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Find Entire Function $f$ such that $f(0)=1$ and $Re(f(x+iy))=x^2-y^2+e^-xcos y$
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Find an entire function $f(z)$ such that $f(0)=1$ and for all $z$, $Re(f(z))=x^2-y^2+e^-xcos y$
An entire function is analytic on all $mathbbC$ and in particular differentiable and therefore fulfils C-R equations
therefore
$u_x=v_y=2x-e^-xcos y$
$v=2xy-e^-xsin y+ c(x)$
$-v_x=-2y-e^-xsin y+c'(x)=-2y-e^-xsin y=u_y$
so $-2y-e^-xsin y+c'(x)=-2y-e^-xsin yRightarrow c'(x)=0 Rightarrow c(x)=k $
So we have $v=2xy-e^-xsin y+k$
and $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+k)$, but $f(0)=0$ so
$1+ik=0iff ik=-1 iff k=i$
So $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+1)$
Is it correct?
complex-analysis
edited Jul 15 at 12:19
Did
242k23208443
asked Jul 15 at 11:02
newhere
759310
add a comment |Â
up vote
1
down vote
favorite
Find an entire function $f(z)$ such that $f(0)=1$ and for all $z$, $Re(f(z))=x^2-y^2+e^-xcos y$
An entire function is analytic on all $mathbbC$ and in particular differentiable and therefore fulfils C-R equations
therefore
$u_x=v_y=2x-e^-xcos y$
$v=2xy-e^-xsin y+ c(x)$
$-v_x=-2y-e^-xsin y+c'(x)=-2y-e^-xsin y=u_y$
so $-2y-e^-xsin y+c'(x)=-2y-e^-xsin yRightarrow c'(x)=0 Rightarrow c(x)=k $
So we have $v=2xy-e^-xsin y+k$
and $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+k)$, but $f(0)=0$ so
$1+ik=0iff ik=-1 iff k=i$
So $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+1)$
Is it correct?
complex-analysis
edited Jul 15 at 12:19
Did
242k23208443
asked Jul 15 at 11:02
newhere
759310
$Re(f(z))=x^2-y^2+e^-xcos y$ implies $Re(f(0)) = 1$, which contradicts $f(0) = 0$.
– Martin R
Jul 15 at 11:49
maybe it is a mistake in the question, I will change it to $1$
– newhere
Jul 15 at 11:55
@newhere Beside the constant for the value of $f(0)$ your calculations are correct! The only thing is that you have to do another small step to write the last function in term of $z$
– Davide Morgante
Jul 15 at 12:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find an entire function $f(z)$ such that $f(0)=1$ and for all $z$, $Re(f(z))=x^2-y^2+e^-xcos y$
An entire function is analytic on all $mathbbC$ and in particular differentiable and therefore fulfils C-R equations
therefore
$u_x=v_y=2x-e^-xcos y$
$v=2xy-e^-xsin y+ c(x)$
$-v_x=-2y-e^-xsin y+c'(x)=-2y-e^-xsin y=u_y$
so $-2y-e^-xsin y+c'(x)=-2y-e^-xsin yRightarrow c'(x)=0 Rightarrow c(x)=k $
So we have $v=2xy-e^-xsin y+k$
and $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+k)$, but $f(0)=0$ so
$1+ik=0iff ik=-1 iff k=i$
So $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+1)$
Is it correct?
complex-analysis
edited Jul 15 at 12:19
Did
242k23208443
asked Jul 15 at 11:02
newhere
759310
Find an entire function $f(z)$ such that $f(0)=1$ and for all $z$, $Re(f(z))=x^2-y^2+e^-xcos y$
An entire function is analytic on all $mathbbC$ and in particular differentiable and therefore fulfils C-R equations
therefore
$u_x=v_y=2x-e^-xcos y$
$v=2xy-e^-xsin y+ c(x)$
$-v_x=-2y-e^-xsin y+c'(x)=-2y-e^-xsin y=u_y$
so $-2y-e^-xsin y+c'(x)=-2y-e^-xsin yRightarrow c'(x)=0 Rightarrow c(x)=k $
So we have $v=2xy-e^-xsin y+k$
and $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+k)$, but $f(0)=0$ so
$1+ik=0iff ik=-1 iff k=i$
So $f(z)=(x^2-y^2+e^-xcos y)+i(2xy-e^-xsin y+1)$
Is it correct?
complex-analysis
edited Jul 15 at 12:19
Did
242k23208443
asked Jul 15 at 11:02
newhere
759310
edited Jul 15 at 12:19
Did
242k23208443
edited Jul 15 at 12:19
Did
242k23208443
edited Jul 15 at 12:19
Did
242k23208443
242k23208443
asked Jul 15 at 11:02
newhere
759310
asked Jul 15 at 11:02
newhere
759310
asked Jul 15 at 11:02
newhere
759310
759310
$Re(f(z))=x^2-y^2+e^-xcos y$ implies $Re(f(0)) = 1$, which contradicts $f(0) = 0$.
– Martin R
Jul 15 at 11:49
maybe it is a mistake in the question, I will change it to $1$
– newhere
Jul 15 at 11:55
@newhere Beside the constant for the value of $f(0)$ your calculations are correct! The only thing is that you have to do another small step to write the last function in term of $z$
– Davide Morgante
Jul 15 at 12:58
add a comment |Â
$Re(f(z))=x^2-y^2+e^-xcos y$ implies $Re(f(0)) = 1$, which contradicts $f(0) = 0$.
