Recover complex function from the real part
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0
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I have $Re f(z) = e^x(xcosy -ysiny) = e^xcosy -e^xysiny$. Therefore,
$$v_x= e^xxcosy+e^xcosy-e^xysiny = e^x(xcosy+cosy-ysiny)$$
$$v_y= -e^xxsiny-e^x(siny-ycosy) = e^x(-xsiny-siny+ycosy)$$
$u_x = v_y = e^x(-xsiny-siny+ycosy)$
$u_y = -v_x = -e^x(xcosy+cosy-ysiny)$
$$u = int u_xdx = -int e^x(xsiny+siny-ycosy)dx = ... = -e^x(xsiny-ycosy)$$
To find $u_y$, I do:
$$u_y = -e^x(xcosy+cosy-ysiny) = -e^x(xsiny-ycosy)$$
What do I do from here?
complex-analysis complex-numbers
asked Nov 22 at 17:53
user3132457
926
add a comment |
up vote
0
down vote
favorite
I have $Re f(z) = e^x(xcosy -ysiny) = e^xcosy -e^xysiny$. Therefore,
$$v_x= e^xxcosy+e^xcosy-e^xysiny = e^x(xcosy+cosy-ysiny)$$
$$v_y= -e^xxsiny-e^x(siny-ycosy) = e^x(-xsiny-siny+ycosy)$$
$u_x = v_y = e^x(-xsiny-siny+ycosy)$
$u_y = -v_x = -e^x(xcosy+cosy-ysiny)$
$$u = int u_xdx = -int e^x(xsiny+siny-ycosy)dx = ... = -e^x(xsiny-ycosy)$$
To find $u_y$, I do:
$$u_y = -e^x(xcosy+cosy-ysiny) = -e^x(xsiny-ycosy)$$
What do I do from here?
complex-analysis complex-numbers
asked Nov 22 at 17:53
user3132457
926
Integrate what? Which side?
– user3132457
Nov 22 at 18:18
I hadn't well seen : you have found an expression for $u$ that deserves to have a "$+k_1(y)$" (an integration constant that is a function of the other variable). Here is the problem... Now if you do the same thing for the integration of $u_y$, you will have a constant $+k_2(x)$. And now you have to match the two expressions in order to get a single result...
– Jean Marie
Nov 22 at 18:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have $Re f(z) = e^x(xcosy -ysiny) = e^xcosy -e^xysiny$. Therefore,
$$v_x= e^xxcosy+e^xcosy-e^xysiny = e^x(xcosy+cosy-ysiny)$$
$$v_y= -e^xxsiny-e^x(siny-ycosy) = e^x(-xsiny-siny+ycosy)$$
$u_x = v_y = e^x(-xsiny-siny+ycosy)$
$u_y = -v_x = -e^x(xcosy+cosy-ysiny)$
$$u = int u_xdx = -int e^x(xsiny+siny-ycosy)dx = ... = -e^x(xsiny-ycosy)$$
To find $u_y$, I do:
$$u_y = -e^x(xcosy+cosy-ysiny) = -e^x(xsiny-ycosy)$$
What do I do from here?
complex-analysis complex-numbers
asked Nov 22 at 17:53
user3132457
926
I have $Re f(z) = e^x(xcosy -ysiny) = e^xcosy -e^xysiny$. Therefore,
$$v_x= e^xxcosy+e^xcosy-e^xysiny = e^x(xcosy+cosy-ysiny)$$
$$v_y= -e^xxsiny-e^x(siny-ycosy) = e^x(-xsiny-siny+ycosy)$$
$u_x = v_y = e^x(-xsiny-siny+ycosy)$
$u_y = -v_x = -e^x(xcosy+cosy-ysiny)$
$$u = int u_xdx = -int e^x(xsiny+siny-ycosy)dx = ... = -e^x(xsiny-ycosy)$$
To find $u_y$, I do:
$$u_y = -e^x(xcosy+cosy-ysiny) = -e^x(xsiny-ycosy)$$
What do I do from here?
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 22 at 17:53
user3132457
926
asked Nov 22 at 17:53
user3132457
926
edited Nov 22 at 18:36
asked Nov 22 at 17:53
user3132457
926
asked Nov 22 at 17:53
user3132457
926
asked Nov 22 at 17:53
user3132457
926
926
Integrate what? Which side?
– user3132457
Nov 22 at 18:18
I hadn't well seen : you have found an expression for $u$ that deserves to have a "$+k_1(y)$" (an integration constant that is a function of the other variable). Here is the problem... Now if you do the same thing for the integration of $u_y$, you will have a constant $+k_2(x)$. And now you have to match the two expressions in order to get a single result...
