Closed paths in the complex plane
I have attached the image above, which is from lecture notes dealing with branch cuts and branch points. My question is: why does $phi$ change by multiples of $2pi$ as we travel around the point $z=1$, on a closed path? How do we show this?
Thanks in advance.
complex-analysis complex-numbers
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
add a comment |
I have attached the image above, which is from lecture notes dealing with branch cuts and branch points. My question is: why does $phi$ change by multiples of $2pi$ as we travel around the point $z=1$, on a closed path? How do we show this?
Thanks in advance.
complex-analysis complex-numbers
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
add a comment |
I have attached the image above, which is from lecture notes dealing with branch cuts and branch points. My question is: why does $phi$ change by multiples of $2pi$ as we travel around the point $z=1$, on a closed path? How do we show this?
Thanks in advance.
complex-analysis complex-numbers
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
I have attached the image above, which is from lecture notes dealing with branch cuts and branch points. My question is: why does $phi$ change by multiples of $2pi$ as we travel around the point $z=1$, on a closed path? How do we show this?
Thanks in advance.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
asked Nov 20 '18 at 22:36
Live Free or π Hard
479213
479213
add a comment |
add a comment |
1 Answer
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This example might be confusing because of the translation. In fact, if $phi$ is allowed to be $any$ real number, then $log$ as written, isn't even a function.
So let's distinguish branch points from "non"-branch points using simpler examples.
Take for example $f(z)=frac{z}{z+1}$ and write $z=rho e^{iphi}.$ Then, of course, since $z=rho e^{i(phi+2pi n)}; nin mathbb Z, $ we can use $any$ of these representations of $z$ that we want. And, substituting any of these into the formula for $f$, we get the $same$ result.
Now consider $f(z)=z^{sqrt 2}$. Take $phi=0, $ and then $phi=2pi$ and $z=(1,0)=e^{i(0)}=e^{i(0+2pi)}.$ Substituting these two representations of $z$ into $f$, we get $1$ and $e^{2pisqrt 2 i}, $ respectively, and these are $textit{different}.$
Now if we consider the line segment in $mathbb C$ from $z=(0,0)$ to $z=(1,0)$ we can regard the second representation of $z$ as having been obtained from the first by a counterclockwise rotation by $2pi$ radians of the segment about $z=0$. And when we do this, we get back to the same point $z$, but the value of $f$ has changed. By definition then, $z=0$ is a branch point.
When we perform this geometric analysis on the first example, we see that the value of $f$ at the rotated vs. unrotated point has not changed, and so $z=0$ is not a branch point for $f$.
So what is the problem with the second example? It lies in the definition of the power function, which uses the Arg function, which is discontinuous on the negative real axis, so that a rotation through $2pi$ radians forces you to "jump" over the discontinuity.
To apply this to your case, translate the whole picture by $1$ unit to the right and use $log$ instead of the power function.
answered Nov 20 '18 at 23:41
Matematleta
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1 Answer
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This example might be confusing because of the translation. In fact, if $phi$ is allowed to be $any$ real number, then $log$ as written, isn't even a function.
So let's distinguish branch points from "non"-branch points using simpler examples.
Take for example $f(z)=frac{z}{z+1}$ and write $z=rho e^{iphi}.$ Then, of course, since $z=rho e^{i(phi+2pi n)}; nin mathbb Z, $ we can use $any$ of these representations of $z$ that we want. And, substituting any of these into the formula for $f$, we get the $same$ result.
Now consider $f(z)=z^{sqrt 2}$. Take $phi=0, $ and then $phi=2pi$ and $z=(1,0)=e^{i(0)}=e^{i(0+2pi)}.$ Substituting these two representations of $z$ into $f$, we get $1$ and $e^{2pisqrt 2 i}, $ respectively, and these are $textit{different}.$
Now if we consider the line segment in $mathbb C$ from $z=(0,0)$ to $z=(1,0)$ we can regard the second representation of $z$ as having been obtained from the first by a counterclockwise rotation by $2pi$ radians of the segment about $z=0$. And when we do this, we get back to the same point $z$, but the value of $f$ has changed. By definition then, $z=0$ is a branch point.
When we perform this geometric analysis on the first example, we see that the value of $f$ at the rotated vs. unrotated point has not changed, and so $z=0$ is not a branch point for $f$.
So what is the problem with the second example? It lies in the definition of the power function, which uses the Arg function, which is discontinuous on the negative real axis, so that a rotation through $2pi$ radians forces you to "jump" over the discontinuity.
To apply this to your case, translate the whole picture by $1$ unit to the right and use $log$ instead of the power function.
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
add a comment |
This example might be confusing because of the translation. In fact, if $phi$ is allowed to be $any$ real number, then $log$ as written, isn't even a function.
So let's distinguish branch points from "non"-branch points using simpler examples.
