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Assume f(z)is analytic in the complex plane and let f be a complex function which is not identically zero.Then,show that Res(1/f(z^3),0)=0.

I know that the residue is calculating for only singularity points but here we don't have any. Could you please give a hint how to deal with this problem?

Thanks

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  • $\begingroup$ Why don't you have singularities? Whenever $f(z)=0$, you have a singularity. $\endgroup$ – Michael Burr May 26 '16 at 11:07
  • $\begingroup$ Because of the assumption $\endgroup$ – Serkan Yaray May 26 '16 at 11:08
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    $\begingroup$ $f$ is analytic, but $1/f$ might not be. For example, consider the case where $f(z)=z$. $\endgroup$ – Michael Burr May 26 '16 at 11:09
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Sketch: Consider the behavior of $f$ at $0$. We can write $f(z)=z^k g(z)$ where $k$ is some nonnegative integer, $g$ is analytic, and $g(0)\not=0$. Then $$ \frac{1}{f(z^3)}=\frac{1}{z^{3k}g(z^3)}=\frac{1}{z^{3k}}\cdot\frac{1}{g(z^3)}. $$ Since $g(0)\not=0$, $\frac{1}{g(z)}$ is an analytic function near $0$. Therefore, $\frac{1}{g(z)}$ has a power series expansion at $0$. In other words: $$ \frac{1}{g(z)}=a_0+a_1z+a_2z^2+a_3z^3+\cdots. $$ Therefore, $$ \frac{1}{g(z^3)}=a_0+a_1z^3+a_2z^6+a_3z^9+\cdots. $$ Hence, $$ \frac{1}{f(z^3)}=\frac{1}{z^{3k}}(a_0+a_1z^3+a_2z^6+a_3z^9+\cdots). $$ The residue at $0$ is the coefficient of $\frac{1}{z}$ in this power series expansion. However, the coefficient of $\frac{1}{z}$ is zero since the exponents are all multiples of $3$.

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