Is there a continous branch in the neighbourhood of 0
Is there a continuous branch in the neighborhood of function $z$ around $0.$ In my opinion no, because as we take a circle and go around it, the argument gains $2pi$ and the value of logarithm changes by $2ipi$. But do I understand it well?
My problem was to determine if there is a continuous branch of a function $f(z)=(z^2-1)^{1/2}$ in infinity, I changed it into $g(z)=1/f(1/z)$ in the 0 neighborhood, which makes it $g(z)=z/(1-z)^{1/2}$, the denominator has continuous branch ofc as I can take the Arg z, but does the numerator have it too?
complex-analysis complex-numbers analytic-functions
edited Nov 27 '18 at 16:13
amWhy
192k28224439
asked Nov 27 '18 at 15:50
ryszard eggink
308110
add a comment |
Is there a continuous branch in the neighborhood of function $z$ around $0.$ In my opinion no, because as we take a circle and go around it, the argument gains $2pi$ and the value of logarithm changes by $2ipi$. But do I understand it well?
My problem was to determine if there is a continuous branch of a function $f(z)=(z^2-1)^{1/2}$ in infinity, I changed it into $g(z)=1/f(1/z)$ in the 0 neighborhood, which makes it $g(z)=z/(1-z)^{1/2}$, the denominator has continuous branch ofc as I can take the Arg z, but does the numerator have it too?
complex-analysis complex-numbers analytic-functions
edited Nov 27 '18 at 16:13
amWhy
192k28224439
asked Nov 27 '18 at 15:50
ryszard eggink
308110
add a comment |
Is there a continuous branch in the neighborhood of function $z$ around $0.$ In my opinion no, because as we take a circle and go around it, the argument gains $2pi$ and the value of logarithm changes by $2ipi$. But do I understand it well?
My problem was to determine if there is a continuous branch of a function $f(z)=(z^2-1)^{1/2}$ in infinity, I changed it into $g(z)=1/f(1/z)$ in the 0 neighborhood, which makes it $g(z)=z/(1-z)^{1/2}$, the denominator has continuous branch ofc as I can take the Arg z, but does the numerator have it too?
complex-analysis complex-numbers analytic-functions
edited Nov 27 '18 at 16:13
amWhy
192k28224439
asked Nov 27 '18 at 15:50
ryszard eggink
308110
Is there a continuous branch in the neighborhood of function $z$ around $0.$ In my opinion no, because as we take a circle and go around it, the argument gains $2pi$ and the value of logarithm changes by $2ipi$. But do I understand it well?
My problem was to determine if there is a continuous branch of a function $f(z)=(z^2-1)^{1/2}$ in infinity, I changed it into $g(z)=1/f(1/z)$ in the 0 neighborhood, which makes it $g(z)=z/(1-z)^{1/2}$, the denominator has continuous branch ofc as I can take the Arg z, but does the numerator have it too?
complex-analysis complex-numbers analytic-functions
complex-analysis complex-numbers analytic-functions
edited Nov 27 '18 at 16:13
amWhy
192k28224439
asked Nov 27 '18 at 15:50
ryszard eggink
308110
edited Nov 27 '18 at 16:13
amWhy
192k28224439
asked Nov 27 '18 at 15:50
ryszard eggink
308110
edited Nov 27 '18 at 16:13
amWhy
192k28224439
edited Nov 27 '18 at 16:13
amWhy
192k28224439
edited Nov 27 '18 at 16:13
amWhy
192k28224439
192k28224439
asked Nov 27 '18 at 15:50
ryszard eggink
308110
asked Nov 27 '18 at 15:50
ryszard eggink
308110
asked Nov 27 '18 at 15:50
ryszard eggink
308110
308110
add a comment |
add a comment |
1 Answer
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I think you are confusing $z$ with $log z$. To the second part, yes there is, provided you cut the plane between the branch points at $pm1$. Circling around $infty$ causes the argument to clock up $pi i$ twice, which cancels. As you say, it looks like $z$ at $infty$. Note, $log sqrt{z^2-1}$ does not have a cts branch around $infty$.
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
I think you are confusing $z$ with $log z$. To the second part, yes there is, provided you cut the plane between the branch points at $pm1$. Circling around $infty$ causes the argument to clock up $pi i$ twice, which cancels. As you say, it looks like $z$ at $infty$. Note, $log sqrt{z^2-1}$ does not have a cts branch around $infty$.
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
add a comment |
I think you are confusing $z$ with $log z$. To the second part, yes there is, provided you cut the plane between the branch points at $pm1$. Circling around $infty$ causes the argument to clock up $pi i$ twice, which cancels. As you say, it looks like $z$ at $infty$. Note, $log sqrt{z^2-1}$ does not have a cts branch around $infty$.
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
add a comment |
I think you are confusing $z$ with $log z$. To the second part, yes there is, provided you cut the plane between the branch points at $pm1$. Circling around $infty$ causes the argument to clock up $pi i$ twice, which cancels. As you say, it looks like $z$ at $infty$. Note, $log sqrt{z^2-1}$ does not have a cts branch around $infty$.
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
I think you are confusing $z$ with $log z$. To the second part, yes there is, provided you cut the plane between the branch points at $pm1$. Circling around $infty$ causes the argument to clock up $pi i$ twice, which cancels. As you say, it looks like $z$ at $infty$. Note, $log sqrt{z^2-1}$ does not have a cts branch around $infty$.
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
answered Nov 27 '18 at 16:11
Richard Martin
1,63618
1,63618
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
add a comment |
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
Is existing branch of z not equivalent to existing branch of Logz? I thought $z=e^{Logz}$
– ryszard eggink
Nov 27 '18 at 16:15
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
No such thing as a branch of $z$ as it's analytic
– Richard Martin
Nov 27 '18 at 16:37
add a comment |
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