Take for example $f(z)=frac{z}{z+1}$ and write $z=rho e^{iphi}.$ Then, of course, since $z=rho e^{i(phi+2pi n)}; nin mathbb Z, $ we can use $any$ of these representations of $z$ that we want. And, substituting any of these into the formula for $f$, we get the $same$ result.
Now consider $f(z)=z^{sqrt 2}$. Take $phi=0, $ and then $phi=2pi$ and $z=(1,0)=e^{i(0)}=e^{i(0+2pi)}.$ Substituting these two representations of $z$ into $f$, we get $1$ and $e^{2pisqrt 2 i}, $ respectively, and these are $textit{different}.$
Now if we consider the line segment in $mathbb C$ from $z=(0,0)$ to $z=(1,0)$ we can regard the second representation of $z$ as having been obtained from the first by a counterclockwise rotation by $2pi$ radians of the segment about $z=0$. And when we do this, we get back to the same point $z$, but the value of $f$ has changed. By definition then, $z=0$ is a branch point.
When we perform this geometric analysis on the first example, we see that the value of $f$ at the rotated vs. unrotated point has not changed, and so $z=0$ is not a branch point for $f$.
So what is the problem with the second example? It lies in the definition of the power function, which uses the Arg function, which is discontinuous on the negative real axis, so that a rotation through $2pi$ radians forces you to "jump" over the discontinuity.
To apply this to your case, translate the whole picture by $1$ unit to the right and use $log$ instead of the power function.
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
add a comment |
This example might be confusing because of the translation. In fact, if $phi$ is allowed to be $any$ real number, then $log$ as written, isn't even a function.
So let's distinguish branch points from "non"-branch points using simpler examples.
Take for example $f(z)=frac{z}{z+1}$ and write $z=rho e^{iphi}.$ Then, of course, since $z=rho e^{i(phi+2pi n)}; nin mathbb Z, $ we can use $any$ of these representations of $z$ that we want. And, substituting any of these into the formula for $f$, we get the $same$ result.
Now consider $f(z)=z^{sqrt 2}$. Take $phi=0, $ and then $phi=2pi$ and $z=(1,0)=e^{i(0)}=e^{i(0+2pi)}.$ Substituting these two representations of $z$ into $f$, we get $1$ and $e^{2pisqrt 2 i}, $ respectively, and these are $textit{different}.$
Now if we consider the line segment in $mathbb C$ from $z=(0,0)$ to $z=(1,0)$ we can regard the second representation of $z$ as having been obtained from the first by a counterclockwise rotation by $2pi$ radians of the segment about $z=0$. And when we do this, we get back to the same point $z$, but the value of $f$ has changed. By definition then, $z=0$ is a branch point.
When we perform this geometric analysis on the first example, we see that the value of $f$ at the rotated vs. unrotated point has not changed, and so $z=0$ is not a branch point for $f$.
So what is the problem with the second example? It lies in the definition of the power function, which uses the Arg function, which is discontinuous on the negative real axis, so that a rotation through $2pi$ radians forces you to "jump" over the discontinuity.
To apply this to your case, translate the whole picture by $1$ unit to the right and use $log$ instead of the power function.
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
This example might be confusing because of the translation. In fact, if $phi$ is allowed to be $any$ real number, then $log$ as written, isn't even a function.
So let's distinguish branch points from "non"-branch points using simpler examples.
Take for example $f(z)=frac{z}{z+1}$ and write $z=rho e^{iphi}.$ Then, of course, since $z=rho e^{i(phi+2pi n)}; nin mathbb Z, $ we can use $any$ of these representations of $z$ that we want. And, substituting any of these into the formula for $f$, we get the $same$ result.
Now consider $f(z)=z^{sqrt 2}$. Take $phi=0, $ and then $phi=2pi$ and $z=(1,0)=e^{i(0)}=e^{i(0+2pi)}.$ Substituting these two representations of $z$ into $f$, we get $1$ and $e^{2pisqrt 2 i}, $ respectively, and these are $textit{different}.$
Now if we consider the line segment in $mathbb C$ from $z=(0,0)$ to $z=(1,0)$ we can regard the second representation of $z$ as having been obtained from the first by a counterclockwise rotation by $2pi$ radians of the segment about $z=0$. And when we do this, we get back to the same point $z$, but the value of $f$ has changed. By definition then, $z=0$ is a branch point.
When we perform this geometric analysis on the first example, we see that the value of $f$ at the rotated vs. unrotated point has not changed, and so $z=0$ is not a branch point for $f$.
So what is the problem with the second example? It lies in the definition of the power function, which uses the Arg function, which is discontinuous on the negative real axis, so that a rotation through $2pi$ radians forces you to "jump" over the discontinuity.
To apply this to your case, translate the whole picture by $1$ unit to the right and use $log$ instead of the power function.
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
answered Nov 20 '18 at 23:41
Matematleta
9,9522918
9,9522918
add a comment |
add a comment |
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