thanks for all your help, I've finally done it, and here is my final answer:
f(z)= (5/3 - 3i) + (10/9 - 3i) (z - 1)/1! + (50/27) [(z-1)^2]/2! + (4/27) [(z - 1)^3]/3! + ...
the problem is that this equation will not be easy to defferenciate it has to be simplified first
or if you could do it as it is please tell me how
thanks
i tried replacing each z wth x+iy but it went into a huge binomial equation i couldn't solve, here is part of my work:
z^4 = (x+iy)^4 = x^4 + (x^3 + iy) + (x^2 - y^2) + (x - iy^3) + y^4
z^4 + (2-3i)*z^3 = x^4 + (3-3i)*(x^3 + x^2 + x - y^2) + (3 + 3i)*(y - y^3) + y^4
and i don't know what...
[SOLVED] compute the taylor's expansion (series)
Homework Statement
compute the taylor's expansion (series) for:
f(z)= [z^4 + (2-3i)*z^3 - 6i*z^2 + 2]/[z(z+2)]
where Zo (Z node or Z(0)) = 1
i need answer as soon as possible please!