Recover complex function from its imaginary part
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The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
$$
left{
begin{array}{c}
G_x = 2x+1 \
G_y = -2y
end{array}
right.
$$
$$
left{
begin{array}{c}
G_{xx} = 2 \
G_{yy} = -2
end{array}
right.
$$
$G_{xx} + G_{yy} = 0$, therefore the function is harmonic.
Next step is
$$
left{
begin{array}{c}
u_x = G_y = -2y \
u_y = -G_x = -2x-1
end{array}
right.
$$
Then,
$$u=int u_x dx=-2xy+h(y)+C$$
$$u_y = -2x-1$$
Therefore,
$$u = -2xy - 2x - 1$$
Finally,
$$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
But this is a wrong answer. Where is my mistake?
complex-analysis complex-numbers
asked Nov 22 at 6:11
user3132457
876
add a comment |
up vote
0
down vote
favorite
The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
$$
left{
begin{array}{c}
G_x = 2x+1 \
G_y = -2y
end{array}
right.
$$
$$
left{
begin{array}{c}
G_{xx} = 2 \
G_{yy} = -2
end{array}
right.
$$
$G_{xx} + G_{yy} = 0$, therefore the function is harmonic.
Next step is
$$
left{
begin{array}{c}
u_x = G_y = -2y \
u_y = -G_x = -2x-1
end{array}
right.
$$
Then,
$$u=int u_x dx=-2xy+h(y)+C$$
$$u_y = -2x-1$$
Therefore,
$$u = -2xy - 2x - 1$$
Finally,
$$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
But this is a wrong answer. Where is my mistake?
complex-analysis complex-numbers
asked Nov 22 at 6:11
user3132457
876
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
$$
left{
begin{array}{c}
G_x = 2x+1 \
G_y = -2y
end{array}
right.
$$
$$
left{
begin{array}{c}
G_{xx} = 2 \
G_{yy} = -2
end{array}
right.
$$
$G_{xx} + G_{yy} = 0$, therefore the function is harmonic.
Next step is
$$
left{
begin{array}{c}
u_x = G_y = -2y \
u_y = -G_x = -2x-1
end{array}
right.
$$
Then,
$$u=int u_x dx=-2xy+h(y)+C$$
$$u_y = -2x-1$$
Therefore,
$$u = -2xy - 2x - 1$$
Finally,
$$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
But this is a wrong answer. Where is my mistake?
complex-analysis complex-numbers
asked Nov 22 at 6:11
user3132457
876
The real part of $z=x+yi$ is given as $Im f(z) = x^2-y_2+x$. Therefore I have
$$
left{
begin{array}{c}
G_x = 2x+1 \
G_y = -2y
end{array}
right.
$$
$$
left{
begin{array}{c}
G_{xx} = 2 \
G_{yy} = -2
end{array}
right.
$$
$G_{xx} + G_{yy} = 0$, therefore the function is harmonic.
Next step is
$$
left{
begin{array}{c}
u_x = G_y = -2y \
u_y = -G_x = -2x-1
end{array}
right.
$$
Then,
$$u=int u_x dx=-2xy+h(y)+C$$
$$u_y = -2x-1$$
Therefore,
$$u = -2xy - 2x - 1$$
Finally,
$$f(z) = u+iv = -2xy-2x-1 + i(x^2-y^2+x) +C= -2xy - 2x-1 + x^2i+y^2i+xi +C= i(x^2-y^2+2xyi) + (-2x-1+xi) + C = iz^2 + (-2x-1+xi) + C$$
But this is a wrong answer. Where is my mistake?
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 22 at 6:11
user3132457
876
asked Nov 22 at 6:11
user3132457
876
asked Nov 22 at 6:11
user3132457
876
asked Nov 22 at 6:11
user3132457
876
asked Nov 22 at 6:11
user3132457
876
876
add a comment |
add a comment |
1 Answer
1
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oldest
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0
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$u = -2xy - 2x - 1$ is wrong !
From $u=-2xy+h(y)+C$ we get
$u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.
answered Nov 22 at 6:21
Fred
43.6k1644
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$u = -2xy - 2x - 1$ is wrong !
From $u=-2xy+h(y)+C$ we get
$u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.
answered Nov 22 at 6:21
Fred
43.6k1644
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
add a comment |
up vote
0
down vote
accepted
$u = -2xy - 2x - 1$ is wrong !
From $u=-2xy+h(y)+C$ we get
$u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.
answered Nov 22 at 6:21
Fred
43.6k1644
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$u = -2xy - 2x - 1$ is wrong !
From $u=-2xy+h(y)+C$ we get
$u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.
answered Nov 22 at 6:21
Fred
43.6k1644
$u = -2xy - 2x - 1$ is wrong !
From $u=-2xy+h(y)+C$ we get
$u_y=-2x+h'(y)=-G_x=-2x-1$, hence $u=-2xy-y+c$.
answered Nov 22 at 6:21
Fred
43.6k1644
answered Nov 22 at 6:21
Fred
43.6k1644
answered Nov 22 at 6:21
Fred
43.6k1644
answered Nov 22 at 6:21
Fred
43.6k1644
43.6k1644
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
add a comment |
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
Thanks, I just don't get why you put $h'(y)$ instead of -1
– user3132457
Nov 22 at 12:19
add a comment |
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