Is there an easy way to show that this function satisfies Cauchy Riemann equations at $z=0$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let
$$ f(z) = left{ beginalign
&e^-frac1z^4 &hspace1mm mboxif hspace1mm z neq 0 \
&0 &hspace1mm mboxif hspace1mm z = 0 \
endalign right. $$
I want to show that this equation satisfies, CR equations at $z=0$
Let $z=x+iy$ where $x,yinmathbbR$. Then
$$frac1z=fracx-iyx^2+y^2.$$
I get the following expression.
$$e^-fracx^4+y^4-6x^2y^2(x^2+y^2)^4left(cosleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)-i sinleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)right)$$
I have shown that this function satisfies CR equation, but is very tedious.
Is there a more elegant and beautiful way? Or Is this the only way possible.
complex-analysis
asked Aug 15 at 3:58
Mathaman Topologius
83
add a comment |Â
up vote
1
down vote
favorite
Let
$$ f(z) = left{ beginalign
&e^-frac1z^4 &hspace1mm mboxif hspace1mm z neq 0 \
&0 &hspace1mm mboxif hspace1mm z = 0 \
endalign right. $$
I want to show that this equation satisfies, CR equations at $z=0$
Let $z=x+iy$ where $x,yinmathbbR$. Then
$$frac1z=fracx-iyx^2+y^2.$$
I get the following expression.
$$e^-fracx^4+y^4-6x^2y^2(x^2+y^2)^4left(cosleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)-i sinleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)right)$$
I have shown that this function satisfies CR equation, but is very tedious.
Is there a more elegant and beautiful way? Or Is this the only way possible.
complex-analysis
asked Aug 15 at 3:58
Mathaman Topologius
83
1
Just an idea: have you tried polar coordinates?
– Andrei
Aug 15 at 4:10
No. Not yet. My notes does not mention anything like that. Let me Google it.
– Mathaman Topologius
Aug 15 at 4:11
1
Polar coordinates CR equations assume $rne 0$. I am not sure if they are applicable here.
– Szeto
Aug 15 at 4:26
Look equations 18 and 19 mathworld.wolfram.com/Cauchy-RiemannEquations.html
– David
Aug 15 at 5:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let
$$ f(z) = left{ beginalign
&e^-frac1z^4 &hspace1mm mboxif hspace1mm z neq 0 \
&0 &hspace1mm mboxif hspace1mm z = 0 \
endalign right. $$
I want to show that this equation satisfies, CR equations at $z=0$
Let $z=x+iy$ where $x,yinmathbbR$. Then
$$frac1z=fracx-iyx^2+y^2.$$
I get the following expression.
$$e^-fracx^4+y^4-6x^2y^2(x^2+y^2)^4left(cosleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)-i sinleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)right)$$
I have shown that this function satisfies CR equation, but is very tedious.
Is there a more elegant and beautiful way? Or Is this the only way possible.
complex-analysis
asked Aug 15 at 3:58
Mathaman Topologius
83
Let
$$ f(z) = left{ beginalign
&e^-frac1z^4 &hspace1mm mboxif hspace1mm z neq 0 \
&0 &hspace1mm mboxif hspace1mm z = 0 \
endalign right. $$
I want to show that this equation satisfies, CR equations at $z=0$
Let $z=x+iy$ where $x,yinmathbbR$. Then
$$frac1z=fracx-iyx^2+y^2.$$
I get the following expression.
$$e^-fracx^4+y^4-6x^2y^2(x^2+y^2)^4left(cosleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)-i sinleft(frac4xy^3-4x^3y(x^2+y^2)^4 right)right)$$
I have shown that this function satisfies CR equation, but is very tedious.
Is there a more elegant and beautiful way? Or Is this the only way possible.
complex-analysis
asked Aug 15 at 3:58
Mathaman Topologius
83
asked Aug 15 at 3:58
Mathaman Topologius
83
asked Aug 15 at 3:58
Mathaman Topologius
83
asked Aug 15 at 3:58
Mathaman Topologius
83
83
1
Just an idea: have you tried polar coordinates?
– Andrei
Aug 15 at 4:10
No. Not yet. My notes does not mention anything like that. Let me Google it.
– Mathaman Topologius
Aug 15 at 4:11
1
Polar coordinates CR equations assume $rne 0$. I am not sure if they are applicable here.
– Szeto
Aug 15 at 4:26
Look equations 18 and 19 mathworld.wolfram.com/Cauchy-RiemannEquations.html
– David
Aug 15 at 5:28
add a comment |Â
1
Just an idea: have you tried polar coordinates?
– Andrei
Aug 15 at 4:10
No. Not yet. My notes does not mention anything like that. Let me Google it.
– Mathaman Topologius
Aug 15 at 4:11
1
Polar coordinates CR equations assume $rne 0$. I am not sure if they are applicable here.
– Szeto
Aug 15 at 4:26
Look equations 18 and 19 mathworld.wolfram.com/Cauchy-RiemannEquations.html
– David
Aug 15 at 5:28
1
1
Just an idea: have you tried polar coordinates?
– Andrei
Aug 15 at 4:10
Just an idea: have you tried polar coordinates?
– Andrei
Aug 15 at 4:10
No. Not yet. My notes does not mention anything like that. Let me Google it.
– Mathaman Topologius
Aug 15 at 4:11
No. Not yet. My notes does not mention anything like that. Let me Google it.
– Mathaman Topologius
Aug 15 at 4:11
1
1
Polar coordinates CR equations assume $rne 0$. I am not sure if they are applicable here.
