Question

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

Answer 1

`1/(z^2+z+1)=2/sqrt(3)sum_(k=0)^oosin((k+1)(2pi)/3)z^k` for `abs z < 1`

Explanation:

`1/(z^2+z+1)=A/(z-z_1)+ (bar A)/(z-bar z_1) = A/(z_1)1/((z/z_1)-1)+bar A/(bar z_1)1/((z/(bar z_1))-1)`

Here `bar z` is the complex conjugate of `z`.

Now with `abs z < abs z_1 = abs( bar z_1)` we have

`1/(z^2+z+1)=-A/(z_1) sum_(k=0)^oo(z/z_1)^k-bar A/(bar z_1)sum_(k=0)^oo(z/(bar z_1))^k`

but

`z_1 = 1/2(-1+i sqrt(3))`

so

`z_1 = abs(z_1) e^(i phi)` where `abs(z_1) = 1` and `phi = (2pi)/3`

and also

`A = abs(A) e^(i theta)`

so

`1/(z^2+z+1)=-abs(A)e^(i(theta-phi))sum_(k=0)^oo z^k e^(-i k phi)-abs(A)e^(i(-theta+phi))sum_(k=0)^oo z^k e^(i k phi)`

or

`1/(z^2+z+1)=-2abs(A)sum_(k=0)^oo cos(theta -(k+1)phi)z^k`

now `absA = 1/sqrt(3)` and `theta = -pi/2`

so finally

`1/(z^2+z+1)=2/sqrt(3)sum_(k=0)^oosin((k+1)(2pi)/3)z^k`

Answer 2

`1/(1+z+z^2) = sum_(n=0)^oo (z^(3n) - z^(3n+1))`

`color(white)(1/(1+z+z^2)) = 1-z+z^3-z^4+z^6-z^7+z^9-z^10+...`

Explanation:

Note that:

`1-z^3 = (1-z)(1+z+z^2)`

`1/(1-t) = sum_(n=0)^oo t^n" "` for `abs(t) < 1`

So we find:

`1/(1+z+z^2) = (1-z)/(1-z^3)`

`color(white)(1/(1+z+z^2)) = (1-z) sum_(n=0)^oo z^(3n)`

`color(white)(1/(1+z+z^2)) = sum_(n=0)^oo (1-z)z^(3n)`

`color(white)(1/(1+z+z^2)) = sum_(n=0)^oo (z^(3n) - z^(3n+1))`

`color(white)(1/(1+z+z^2)) = 1-z+z^3-z^4+z^6-z^7+z^9-z^10+...`

when `abs(z) < 1`