Mixing
If $f$ and $1/f$ are harmonic then $f$ is holomorphic or antiholomorphic
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I have this problem.
Let $f:Dto mathbbC$ be a function such that $f$ and $1/f$ are harmonic (Their real and imaginary parts are harmonic). Then $f$ is holomorphic or antiholomorphic.
I tried to solve it by computing the laplacian of real and imaginary parts, but it becomes very cumbersome. Is there a better way?
complex-analysis
asked Jul 20 at 17:04
Nell
33819
add a comment |Â
up vote
4
down vote
favorite
I have this problem.
Let $f:Dto mathbbC$ be a function such that $f$ and $1/f$ are harmonic (Their real and imaginary parts are harmonic). Then $f$ is holomorphic or antiholomorphic.
I tried to solve it by computing the laplacian of real and imaginary parts, but it becomes very cumbersome. Is there a better way?
complex-analysis
asked Jul 20 at 17:04
Nell
33819
1
It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $fracpartialpartial overlinezfracpartialpartial zf=0$. Then, compute $0=fracpartialpartial overlinezfracpartialpartial z(frac1zcirc f)$. The chain rule gives you $=fracpartialpartial overlinez(-frac1f^2)cdot fracpartial fpartial z+left(-frac1f^2right)cdot fracpartialpartial overlinezfracpartialpartial zf$ ...
– user577471
Jul 20 at 17:18
So, either $fracpartial fpartial z=0$, in which case $f$ is anti-holomorphic, or $fracpartialpartialoverlinezleft(-frac1f^2right)=0$. Do another chain rule to get $2cdot frac1f^3fracpartial fpartialoverlinez=0$, which gives you that $f$ is holomorphic.
– user577471
Jul 20 at 17:21
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have this problem.
Let $f:Dto mathbbC$ be a function such that $f$ and $1/f$ are harmonic (Their real and imaginary parts are harmonic). Then $f$ is holomorphic or antiholomorphic.
I tried to solve it by computing the laplacian of real and imaginary parts, but it becomes very cumbersome. Is there a better way?
complex-analysis
asked Jul 20 at 17:04
Nell
33819
I have this problem.
Let $f:Dto mathbbC$ be a function such that $f$ and $1/f$ are harmonic (Their real and imaginary parts are harmonic). Then $f$ is holomorphic or antiholomorphic.
I tried to solve it by computing the laplacian of real and imaginary parts, but it becomes very cumbersome. Is there a better way?
complex-analysis
asked Jul 20 at 17:04
Nell
33819
asked Jul 20 at 17:04
Nell
33819
asked Jul 20 at 17:04
Nell
33819
asked Jul 20 at 17:04
Nell
33819
33819
1
It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $fracpartialpartial overlinezfracpartialpartial zf=0$. Then, compute $0=fracpartialpartial overlinezfracpartialpartial z(frac1zcirc f)$. The chain rule gives you $=fracpartialpartial overlinez(-frac1f^2)cdot fracpartial fpartial z+left(-frac1f^2right)cdot fracpartialpartial overlinezfracpartialpartial zf$ ...
– user577471
Jul 20 at 17:18
So, either $fracpartial fpartial z=0$, in which case $f$ is anti-holomorphic, or $fracpartialpartialoverlinezleft(-frac1f^2right)=0$. Do another chain rule to get $2cdot frac1f^3fracpartial fpartialoverlinez=0$, which gives you that $f$ is holomorphic.
– user577471
Jul 20 at 17:21
add a comment |Â
1
It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $fracpartialpartial overlinezfracpartialpartial zf=0$. Then, compute $0=fracpartialpartial overlinezfracpartialpartial z(frac1zcirc f)$. The chain rule gives you $=fracpartialpartial overlinez(-frac1f^2)cdot fracpartial fpartial z+left(-frac1f^2right)cdot fracpartialpartial overlinezfracpartialpartial zf$ ...
– user577471
Jul 20 at 17:18
So, either $fracpartial fpartial z=0$, in which case $f$ is anti-holomorphic, or $fracpartialpartialoverlinezleft(-frac1f^2right)=0$. Do another chain rule to get $2cdot frac1f^3fracpartial fpartialoverlinez=0$, which gives you that $f$ is holomorphic.
