# How to find the Taylor series for the function 1/(1+z+z^(2)) about the point z=0 ?

##### 2 Answers
Apr 13, 2017

$\frac{1}{{z}^{2} + z + 1} = \frac{2}{\sqrt{3}} {\sum}_{k = 0}^{\infty} \sin \left(\left(k + 1\right) \frac{2 \pi}{3}\right) {z}^{k}$ for $\left\mid z \right\mid < 1$

#### Explanation:

$\frac{1}{{z}^{2} + z + 1} = \frac{A}{z - {z}_{1}} + \frac{\overline{A}}{z - {\overline{z}}_{1}} = \frac{A}{{z}_{1}} \frac{1}{\left(\frac{z}{z} _ 1\right) - 1} + \frac{\overline{A}}{{\overline{z}}_{1}} \frac{1}{\left(\frac{z}{{\overline{z}}_{1}}\right) - 1}$

Here $\overline{z}$ is the complex conjugate of $z$.

Now with $\left\mid z \right\mid < {\left\mid z \right\mid}_{1} = \left\mid {\overline{z}}_{1} \right\mid$ we have

$\frac{1}{{z}^{2} + z + 1} = - \frac{A}{{z}_{1}} {\sum}_{k = 0}^{\infty} {\left(\frac{z}{z} _ 1\right)}^{k} - \frac{\overline{A}}{{\overline{z}}_{1}} {\sum}_{k = 0}^{\infty} {\left(\frac{z}{{\overline{z}}_{1}}\right)}^{k}$

but

${z}_{1} = \frac{1}{2} \left(- 1 + i \sqrt{3}\right)$

so

${z}_{1} = \left\mid {z}_{1} \right\mid {e}^{i \phi}$ where $\left\mid {z}_{1} \right\mid = 1$ and $\phi = \frac{2 \pi}{3}$

and also

$A = \left\mid A \right\mid {e}^{i \theta}$

so

$\frac{1}{{z}^{2} + z + 1} = - \left\mid A \right\mid {e}^{i \left(\theta - \phi\right)} {\sum}_{k = 0}^{\infty} {z}^{k} {e}^{- i k \phi} - \left\mid A \right\mid {e}^{i \left(- \theta + \phi\right)} {\sum}_{k = 0}^{\infty} {z}^{k} {e}^{i k \phi}$

or

$\frac{1}{{z}^{2} + z + 1} = - 2 \left\mid A \right\mid {\sum}_{k = 0}^{\infty} \cos \left(\theta - \left(k + 1\right) \phi\right) {z}^{k}$

now $\left\mid A \right\mid = \frac{1}{\sqrt{3}}$ and $\theta = - \frac{\pi}{2}$

so finally

$\frac{1}{{z}^{2} + z + 1} = \frac{2}{\sqrt{3}} {\sum}_{k = 0}^{\infty} \sin \left(\left(k + 1\right) \frac{2 \pi}{3}\right) {z}^{k}$

Apr 13, 2017

$\frac{1}{1 + z + {z}^{2}} = {\sum}_{n = 0}^{\infty} \left({z}^{3 n} - {z}^{3 n + 1}\right)$

$\textcolor{w h i t e}{\frac{1}{1 + z + {z}^{2}}} = 1 - z + {z}^{3} - {z}^{4} + {z}^{6} - {z}^{7} + {z}^{9} - {z}^{10} + \ldots$

#### Explanation:

Note that:

$1 - {z}^{3} = \left(1 - z\right) \left(1 + z + {z}^{2}\right)$

$\frac{1}{1 - t} = {\sum}_{n = 0}^{\infty} {t}^{n} \text{ }$ for $\left\mid t \right\mid < 1$

So we find:

$\frac{1}{1 + z + {z}^{2}} = \frac{1 - z}{1 - {z}^{3}}$

$\textcolor{w h i t e}{\frac{1}{1 + z + {z}^{2}}} = \left(1 - z\right) {\sum}_{n = 0}^{\infty} {z}^{3 n}$

$\textcolor{w h i t e}{\frac{1}{1 + z + {z}^{2}}} = {\sum}_{n = 0}^{\infty} \left(1 - z\right) {z}^{3 n}$

$\textcolor{w h i t e}{\frac{1}{1 + z + {z}^{2}}} = {\sum}_{n = 0}^{\infty} \left({z}^{3 n} - {z}^{3 n + 1}\right)$

$\textcolor{w h i t e}{\frac{1}{1 + z + {z}^{2}}} = 1 - z + {z}^{3} - {z}^{4} + {z}^{6} - {z}^{7} + {z}^{9} - {z}^{10} + \ldots$

when $\left\mid z \right\mid < 1$