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Can you help me solve the following problem?

Let $f$ be a holomorphic function on $D_{r}(0), $ the disc of radius $r>1$ with center on the origin. Calculate the following integral:

$$\int_{|z|=1} (2\pm(z + z^{-1}))\frac{f(z)}{z}\mathbb dz$$

Solution:

Observe that $(2\pm(z + z^{-1}))\frac{f(z)}{z}= \frac{\left( 2z \pm z^{2} \pm 1 \right)f(z)}{z^{2}}$. Let $h(z)= \left( 2z \pm z^{2} \pm 1 \right)f(z)$. Then this function is holomorphic in $D_{r}(0)$ since $f$ is and it is multiplied by a polynomial. Then by the Cauchy's Integral Formula:

$$\int_{|z|=1} (2\pm(z + z^{-1}))\frac{f(z)}{z}\mathbb dz =\int_{|z|=1} \frac{h(z)}{z^{2}} \mathbb dz = 2\pi i \frac{h'(0)}{1!}$$

But $h'(0) = 2f(0) \pm f'(0)$.

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1 Answer 1

up vote 5 down vote accepted

Hint: Use Cauchy integral formula: If $f$ is holomorphic in $|z-a|<r$ and continuous in $|z-a|\leq r$ then $$\frac{1}{2\pi i}\underset{|z-a|=r}{\int}\frac{f(z)}{(z-a)^{k+1}}dz=\frac{f^{(k)}(a)}{k!}$$ Also, observe that if $f(z)$ is holomorphic in $D_r(0)$ then so is $p(z)f(z)$ for any polynomial $p(z)$

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