Find Singularity Type $fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$

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Let $f(z),g(z)$ have pole of order $1$ at $z_0$, $h(z)$ pole of order $2$ and $r(z)$ removable singularity at $z_0$



What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$



So I set $f(z)=fraca(z)z-z_0,g(z)=fracb(z)z-z_0,h(z)=fracc(z)(z-z_0)^2$ can ignore $r(z)$?



$$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)=fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)=fraca^2(z)+b(z)(z-z_0)(z-z_0)^2*frac(z-z_0)^2c(z)+r(z)(z-z_0)^2=fraca^2(z)+b(z)(z-z_0)c(z)+r(z)(z-z_0)^2$$



Plug in $z=z_0$ we get:



$$fraca^2(z_0)+b(z)*0c(z_0)+0$$



But $c(z_0)$ is analytic




complex-analysis




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  • "What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$" That doesn't make sense.
    – zhw.
    Jul 16 at 20:11














up vote
0
down vote

favorite












Let $f(z),g(z)$ have pole of order $1$ at $z_0$, $h(z)$ pole of order $2$ and $r(z)$ removable singularity at $z_0$



What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$



So I set $f(z)=fraca(z)z-z_0,g(z)=fracb(z)z-z_0,h(z)=fracc(z)(z-z_0)^2$ can ignore $r(z)$?



$$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)=fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)=fraca^2(z)+b(z)(z-z_0)(z-z_0)^2*frac(z-z_0)^2c(z)+r(z)(z-z_0)^2=fraca^2(z)+b(z)(z-z_0)c(z)+r(z)(z-z_0)^2$$



Plug in $z=z_0$ we get:



$$fraca^2(z_0)+b(z)*0c(z_0)+0$$



But $c(z_0)$ is analytic




complex-analysis




share|cite|improve this question





















  • "What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$" That doesn't make sense.
    – zhw.
    Jul 16 at 20:11












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f(z),g(z)$ have pole of order $1$ at $z_0$, $h(z)$ pole of order $2$ and $r(z)$ removable singularity at $z_0$



What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$



So I set $f(z)=fraca(z)z-z_0,g(z)=fracb(z)z-z_0,h(z)=fracc(z)(z-z_0)^2$ can ignore $r(z)$?



$$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)=fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)=fraca^2(z)+b(z)(z-z_0)(z-z_0)^2*frac(z-z_0)^2c(z)+r(z)(z-z_0)^2=fraca^2(z)+b(z)(z-z_0)c(z)+r(z)(z-z_0)^2$$



Plug in $z=z_0$ we get:



$$fraca^2(z_0)+b(z)*0c(z_0)+0$$



But $c(z_0)$ is analytic




complex-analysis




share|cite|improve this question













Let $f(z),g(z)$ have pole of order $1$ at $z_0$, $h(z)$ pole of order $2$ and $r(z)$ removable singularity at $z_0$



What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$



So I set $f(z)=fraca(z)z-z_0,g(z)=fracb(z)z-z_0,h(z)=fracc(z)(z-z_0)^2$ can ignore $r(z)$?



$$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)=fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)=fraca^2(z)+b(z)(z-z_0)(z-z_0)^2*frac(z-z_0)^2c(z)+r(z)(z-z_0)^2=fraca^2(z)+b(z)(z-z_0)c(z)+r(z)(z-z_0)^2$$



Plug in $z=z_0$ we get:



$$fraca^2(z_0)+b(z)*0c(z_0)+0$$



But $c(z_0)$ is analytic





complex-analysis





share|cite|improve this question












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edited Jul 16 at 16:15
























asked Jul 16 at 15:58









newhere

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  • "What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$" That doesn't make sense.
    – zhw.
    Jul 16 at 20:11
















  • "What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$" That doesn't make sense.
    – zhw.
    Jul 16 at 20:11















"What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$" That doesn't make sense.
– zhw.
Jul 16 at 20:11




"What is singularity of $$fracf(z_0)^2+g(z_0)h(z_0)+r(z_0)$$" That doesn't make sense.
– zhw.
Jul 16 at 20:11










1 Answer
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Your approach is fine. Note thatbeginalignlim_zto z_0fracf^2(z)+g(z)h(z)+r(z)&=lim_zto z_0fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)\&=lim_zto z_0fraca^2(z)+(z-z_0)b(z)c(z)+(z-z_0)^2c(z)\&=fraca^2(z_0)c(z_0)inmathbbCendalignand that therefore your function has a removable singularity at $z_0$ if $z_0$ is not a zero of $c$ and a pole otherwise (whose order is the order of $z_0$ as a zero of $c$).






