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Annoying q

Im getting really bugged by this question:

Let g and h be analytic in the open disc [itex]\{z \in \mathbb{C} : |z-a| < r \}[/itex], r>0and let [itex]f(z)=\frac{g(z)}{h(z)}[/itex]

if [itex]g(a) \neq 0, h(a)=0, h'(a) \neq 0[/itex] show that f has a pole at z=a and find the corresponding residue of f at a.


Now I initially though we had to Laurent expand the functions g and h but that got really complicated very quickly and i couldn't find anyway of sorting stuff out.
then i tried taylor expanding them which was nicer but still didn't rearrange well.

so my 2 questions are:
(i) how do i do the above problem, and
(ii) am i correct in saying that you can't taylor expand a complex function you can only laurent expand it and if the negative coefficients turn out to be 0 then it reduces to a taylor expansion or are we allowed to do taylor expansions? if we are allowed to, aren't we ignoring the negative terms - is this allowed?
 
Try proving first h(z)=zq(z) for some analytic q s.t. q(a) neq 0. [Here q(a)=??]
 
q(a)=g(a)/(a f(a))??? why are we doing this though?
 

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