How to write g(z)= (z-2)/((z+3)(z+2)) as a power series at the point z=1?

Question

z is a complex number
z=x+iy and x,y are Real numbers

Answer 1

g(x,y) = [(x+ iy)- 2]/{ [(x+ iy)+ 3][(x+ iy)+ 2] }
__ = (x- 2+ iy)/{ [(x+ 3)+ iy][(x+ 2)+ iy] }
__ = (x- 2+ iy)/{ [(x+ 3)(x+ 2)- y]+ iy(x+ 3)(x+ 2) }
__ = (x- 2+ iy)/[ (x²+ 5x+ 6- y)+ iy(x²+ 5x+ 6) ]
__ = (x- 2+ iy)[ (x²+ 5x+ 6- y)+ iy(x²+ 5x+ 6) ]/{ (x²+ 5x+ 6- y)²+ [y(x²+ 5x+ 6)]² }
__ = { [(x- 2)(x²+ 5x+ 6- y)- y²(x²+ 5x+ 6)]+ iy[(x-2)(x²+ 5x+ 6)+ (x²+ 5x+ 6- y)] }/{ (x²+ 5x+ 6- y)²+ [y(x²+ 5x+ 6)]² }

At z=1, x=1, so
g(1,y) = {[(-1)(12- y)- y²(12)]+ iy[(-1)(12)+ (12- y)] }/[(12- y)²+ 144y²]
g(1,y) = [(-12y²+ y- 12)- iy²]/(145y²- 24y+ 144)

Now apply Taylor series expansion over g(1,y) for y and you're golden.

Answer 2

This should be interesting. I could see a power series
at z= -3, or -2 where you have vertical asymptotes.
And as z gets very large in a positive direction g(z)
approaches zero from the positive side of the z-axis and conversely when z goes large in a negative direction. So the z axis is an asymptote also.
But when z=1, g(z)=-(1/12) which is no big deal.
Come on you math majors, is this guy all wet or am I?

Answer 3

Just apply Taylor Series.
In this case you do not need a Laurent expansion

Answer 4

it doesnt equal one...