3D-plots of complex functions
$begingroup$
Commonly it's believed that one cannot fully visualize a complex function $f:mathbb{C}rightarrow mathbb{C}$ because the full plot would have to be 4-dimensional. But is this true? And why would the following surface (embedded in $mathbb{R}^3$) not count as a full plot of $f$?
Consider complex polynomials $f(z)$ (which are prototypes of entire functions). For each $r$ you can plot the circle $C_r = {r e^{ivarphi} | 0 leq varphi < 2pi}$ which is a closed curve in $mathbb{R}^2$. Adding $r$ as a third dimension you get an intricated surface in $mathbb{R}^3$ which can be considered a plot of $f$:
source
For a given point $(u,v,r)$ on the surface – which indicates a pair $(z,f(z))$ – you can tell $f(z) = (u,v)$ and $z = re^{ivarphi}$ for some $0 leq varphi < 2pi$. But you cannot tell $varphi$, that's the missing information (dimension). But in the two branched plot of the real square root you cannot tell what the square root of $4$ is, neither: $+2$ or $-2$?
Nevertheless the plot gives you a unique picture of the function – I guess there are no two polynomials with the same plot, are there?
My questions is:
What's the name of these surfaces? (Are they some kind of "Riemannian surfaces"?)
If they are not Riemannian surfaces: How are they related to those?
- Why do they seem to be not so prominent as they seem to deserve (at least in my opinion)?
They reveal a lot about a complex function, and they are rather easy to grasp - at least easier than domain-colored plots, at least for the beginner. But you rarely find them, even in the visualization of complex functions literature, do you?
functional-analysis riemann-surfaces visualization
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
$endgroup$
add a comment |
$begingroup$
Commonly it's believed that one cannot fully visualize a complex function $f:mathbb{C}rightarrow mathbb{C}$ because the full plot would have to be 4-dimensional. But is this true? And why would the following surface (embedded in $mathbb{R}^3$) not count as a full plot of $f$?
Consider complex polynomials $f(z)$ (which are prototypes of entire functions). For each $r$ you can plot the circle $C_r = {r e^{ivarphi} | 0 leq varphi < 2pi}$ which is a closed curve in $mathbb{R}^2$. Adding $r$ as a third dimension you get an intricated surface in $mathbb{R}^3$ which can be considered a plot of $f$:
source
For a given point $(u,v,r)$ on the surface – which indicates a pair $(z,f(z))$ – you can tell $f(z) = (u,v)$ and $z = re^{ivarphi}$ for some $0 leq varphi < 2pi$. But you cannot tell $varphi$, that's the missing information (dimension). But in the two branched plot of the real square root you cannot tell what the square root of $4$ is, neither: $+2$ or $-2$?
Nevertheless the plot gives you a unique picture of the function – I guess there are no two polynomials with the same plot, are there?
My questions is:
What's the name of these surfaces? (Are they some kind of "Riemannian surfaces"?)
If they are not Riemannian surfaces: How are they related to those?
- Why do they seem to be not so prominent as they seem to deserve (at least in my opinion)?
They reveal a lot about a complex function, and they are rather easy to grasp - at least easier than domain-colored plots, at least for the beginner. But you rarely find them, even in the visualization of complex functions literature, do you?
functional-analysis riemann-surfaces visualization
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
$endgroup$
1
$begingroup$
See also math.stackexchange.com/questions/10627/… and math.stackexchange.com/questions/607436/…
$endgroup$
– lhf
Jan 8 at 13:43
add a comment |
$begingroup$
Commonly it's believed that one cannot fully visualize a complex function $f:mathbb{C}rightarrow mathbb{C}$ because the full plot would have to be 4-dimensional. But is this true? And why would the following surface (embedded in $mathbb{R}^3$) not count as a full plot of $f$?