– user577471
Jul 20 at 17:21
1
1
It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $fracpartialpartial overlinezfracpartialpartial zf=0$. Then, compute $0=fracpartialpartial overlinezfracpartialpartial z(frac1zcirc f)$. The chain rule gives you $=fracpartialpartial overlinez(-frac1f^2)cdot fracpartial fpartial z+left(-frac1f^2right)cdot fracpartialpartial overlinezfracpartialpartial zf$ ...
– user577471
Jul 20 at 17:18
It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $fracpartialpartial overlinezfracpartialpartial zf=0$. Then, compute $0=fracpartialpartial overlinezfracpartialpartial z(frac1zcirc f)$. The chain rule gives you $=fracpartialpartial overlinez(-frac1f^2)cdot fracpartial fpartial z+left(-frac1f^2right)cdot fracpartialpartial overlinezfracpartialpartial zf$ ...
– user577471
Jul 20 at 17:18
So, either $fracpartial fpartial z=0$, in which case $f$ is anti-holomorphic, or $fracpartialpartialoverlinezleft(-frac1f^2right)=0$. Do another chain rule to get $2cdot frac1f^3fracpartial fpartialoverlinez=0$, which gives you that $f$ is holomorphic.
– user577471
Jul 20 at 17:21
So, either $fracpartial fpartial z=0$, in which case $f$ is anti-holomorphic, or $fracpartialpartialoverlinezleft(-frac1f^2right)=0$. Do another chain rule to get $2cdot frac1f^3fracpartial fpartialoverlinez=0$, which gives you that $f$ is holomorphic.
– user577471
Jul 20 at 17:21
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
Possible idea. Let $f=f(x,y)$. If $f$ and $1/f$ are harmonic then
$$
fracpartial^2 fpartial x^2 + fracpartial^2 fpartial y^2 = 0
$$
$$
fracpartial^2 1/fpartial x^2 + fracpartial^2 1/fpartial y^2 = 0
$$
where
$$
fracpartial 1/fpartial x = -fracpartial fpartial x frac1f^2 qquad fracpartial^2 1/fpartial x^2 = -fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3
$$
$$
fracpartial 1/fpartial y = -fracpartial fpartial y frac1f^2 qquad fracpartial^2 1/fpartial y^2 = -fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3
$$
So
$$
-fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3-fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3 = frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)
$$
$$
frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)=0
$$
Since $f not equiv 0$, we have
$$
0=left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2 =left(fracpartial fpartial x+fracpartial f partial yiright)left(fracpartial fpartial x-fracpartial f partial yiright)
$$
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
add a comment |Â
up vote
2
down vote
Let $fracmathrmdmathrmdz=tfrac12(fracmathrmdmathrmdx-mathrmifracmathrmdmathrmdy)$ and $fracmathrmdmathrmdoverlinez=tfrac12(fracmathrmdmathrmdx+mathrmifracmathrmdmathrmdy)$ as usual. Then $$fracmathrmd^2mathrmdz , mathrmdoverlinez = frac14left(fracmathrmd^2mathrmdx^2+ fracmathrmd^2mathrmdy^2right)$$ and $f$ is holomorphic if $fracmathrmdfmathrmdoverlinez=0$ or anti-holomorphic if $fracmathrmdfmathrmdz=0$. Now under the given conditions $$0 = fracmathrmd^2 f^-1mathrmdz , mathrmdoverlinez = 2 f^-3 fracmathrmdfmathrmdz fracmathrmdfmathrmdoverlinez.$$
answered Jul 20 at 17:30
WimC
23.7k22860
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Possible idea. Let $f=f(x,y)$. If $f$ and $1/f$ are harmonic then
$$
fracpartial^2 fpartial x^2 + fracpartial^2 fpartial y^2 = 0
$$
$$
fracpartial^2 1/fpartial x^2 + fracpartial^2 1/fpartial y^2 = 0
$$
where
$$
fracpartial 1/fpartial x = -fracpartial fpartial x frac1f^2 qquad fracpartial^2 1/fpartial x^2 = -fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3
$$
$$
fracpartial 1/fpartial y = -fracpartial fpartial y frac1f^2 qquad fracpartial^2 1/fpartial y^2 = -fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3
$$
So
$$
-fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3-fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3 = frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)
$$
$$
frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)=0
$$
Since $f not equiv 0$, we have
$$
0=left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2 =left(fracpartial fpartial x+fracpartial f partial yiright)left(fracpartial fpartial x-fracpartial f partial yiright)
$$
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
add a comment |Â
up vote
4
down vote
Possible idea. Let $f=f(x,y)$. If $f$ and $1/f$ are harmonic then
$$
fracpartial^2 fpartial x^2 + fracpartial^2 fpartial y^2 = 0
$$
$$
fracpartial^2 1/fpartial x^2 + fracpartial^2 1/fpartial y^2 = 0
$$
where
$$
fracpartial 1/fpartial x = -fracpartial fpartial x frac1f^2 qquad fracpartial^2 1/fpartial x^2 = -fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3
$$
$$
fracpartial 1/fpartial y = -fracpartial fpartial y frac1f^2 qquad fracpartial^2 1/fpartial y^2 = -fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3
$$
So
$$
-fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3-fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3 = frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)
$$
$$
frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)=0
$$
Since $f not equiv 0$, we have
$$
0=left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2 =left(fracpartial fpartial x+fracpartial f partial yiright)left(fracpartial fpartial x-fracpartial f partial yiright)
$$
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Possible idea. Let $f=f(x,y)$. If $f$ and $1/f$ are harmonic then
$$
fracpartial^2 fpartial x^2 + fracpartial^2 fpartial y^2 = 0
$$
$$
fracpartial^2 1/fpartial x^2 + fracpartial^2 1/fpartial y^2 = 0
$$
where
$$
fracpartial 1/fpartial x = -fracpartial fpartial x frac1f^2 qquad fracpartial^2 1/fpartial x^2 = -fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3
$$
$$
fracpartial 1/fpartial y = -fracpartial fpartial y frac1f^2 qquad fracpartial^2 1/fpartial y^2 = -fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3
$$
So
$$
-fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3-fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3 = frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)
$$
$$
frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)=0
$$
Since $f not equiv 0$, we have
$$
0=left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2 =left(fracpartial fpartial x+fracpartial f partial yiright)left(fracpartial fpartial x-fracpartial f partial yiright)
$$
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
Possible idea. Let $f=f(x,y)$. If $f$ and $1/f$ are harmonic then
$$
fracpartial^2 fpartial x^2 + fracpartial^2 fpartial y^2 = 0
$$
$$
fracpartial^2 1/fpartial x^2 + fracpartial^2 1/fpartial y^2 = 0
$$
where
$$
fracpartial 1/fpartial x = -fracpartial fpartial x frac1f^2 qquad fracpartial^2 1/fpartial x^2 = -fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3
$$
$$
fracpartial 1/fpartial y = -fracpartial fpartial y frac1f^2 qquad fracpartial^2 1/fpartial y^2 = -fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3
$$
So
$$
-fracpartial^2 fpartial x^2 frac1f^2 +2left(fracpartial fpartial xright)^2frac1f^3-fracpartial^2 fpartial y^2 frac1f^2 +2left(fracpartial fpartial yright)^2frac1f^3 = frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)
$$
$$
frac2f^3left(left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2right)=0
$$
Since $f not equiv 0$, we have
$$
0=left(fracpartial fpartial xright)^2+left(fracpartial f partial yright)^2 =left(fracpartial fpartial x+fracpartial f partial yiright)left(fracpartial fpartial x-fracpartial f partial yiright)
$$
