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

2 Answers
Apr 13, 2017

#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#

Apr 13, 2017

#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#