Complex series?

Question

How do you expand f(z)=log(z) in a taylor series about Zo=-1?

Answer 1

(z-1) - (1/2)(z-1)^2 + (1/3)(z-1)^3 - (1/4)(z-1)^4 + 1/5(z-1)^5 - .... etc. , if z0 = 1

i π - (z+1) - (1/2)(z+1)^2 - (1/3)(z+1)^3 - (1/4)(z+1)^4 - .... etc. if z0 = -1.

Log(z) is complex where z < 0. Let z = -1. Then

Log(-1) = i π

-1 = e^(i π)

which is Euler's formula

And, again, if we have Log(-z), where z is positive, this can be expressed as

Log(-z) = Log(-1) + Log(z)

so you can see how the two series relate.

Answer 2

f(Z) = fZo)+f'(Zo) (Z-Zo)+..
f(-1) = log[1 f'(z) =1/z = -1
f''(z)=-1/z^2 =-1,
f'''(z) = 2/z^3 =-2
f''''(z)=-6/z^4 =-6 So it seams that the derivatives are -(n-1)!