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
answered Jul 20 at 17:18
Rafael Gonzalez Lopez
652112
652112
add a comment |Â
add a comment |Â
up vote
2
down vote
Let $fracmathrmdmathrmdz=tfrac12(fracmathrmdmathrmdx-mathrmifracmathrmdmathrmdy)$ and $fracmathrmdmathrmdoverlinez=tfrac12(fracmathrmdmathrmdx+mathrmifracmathrmdmathrmdy)$ as usual. Then $$fracmathrmd^2mathrmdz , mathrmdoverlinez = frac14left(fracmathrmd^2mathrmdx^2+ fracmathrmd^2mathrmdy^2right)$$ and $f$ is holomorphic if $fracmathrmdfmathrmdoverlinez=0$ or anti-holomorphic if $fracmathrmdfmathrmdz=0$. Now under the given conditions $$0 = fracmathrmd^2 f^-1mathrmdz , mathrmdoverlinez = 2 f^-3 fracmathrmdfmathrmdz fracmathrmdfmathrmdoverlinez.$$
answered Jul 20 at 17:30
WimC
23.7k22860
add a comment |Â
up vote
2
down vote
Let $fracmathrmdmathrmdz=tfrac12(fracmathrmdmathrmdx-mathrmifracmathrmdmathrmdy)$ and $fracmathrmdmathrmdoverlinez=tfrac12(fracmathrmdmathrmdx+mathrmifracmathrmdmathrmdy)$ as usual. Then $$fracmathrmd^2mathrmdz , mathrmdoverlinez = frac14left(fracmathrmd^2mathrmdx^2+ fracmathrmd^2mathrmdy^2right)$$ and $f$ is holomorphic if $fracmathrmdfmathrmdoverlinez=0$ or anti-holomorphic if $fracmathrmdfmathrmdz=0$. Now under the given conditions $$0 = fracmathrmd^2 f^-1mathrmdz , mathrmdoverlinez = 2 f^-3 fracmathrmdfmathrmdz fracmathrmdfmathrmdoverlinez.$$
answered Jul 20 at 17:30
WimC
23.7k22860
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $fracmathrmdmathrmdz=tfrac12(fracmathrmdmathrmdx-mathrmifracmathrmdmathrmdy)$ and $fracmathrmdmathrmdoverlinez=tfrac12(fracmathrmdmathrmdx+mathrmifracmathrmdmathrmdy)$ as usual. Then $$fracmathrmd^2mathrmdz , mathrmdoverlinez = frac14left(fracmathrmd^2mathrmdx^2+ fracmathrmd^2mathrmdy^2right)$$ and $f$ is holomorphic if $fracmathrmdfmathrmdoverlinez=0$ or anti-holomorphic if $fracmathrmdfmathrmdz=0$. Now under the given conditions $$0 = fracmathrmd^2 f^-1mathrmdz , mathrmdoverlinez = 2 f^-3 fracmathrmdfmathrmdz fracmathrmdfmathrmdoverlinez.$$
answered Jul 20 at 17:30
WimC
23.7k22860
Let $fracmathrmdmathrmdz=tfrac12(fracmathrmdmathrmdx-mathrmifracmathrmdmathrmdy)$ and $fracmathrmdmathrmdoverlinez=tfrac12(fracmathrmdmathrmdx+mathrmifracmathrmdmathrmdy)$ as usual. Then $$fracmathrmd^2mathrmdz , mathrmdoverlinez = frac14left(fracmathrmd^2mathrmdx^2+ fracmathrmd^2mathrmdy^2right)$$ and $f$ is holomorphic if $fracmathrmdfmathrmdoverlinez=0$ or anti-holomorphic if $fracmathrmdfmathrmdz=0$. Now under the given conditions $$0 = fracmathrmd^2 f^-1mathrmdz , mathrmdoverlinez = 2 f^-3 fracmathrmdfmathrmdz fracmathrmdfmathrmdoverlinez.$$
answered Jul 20 at 17:30
WimC
23.7k22860
edited Jul 20 at 17:36
answered Jul 20 at 17:30
WimC
23.7k22860
answered Jul 20 at 17:30
WimC
23.7k22860
answered Jul 20 at 17:30
WimC
23.7k22860
23.7k22860
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857840%2fif-f-and-1-f-are-harmonic-then-f-is-holomorphic-or-antiholomorphic%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
It is a cleaner trip if you write the computation using the Wirtinger derivatives. That $f$ is harmonic means that $fracpartialpartial overlinezfracpartialpartial zf=0$. Then, compute $0=fracpartialpartial overlinezfracpartialpartial z(frac1zcirc f)$. The chain rule gives you $=fracpartialpartial overlinez(-frac1f^2)cdot fracpartial fpartial z+left(-frac1f^2right)cdot fracpartialpartial overlinezfracpartialpartial zf$ ...
– user577471
Jul 20 at 17:18
So, either $fracpartial fpartial z=0$, in which case $f$ is anti-holomorphic, or $fracpartialpartialoverlinezleft(-frac1f^2right)=0$. Do another chain rule to get $2cdot frac1f^3fracpartial fpartialoverlinez=0$, which gives you that $f$ is holomorphic.
– user577471
Jul 20 at 17:21