivier Oloa

120,989

score 1 · Answer 2 · answered Jul 13 '15 at 15:10

An alternative approach to the one given by Olivier Oloa above can be to use an ansatz:

$\frac1{(1+z)^2}=a+bz+O(z^2), \quad |z|<1, \quad a,b\in\mathbb{C}\tag1$

Multiplying each side by the factor $(1+z)^2$, we get:

$1=(1+z)^2(a+bz+O(z^2)), \quad |z|<1, \tag2$

Now we expand, and simplify:

$(2)\Leftrightarrow ... \Leftrightarrow 1=a+(2a+b)z+O(z^2), \quad |z|<1, \tag3$

Identification in $(3)$ now gives that $a=1$ and $b=-2$, as desired.

Prove $\frac{1}{(1+z)^2}=1-2z + \mathcal{O}(z^2)$ as $z \to 0$

2 Answers2