Mixing
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Complex Conformal Mapping
$begingroup$
I'm trying to prove that the mapping
$Z=z^m$
for $m geq dfrac 12$, transforms the region $r geq 0$, $0leqthetaleqdfrac{pi}{m}$ into the upper half plane $Ygeq 0$ where $Z=X+iY.$
My attempt was to write
$z=re^{itheta}$ so that
$$z^m = r^mbigg[cos(theta m) + isin(theta m)bigg],$$
and the upper-half region would be controlled by $operatorname{Im}(z)$, so
$$operatorname{Im}(z^m) = r^m sin(pi)=r^mk$$
so that $k in [0,1].$
I feel like I am missing why $m$ must be $ geq dfrac 12.$
Any help is appreciated.
complex-analysis functions complex-numbers
edited Jan 8 at 14:04
amWhy
1
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that the mapping
$Z=z^m$
for $m geq dfrac 12$, transforms the region $r geq 0$, $0leqthetaleqdfrac{pi}{m}$ into the upper half plane $Ygeq 0$ where $Z=X+iY.$
My attempt was to write
$z=re^{itheta}$ so that
$$z^m = r^mbigg[cos(theta m) + isin(theta m)bigg],$$
and the upper-half region would be controlled by $operatorname{Im}(z)$, so
$$operatorname{Im}(z^m) = r^m sin(pi)=r^mk$$
so that $k in [0,1].$
I feel like I am missing why $m$ must be $ geq dfrac 12.$
Any help is appreciated.
complex-analysis functions complex-numbers
edited Jan 8 at 14:04
amWhy
1
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
$endgroup$
$begingroup$
I think it has to do with your angle range: $0lethetalepi/m.$ If, say, $m=1/4,$ then you'd be double-covering the complex plane with the inequality $0lethetale4pi.$
$endgroup$
– Adrian Keister
Jan 8 at 14:14
$begingroup$
Yes my thoughts! Thanks for your comment Adrian!
$endgroup$
– kroneckerdel69
Jan 8 at 15:03
add a comment |
$begingroup$
I'm trying to prove that the mapping
$Z=z^m$
for $m geq dfrac 12$, transforms the region $r geq 0$, $0leqthetaleqdfrac{pi}{m}$ into the upper half plane $Ygeq 0$ where $Z=X+iY.$
My attempt was to write
$z=re^{itheta}$ so that
$$z^m = r^mbigg[cos(theta m) + isin(theta m)bigg],$$
and the upper-half region would be controlled by $operatorname{Im}(z)$, so
$$operatorname{Im}(z^m) = r^m sin(pi)=r^mk$$
so that $k in [0,1].$
I feel like I am missing why $m$ must be $ geq dfrac 12.$
Any help is appreciated.
complex-analysis functions complex-numbers
edited Jan 8 at 14:04
amWhy
1
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
$endgroup$
I'm trying to prove that the mapping
$Z=z^m$
for $m geq dfrac 12$, transforms the region $r geq 0$, $0leqthetaleqdfrac{pi}{m}$ into the upper half plane $Ygeq 0$ where $Z=X+iY.$
My attempt was to write
$z=re^{itheta}$ so that
$$z^m = r^mbigg[cos(theta m) + isin(theta m)bigg],$$
and the upper-half region would be controlled by $operatorname{Im}(z)$, so
$$operatorname{Im}(z^m) = r^m sin(pi)=r^mk$$
so that $k in [0,1].$
I feel like I am missing why $m$ must be $ geq dfrac 12.$
Any help is appreciated.
complex-analysis functions complex-numbers
complex-analysis functions complex-numbers
edited Jan 8 at 14:04
amWhy
1
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
edited Jan 8 at 14:04
amWhy
1
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
edited Jan 8 at 14:04
amWhy
1
edited Jan 8 at 14:04
amWhy
1
edited Jan 8 at 14:04
amWhy
1
1
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
asked Jan 8 at 13:02
kroneckerdel69kroneckerdel69
258
258
$begingroup$
I think it has to do with your angle range: $0lethetalepi/m.$ If, say, $m=1/4,$ then you'd be double-covering the complex plane with the inequality $0lethetale4pi.$
$endgroup$
– Adrian Keister
Jan 8 at 14:14
$begingroup$
Yes my thoughts! Thanks for your comment Adrian!
$endgroup$
– kroneckerdel69
Jan 8 at 15:03
add a comment |
$begingroup$
I think it has to do with your angle range: $0lethetalepi/m.$ If, say, $m=1/4,$ then you'd be double-covering the complex plane with the inequality $0lethetale4pi.$
$endgroup$
– Adrian Keister
Jan 8 at 14:14
$begingroup$
Yes my thoughts! Thanks for your comment Adrian!
$endgroup$
– kroneckerdel69
Jan 8 at 15:03
$begingroup$
I think it has to do with your angle range: $0lethetalepi/m.$ If, say, $m=1/4,$ then you'd be double-covering the complex plane with the inequality $0lethetale4pi.$
$endgroup$
– Adrian Keister
Jan 8 at 14:14
$begingroup$
I think it has to do with your angle range: $0lethetalepi/m.$ If, say, $m=1/4,$ then you'd be double-covering the complex plane with the inequality $0lethetale4pi.$
$endgroup$
– Adrian Keister
Jan 8 at 14:14
$begingroup$
Yes my thoughts! Thanks for your comment Adrian!
$endgroup$
– kroneckerdel69
Jan 8 at 15:03
$begingroup$
Yes my thoughts! Thanks for your comment Adrian!
$endgroup$
– kroneckerdel69
Jan 8 at 15:03
add a comment |
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$begingroup$
I think it has to do with your angle range: $0lethetalepi/m.$ If, say, $m=1/4,$ then you'd be double-covering the complex plane with the inequality $0lethetale4pi.$
$endgroup$
– Adrian Keister
Jan 8 at 14:14
$begingroup$
Yes my thoughts! Thanks for your comment Adrian!
$endgroup$
– kroneckerdel69
Jan 8 at 15:03