– Szeto
Aug 15 at 4:26
Polar coordinates CR equations assume $rne 0$. I am not sure if they are applicable here.
– Szeto
Aug 15 at 4:26
Look equations 18 and 19 mathworld.wolfram.com/Cauchy-RiemannEquations.html
– David
Aug 15 at 5:28
Look equations 18 and 19 mathworld.wolfram.com/Cauchy-RiemannEquations.html
– David
Aug 15 at 5:28
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
For $z = x in Bbb R$, $x ne 0$ is $f(z) = e^-1/x^4$ purely real,
so that
$$
beginaligned
u(x, 0) &= e^-1/x^4 \ v(x, 0) &= 0
endaligned
$$
and therefore
$$
beginaligned
u_x(0, 0) &= lim_x to 0 fracu(x, 0)-u(0, 0)x - 0
= lim_x to 0 frace^-1/x^4x
= lim_t to infty fracte^t^4 = 0 \
v_x(0, 0) &= 0 , .
endaligned
$$
In a similar way you can show that
$$
u_y(0, 0) = v_y(0, 0) = 0 , .
$$
All partial derivatives are zero at $z=0$, so that the Cauchy-Riemann
equations are satisfied at that point.
answered Aug 15 at 5:43
Martin R
23.9k32743
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For $z = x in Bbb R$, $x ne 0$ is $f(z) = e^-1/x^4$ purely real,
so that
$$
beginaligned
u(x, 0) &= e^-1/x^4 \ v(x, 0) &= 0
endaligned
$$
and therefore
$$
beginaligned
u_x(0, 0) &= lim_x to 0 fracu(x, 0)-u(0, 0)x - 0
= lim_x to 0 frace^-1/x^4x
= lim_t to infty fracte^t^4 = 0 \
v_x(0, 0) &= 0 , .
endaligned
$$
In a similar way you can show that
$$
u_y(0, 0) = v_y(0, 0) = 0 , .
$$
All partial derivatives are zero at $z=0$, so that the Cauchy-Riemann
equations are satisfied at that point.
answered Aug 15 at 5:43
Martin R
23.9k32743
add a comment |Â
up vote
1
down vote
For $z = x in Bbb R$, $x ne 0$ is $f(z) = e^-1/x^4$ purely real,
so that
$$
beginaligned
u(x, 0) &= e^-1/x^4 \ v(x, 0) &= 0
endaligned
$$
and therefore
$$
beginaligned
u_x(0, 0) &= lim_x to 0 fracu(x, 0)-u(0, 0)x - 0
= lim_x to 0 frace^-1/x^4x
= lim_t to infty fracte^t^4 = 0 \
v_x(0, 0) &= 0 , .
endaligned
$$
In a similar way you can show that
$$
u_y(0, 0) = v_y(0, 0) = 0 , .
$$
All partial derivatives are zero at $z=0$, so that the Cauchy-Riemann
equations are satisfied at that point.
answered Aug 15 at 5:43
Martin R
23.9k32743
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $z = x in Bbb R$, $x ne 0$ is $f(z) = e^-1/x^4$ purely real,
so that
$$
beginaligned
u(x, 0) &= e^-1/x^4 \ v(x, 0) &= 0
endaligned
$$
and therefore
$$
beginaligned
u_x(0, 0) &= lim_x to 0 fracu(x, 0)-u(0, 0)x - 0
= lim_x to 0 frace^-1/x^4x
= lim_t to infty fracte^t^4 = 0 \
v_x(0, 0) &= 0 , .
endaligned
$$
In a similar way you can show that
$$
u_y(0, 0) = v_y(0, 0) = 0 , .
$$
All partial derivatives are zero at $z=0$, so that the Cauchy-Riemann
equations are satisfied at that point.
answered Aug 15 at 5:43
Martin R
23.9k32743
For $z = x in Bbb R$, $x ne 0$ is $f(z) = e^-1/x^4$ purely real,
so that
$$
beginaligned
u(x, 0) &= e^-1/x^4 \ v(x, 0) &= 0
endaligned
$$
and therefore
$$
beginaligned
u_x(0, 0) &= lim_x to 0 fracu(x, 0)-u(0, 0)x - 0
= lim_x to 0 frace^-1/x^4x
= lim_t to infty fracte^t^4 = 0 \
v_x(0, 0) &= 0 , .
endaligned
$$
In a similar way you can show that
$$
u_y(0, 0) = v_y(0, 0) = 0 , .
$$
All partial derivatives are zero at $z=0$, so that the Cauchy-Riemann
equations are satisfied at that point.
answered Aug 15 at 5:43
Martin R
23.9k32743
answered Aug 15 at 5:43
Martin R
23.9k32743
answered Aug 15 at 5:43
Martin R
23.9k32743
answered Aug 15 at 5:43
Martin R
23.9k32743
23.9k32743
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%2f2883176%2fis-there-an-easy-way-to-show-that-this-function-satisfies-cauchy-riemann-equatio%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
Just an idea: have you tried polar coordinates?
– Andrei
Aug 15 at 4:10
No. Not yet. My notes does not mention anything like that. Let me Google it.
– Mathaman Topologius
Aug 15 at 4:11
1
Polar coordinates CR equations assume $rne 0$. I am not sure if they are applicable here.
– Szeto
Aug 15 at 4:26
Look equations 18 and 19 mathworld.wolfram.com/Cauchy-RiemannEquations.html
– David
Aug 15 at 5:28