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  • Why removable singularity? as $c(z_0)$ is analytic
    – newhere
    Jul 16 at 16:13










  • @newhere Right. I've edited my answer.
    – José Carlos Santos
    Jul 16 at 16:17










Your Answer




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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Your approach is fine. Note thatbeginalignlim_zto z_0fracf^2(z)+g(z)h(z)+r(z)&=lim_zto z_0fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)\&=lim_zto z_0fraca^2(z)+(z-z_0)b(z)c(z)+(z-z_0)^2c(z)\&=fraca^2(z_0)c(z_0)inmathbbCendalignand that therefore your function has a removable singularity at $z_0$ if $z_0$ is not a zero of $c$ and a pole otherwise (whose order is the order of $z_0$ as a zero of $c$).






share|cite|improve this answer























  • Why removable singularity? as $c(z_0)$ is analytic
    – newhere
    Jul 16 at 16:13










  • @newhere Right. I've edited my answer.
    – José Carlos Santos
    Jul 16 at 16:17














up vote
1
down vote



accepted










Your approach is fine. Note thatbeginalignlim_zto z_0fracf^2(z)+g(z)h(z)+r(z)&=lim_zto z_0fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)\&=lim_zto z_0fraca^2(z)+(z-z_0)b(z)c(z)+(z-z_0)^2c(z)\&=fraca^2(z_0)c(z_0)inmathbbCendalignand that therefore your function has a removable singularity at $z_0$ if $z_0$ is not a zero of $c$ and a pole otherwise (whose order is the order of $z_0$ as a zero of $c$).






share|cite|improve this answer























  • Why removable singularity? as $c(z_0)$ is analytic
    – newhere
    Jul 16 at 16:13










  • @newhere Right. I've edited my answer.
    – José Carlos Santos
    Jul 16 at 16:17












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Your approach is fine. Note thatbeginalignlim_zto z_0fracf^2(z)+g(z)h(z)+r(z)&=lim_zto z_0fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)\&=lim_zto z_0fraca^2(z)+(z-z_0)b(z)c(z)+(z-z_0)^2c(z)\&=fraca^2(z_0)c(z_0)inmathbbCendalignand that therefore your function has a removable singularity at $z_0$ if $z_0$ is not a zero of $c$ and a pole otherwise (whose order is the order of $z_0$ as a zero of $c$).






share|cite|improve this answer















Your approach is fine. Note thatbeginalignlim_zto z_0fracf^2(z)+g(z)h(z)+r(z)&=lim_zto z_0fracfraca^2(z)(z-z_0)^2+fracb(z)z-z_0fracc(z)(z-z_0)^2+r(z)\&=lim_zto z_0fraca^2(z)+(z-z_0)b(z)c(z)+(z-z_0)^2c(z)\&=fraca^2(z_0)c(z_0)inmathbbCendalignand that therefore your function has a removable singularity at $z_0$ if $z_0$ is not a zero of $c$ and a pole otherwise (whose order is the order of $z_0$ as a zero of $c$).







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 16 at 16:17


























answered Jul 16 at 16:12









José Carlos Santos

114k1698177




114k1698177











  • Why removable singularity? as $c(z_0)$ is analytic
    – newhere
    Jul 16 at 16:13










  • @newhere Right. I've edited my answer.
    – José Carlos Santos
    Jul 16 at 16:17
















  • Why removable singularity? as $c(z_0)$ is analytic
    – newhere
    Jul 16 at 16:13










  • @newhere Right. I've edited my answer.
    – José Carlos Santos
    Jul 16 at 16:17















Why removable singularity? as $c(z_0)$ is analytic
– newhere
Jul 16 at 16:13




Why removable singularity? as $c(z_0)$ is analytic
– newhere
Jul 16 at 16:13












@newhere Right. I've edited my answer.
– José Carlos Santos
Jul 16 at 16:17




@newhere Right. I've edited my answer.
– José Carlos Santos
Jul 16 at 16:17












 

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