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Find the saddle points $z_{1},z_{2}$ of $f(z)=\frac{(z - 1)^{2}(z + 1)}{z^{2}}$

Does anyone can help me with this problem?

$z_{0}$ is a saddle point of an analytic function f if and only if $f´(z_{0})=0$ and $f(z_{0})=0$.

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What have you tried? –  Antonio Vargas May 4 '12 at 5:36
    
Maybe you could try looking at the nicely factored formula for $f$ to guess where $f(z)=0$? There are only two such points in $\mathbb{C}$. Then try differentiating the formula using the product and quotient rules. Then check to see if any of the solutions to $f(z)=0$ satisfy $f'(z) = 0$ also. Just as a hint, you can figure all of this out from the formula for $f$ without having to do any work. –  copper.hat May 4 '12 at 6:59
    
@Breton: Saddle point $z_0 $ is where $f'(z_0)=0$ as well as $f''(z_0)=0$. there is problem in your definition –  Aang Jul 12 '12 at 7:49

1 Answer 1

I've never seen that definition of a saddle point before. There must be a mistake somewhere, since there's only one such point for that function. A hint for finding it: Before differentiating, multiply through by $z^2$; then the derivative of the left-hand side is $0$ at the desired point(s), and you can easily check for which zeros of $f$ the derivative of the right-hand side is $0$.

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