Proof that statements

i)If $f_1$ and $f_2$ have residues $r_1$ and $r_2$, show that the residue of $f_1+f_2$ at $z_0$ is $r_1+r_2$.

ii)If $f_1$ and $f_2$ have simple poles at $z_0$ show that $f_1f_2$ has a second-order pole at $z_0$. Derive a formule for the residue

I get a little lost in these exercises involving demonstrations, what I tried

i)If $f_1$ have residue $r_1$ then $lim_{z\rightarrow z_0}f_1=r_1$ and similarly, if $f_2$ have residue $r_2$ then $lim_{z\rightarrow z_0}f_2=r_2$, from that I have that $$lim_{z\rightarrow z_0}f_1+f_2=lim_{z\rightarrow z_0}f_1+lim_{z\rightarrow z_0}f_2=r_1+r_2$$

ii)Here I do not quite know what to do, what I tried was, if $f_1f_2$ have pole of second order then $lim_{z\rightarrow z_0}(z-z_0)^2f_1f_2$ exists that I did $$lim_{z\rightarrow z_0}(z-z_0)^2f_1f_2=lim_{z\rightarrow z_0}(z-z_0)f_1*lim_{z\rightarrow z_0}(z-z_0)f_2$$ then $f_1$ and $f_2$ have simple poles.

Someone can correct me?

  • $\begingroup$ -1 for continued abuse of the "self-learning" tag. $\endgroup$ – Henning Makholm May 31 '15 at 18:31
  • $\begingroup$ @HenningMakholm Why I'm abusing of the tag? $\endgroup$ – Roland May 31 '15 at 18:57
  • $\begingroup$ x @askazy: You keep adding it to question that are not about self-learning. $\endgroup$ – Henning Makholm May 31 '15 at 19:01
  • $\begingroup$ @HenningMakholm The process of studying mathematics without formal instruction. It is for self-taught? $\endgroup$ – Roland May 31 '15 at 19:12
  • 1
    $\begingroup$ x @askazy: And none of your question is about the process of studying mathematics without formal instruction. For all I know you may well be doing that, but that doesn't make it what your questions are about. $\endgroup$ – Henning Makholm May 31 '15 at 19:14

It is rather:

i) If $f_1$ has residue $r_1$ then $\lim_{z\rightarrow z_0}(z-z_0)f_1=r_1$ and similarly, if $f_2$ has residue $r_2$ then $\lim_{z\rightarrow z_0}(z-z_0)f_2=r_2$, from that we have that $$\lim_{z\rightarrow z_0}(z-z_0)\times (f_1+f_2)=\lim_{z\rightarrow z_0}(z-z_0)f_1+\lim_{z\rightarrow z_0}(z-z_0)f_2=r_1+r_2.$$


ii) We have $$\lim_{z\rightarrow z_0}(z-z_0)^2f_1f_2=\lim_{z\rightarrow z_0}(z-z_0)f_1\times\lim_{z\rightarrow z_0}(z-z_0)f_2=r_1r_2$$ then $f_1f_2$ has a pole of order $2$ with a residue equal to $r_1r_2$.


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