– Jean Marie
Nov 22 at 18:27
add a comment |
Integrate what? Which side?
– user3132457
Nov 22 at 18:18
I hadn't well seen : you have found an expression for $u$ that deserves to have a "$+k_1(y)$" (an integration constant that is a function of the other variable). Here is the problem... Now if you do the same thing for the integration of $u_y$, you will have a constant $+k_2(x)$. And now you have to match the two expressions in order to get a single result...
– Jean Marie
Nov 22 at 18:27
Integrate what? Which side?
– user3132457
Nov 22 at 18:18
Integrate what? Which side?
– user3132457
Nov 22 at 18:18
I hadn't well seen : you have found an expression for $u$ that deserves to have a "$+k_1(y)$" (an integration constant that is a function of the other variable). Here is the problem... Now if you do the same thing for the integration of $u_y$, you will have a constant $+k_2(x)$. And now you have to match the two expressions in order to get a single result...
– Jean Marie
Nov 22 at 18:27
I hadn't well seen : you have found an expression for $u$ that deserves to have a "$+k_1(y)$" (an integration constant that is a function of the other variable). Here is the problem... Now if you do the same thing for the integration of $u_y$, you will have a constant $+k_2(x)$. And now you have to match the two expressions in order to get a single result...
– Jean Marie
Nov 22 at 18:27
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
First of all, we observe that
$$
mathrm{e}^x(xcos y-ysin y)=mathrm{Re},big(mathrm{e}^x(cos y+isin y)cdot (x+iy)big)=mathrm{Re},(mathrm{e}^zz).
$$
Next, use the fact that, if $g$ is analytic in some region and $,mathrm{Re}(g)=0$, then $gequiv ic$, where $c$ real constant.
Hence the function you are looking for is $mathrm{e}^zz+ic$.
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
add a comment |
up vote
1
down vote
You know that$$v_x=-u_y=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$$and$$v_y=u_x=e^xbigl(cos(y)+xcos(y)-ysin(y)bigr).$$Integrating, you get that$$v(x,y)=e^xbigl(ycos(y)+xsin(y)bigr)+k$$Thereforebegin{align}f(x+yi)&=u(x,y)+v(x,y)i\&=e^xbigl(xcos(y)-ysin(y)bigr)+e^xbigl(ycos(y)+xsin(y)bigr)i+ki\&=e^xbigl((cos(y)+sin(y)ibigr)(x+yi)+ki\&=e^{x+yi}(x+yi)+kiend{align}and so $f(z)=ze^z+ki$.
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
|
show 2 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First of all, we observe that
$$
mathrm{e}^x(xcos y-ysin y)=mathrm{Re},big(mathrm{e}^x(cos y+isin y)cdot (x+iy)big)=mathrm{Re},(mathrm{e}^zz).
$$
Next, use the fact that, if $g$ is analytic in some region and $,mathrm{Re}(g)=0$, then $gequiv ic$, where $c$ real constant.
Hence the function you are looking for is $mathrm{e}^zz+ic$.
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
add a comment |
up vote
3
down vote
accepted
First of all, we observe that
$$
mathrm{e}^x(xcos y-ysin y)=mathrm{Re},big(mathrm{e}^x(cos y+isin y)cdot (x+iy)big)=mathrm{Re},(mathrm{e}^zz).
$$
Next, use the fact that, if $g$ is analytic in some region and $,mathrm{Re}(g)=0$, then $gequiv ic$, where $c$ real constant.
Hence the function you are looking for is $mathrm{e}^zz+ic$.
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First of all, we observe that
$$
mathrm{e}^x(xcos y-ysin y)=mathrm{Re},big(mathrm{e}^x(cos y+isin y)cdot (x+iy)big)=mathrm{Re},(mathrm{e}^zz).
$$
Next, use the fact that, if $g$ is analytic in some region and $,mathrm{Re}(g)=0$, then $gequiv ic$, where $c$ real constant.
Hence the function you are looking for is $mathrm{e}^zz+ic$.
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
First of all, we observe that
$$
mathrm{e}^x(xcos y-ysin y)=mathrm{Re},big(mathrm{e}^x(cos y+isin y)cdot (x+iy)big)=mathrm{Re},(mathrm{e}^zz).
$$
Next, use the fact that, if $g$ is analytic in some region and $,mathrm{Re}(g)=0$, then $gequiv ic$, where $c$ real constant.