Consider complex polynomials $f(z)$ (which are prototypes of entire functions). For each $r$ you can plot the circle $C_r = {r e^{ivarphi} | 0 leq varphi < 2pi}$ which is a closed curve in $mathbb{R}^2$. Adding $r$ as a third dimension you get an intricated surface in $mathbb{R}^3$ which can be considered a plot of $f$:
source
For a given point $(u,v,r)$ on the surface – which indicates a pair $(z,f(z))$ – you can tell $f(z) = (u,v)$ and $z = re^{ivarphi}$ for some $0 leq varphi < 2pi$. But you cannot tell $varphi$, that's the missing information (dimension). But in the two branched plot of the real square root you cannot tell what the square root of $4$ is, neither: $+2$ or $-2$?
Nevertheless the plot gives you a unique picture of the function – I guess there are no two polynomials with the same plot, are there?
My questions is:
What's the name of these surfaces? (Are they some kind of "Riemannian surfaces"?)
If they are not Riemannian surfaces: How are they related to those?
- Why do they seem to be not so prominent as they seem to deserve (at least in my opinion)?
They reveal a lot about a complex function, and they are rather easy to grasp - at least easier than domain-colored plots, at least for the beginner. But you rarely find them, even in the visualization of complex functions literature, do you?
functional-analysis riemann-surfaces visualization
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
$endgroup$
Commonly it's believed that one cannot fully visualize a complex function $f:mathbb{C}rightarrow mathbb{C}$ because the full plot would have to be 4-dimensional. But is this true? And why would the following surface (embedded in $mathbb{R}^3$) not count as a full plot of $f$?
Consider complex polynomials $f(z)$ (which are prototypes of entire functions). For each $r$ you can plot the circle $C_r = {r e^{ivarphi} | 0 leq varphi < 2pi}$ which is a closed curve in $mathbb{R}^2$. Adding $r$ as a third dimension you get an intricated surface in $mathbb{R}^3$ which can be considered a plot of $f$:
source
For a given point $(u,v,r)$ on the surface – which indicates a pair $(z,f(z))$ – you can tell $f(z) = (u,v)$ and $z = re^{ivarphi}$ for some $0 leq varphi < 2pi$. But you cannot tell $varphi$, that's the missing information (dimension). But in the two branched plot of the real square root you cannot tell what the square root of $4$ is, neither: $+2$ or $-2$?
Nevertheless the plot gives you a unique picture of the function – I guess there are no two polynomials with the same plot, are there?
My questions is:
What's the name of these surfaces? (Are they some kind of "Riemannian surfaces"?)
If they are not Riemannian surfaces: How are they related to those?
- Why do they seem to be not so prominent as they seem to deserve (at least in my opinion)?
They reveal a lot about a complex function, and they are rather easy to grasp - at least easier than domain-colored plots, at least for the beginner. But you rarely find them, even in the visualization of complex functions literature, do you?
functional-analysis riemann-surfaces visualization
functional-analysis riemann-surfaces visualization
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
edited Jan 11 at 15:29
Hans Stricker
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
asked Jan 8 at 13:16
Hans StrickerHans Stricker
6,36343990
6,36343990
1
$begingroup$
See also math.stackexchange.com/questions/10627/… and math.stackexchange.com/questions/607436/…
$endgroup$
– lhf
Jan 8 at 13:43
add a comment |
1
$begingroup$
See also math.stackexchange.com/questions/10627/… and math.stackexchange.com/questions/607436/…
$endgroup$
– lhf
Jan 8 at 13:43
1
1
$begingroup$
See also math.stackexchange.com/questions/10627/… and math.stackexchange.com/questions/607436/…
$endgroup$
– lhf
Jan 8 at 13:43
$begingroup$
See also math.stackexchange.com/questions/10627/… and math.stackexchange.com/questions/607436/…
$endgroup$
– lhf
Jan 8 at 13:43
add a comment |
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$begingroup$
See also math.stackexchange.com/questions/10627/… and math.stackexchange.com/questions/607436/…
$endgroup$
– lhf
Jan 8 at 13:43