Hence the function you are looking for is $mathrm{e}^zz+ic$.
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
answered Nov 22 at 18:48
Yiorgos S. Smyrlis
62.3k1383162
62.3k1383162
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
add a comment |
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
I don't get why $Re(g)=0$.
– user3132457
Nov 25 at 18:39
add a comment |
up vote
1
down vote
You know that$$v_x=-u_y=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$$and$$v_y=u_x=e^xbigl(cos(y)+xcos(y)-ysin(y)bigr).$$Integrating, you get that$$v(x,y)=e^xbigl(ycos(y)+xsin(y)bigr)+k$$Thereforebegin{align}f(x+yi)&=u(x,y)+v(x,y)i\&=e^xbigl(xcos(y)-ysin(y)bigr)+e^xbigl(ycos(y)+xsin(y)bigr)i+ki\&=e^xbigl((cos(y)+sin(y)ibigr)(x+yi)+ki\&=e^{x+yi}(x+yi)+kiend{align}and so $f(z)=ze^z+ki$.
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
|
show 2 more comments
up vote
1
down vote
You know that$$v_x=-u_y=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$$and$$v_y=u_x=e^xbigl(cos(y)+xcos(y)-ysin(y)bigr).$$Integrating, you get that$$v(x,y)=e^xbigl(ycos(y)+xsin(y)bigr)+k$$Thereforebegin{align}f(x+yi)&=u(x,y)+v(x,y)i\&=e^xbigl(xcos(y)-ysin(y)bigr)+e^xbigl(ycos(y)+xsin(y)bigr)i+ki\&=e^xbigl((cos(y)+sin(y)ibigr)(x+yi)+ki\&=e^{x+yi}(x+yi)+kiend{align}and so $f(z)=ze^z+ki$.
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
|
show 2 more comments
up vote
1
down vote
up vote
1
down vote
You know that$$v_x=-u_y=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$$and$$v_y=u_x=e^xbigl(cos(y)+xcos(y)-ysin(y)bigr).$$Integrating, you get that$$v(x,y)=e^xbigl(ycos(y)+xsin(y)bigr)+k$$Thereforebegin{align}f(x+yi)&=u(x,y)+v(x,y)i\&=e^xbigl(xcos(y)-ysin(y)bigr)+e^xbigl(ycos(y)+xsin(y)bigr)i+ki\&=e^xbigl((cos(y)+sin(y)ibigr)(x+yi)+ki\&=e^{x+yi}(x+yi)+kiend{align}and so $f(z)=ze^z+ki$.
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
You know that$$v_x=-u_y=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$$and$$v_y=u_x=e^xbigl(cos(y)+xcos(y)-ysin(y)bigr).$$Integrating, you get that$$v(x,y)=e^xbigl(ycos(y)+xsin(y)bigr)+k$$Thereforebegin{align}f(x+yi)&=u(x,y)+v(x,y)i\&=e^xbigl(xcos(y)-ysin(y)bigr)+e^xbigl(ycos(y)+xsin(y)bigr)i+ki\&=e^xbigl((cos(y)+sin(y)ibigr)(x+yi)+ki\&=e^{x+yi}(x+yi)+kiend{align}and so $f(z)=ze^z+ki$.
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
edited Nov 22 at 18:51
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
answered Nov 22 at 18:29
José Carlos Santos
147k22117217
147k22117217
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
|
show 2 more comments
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
sorry, the title is wrong, should be real part
– user3132457
Nov 22 at 18:37
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
I've edited my answer.
– José Carlos Santos
Nov 22 at 18:51
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
How do you integrate in step 3? I get $$v(x,y)=-e^xbigl(ycos(y)+xsin(y)bigr)+k$$
– user3132457
Nov 25 at 18:37
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
@user3132457 That function is such that $v_x=-e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$. But what you should be aiming at is $v_x=e^xbigl(ycos(y)+sin(y)+xsin(y)bigr)$.
– José Carlos Santos
Nov 25 at 18:46
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
Hmm, I'm confused. In my original post, $v_x$ isn't negative but in your calculations it is.
– user3132457
Nov 25 at 18:51
|
show 2 more comments
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Integrate what? Which side?
– user3132457
Nov 22 at 18:18
I hadn't well seen : you have found an expression for $u$ that deserves to have a "$+k_1(y)$" (an integration constant that is a function of the other variable). Here is the problem... Now if you do the same thing for the integration of $u_y$, you will have a constant $+k_2(x)$. And now you have to match the two expressions in order to get a single result...
– Jean Marie
Nov 22 at 18:27