– Martin R
Jul 15 at 11:49
maybe it is a mistake in the question, I will change it to $1$
– newhere
Jul 15 at 11:55
@newhere Beside the constant for the value of $f(0)$ your calculations are correct! The only thing is that you have to do another small step to write the last function in term of $z$
– Davide Morgante
Jul 15 at 12:58
$Re(f(z))=x^2-y^2+e^-xcos y$ implies $Re(f(0)) = 1$, which contradicts $f(0) = 0$.
– Martin R
Jul 15 at 11:49
$Re(f(z))=x^2-y^2+e^-xcos y$ implies $Re(f(0)) = 1$, which contradicts $f(0) = 0$.
– Martin R
Jul 15 at 11:49
maybe it is a mistake in the question, I will change it to $1$
– newhere
Jul 15 at 11:55
maybe it is a mistake in the question, I will change it to $1$
– newhere
Jul 15 at 11:55
@newhere Beside the constant for the value of $f(0)$ your calculations are correct! The only thing is that you have to do another small step to write the last function in term of $z$
– Davide Morgante
Jul 15 at 12:58
@newhere Beside the constant for the value of $f(0)$ your calculations are correct! The only thing is that you have to do another small step to write the last function in term of $z$
– Davide Morgante
Jul 15 at 12:58
add a comment |Â
2 Answers
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up vote
2
down vote
accepted
If the real part is $u(x,y)$ we have that:
$$f(z)=2uleft(fracz2, -fraciz2right)-u(0,0)$$
$$f(z)=2left(fracz^24-frac(iz)^24+expleft(-fracz2right)cosleft(-fraciz2right)right)-1$$
$$f(z)=z^2+2expleft(-fracz2right)coshleft(fracz2right)-1$$
$$f(z)=z^2+exp(-z)$$
But with this function, $f(0)=1$. Are you sure about $f(0)=0$?
Thanks to Davide Morgante, you can find out more about this way in the second paragraph of this paper or in this question.
answered Jul 15 at 11:11
Botond
3,8882632
1
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
1
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
1
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
1
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
1
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
 |Â
show 30 more comments
up vote
2
down vote
We know that $Re(f(z)) = u(x,y)$, to be the real part of an analytic function this has to be harmonic, so: $$nabla^2 u(x,y) = 0$$ which is easy to prove.
To find a complex part such that $f(z)$ will be analytic we integrate the differential form: $$dv = fracpartial vpartial xdx + fracpartial vpartial ydy = -fracpartial upartial ydx + fracpartial upartial xdy$$ for the Cauchy-Riemann conditions. To integrate the differential form we choose the spline connecting the points $(0,0)rightarrow (x,0)rightarrow (x,y)$, then $$v(x,y)-v(0,0) = int_gamma-fracpartial upartial ydx + fracpartial upartial xdy\
= -undersety'=0int_0^xfracpartial u(x',y')partial y'dx'+undersetx'=xint_0^yfracpartial u(x',y')partial x'dy'\
= undersety'=0int_0^x(2y'+e^-x'sin(y'))dx'+undersetx'=xint_0^y(2x'+e^-x'cos(y'))dy' \
= left.(2x'y'-e^-x'sin(y'))right|_0,;y'=0^x + left.(2x'y'-e^-x'sin(y'))right|_0, ;x'=x^y \
= 0 + 2xy-e^-xsin(y) = 2xy-e^-xsin(y)$$
If this is a good answer, we could see if $$nabla^2v(x,y)=0$$ which is true.
For to be $f(0) = f(0+i0)=1$ we have $v(0,0)=0$ because $u(0,0) = 1$. So in the end $$f(z) = (x^2-y^2+e^-xcos(y))+i(2xy-e^-xsin(y))$$ which can be rewritten knowing that $$(x+iy)^2 = x^2-y^2+2ixy= z^2;;;e^-x(cos(y)-isin(y)) = e^-xe^-iy = e^-z$$
$$f(z) = z^2+e^-z$$
Edit
I've edited the answer to take count of the correction $f(0)=1$ but if the initial condition was $f(0)=0$ I think that you could cheat using $v(0,0)=i$, this would give you the right answer, but an integral in $mathbbR$ cannot give you a complex result
answered Jul 15 at 11:44
Davide Morgante
1,904220
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If the real part is $u(x,y)$ we have that:
$$f(z)=2uleft(fracz2, -fraciz2right)-u(0,0)$$
$$f(z)=2left(fracz^24-frac(iz)^24+expleft(-fracz2right)cosleft(-fraciz2right)right)-1$$
$$f(z)=z^2+2expleft(-fracz2right)coshleft(fracz2right)-1$$
$$f(z)=z^2+exp(-z)$$
But with this function, $f(0)=1$. Are you sure about $f(0)=0$?
Thanks to Davide Morgante, you can find out more about this way in the second paragraph of this paper or in this question.
answered Jul 15 at 11:11
Botond
3,8882632
1
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
1
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
1
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
1
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
1
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
 |Â
show 30 more comments
up vote
2
down vote
accepted
If the real part is $u(x,y)$ we have that:
$$f(z)=2uleft(fracz2, -fraciz2right)-u(0,0)$$
$$f(z)=2left(fracz^24-frac(iz)^24+expleft(-fracz2right)cosleft(-fraciz2right)right)-1$$
$$f(z)=z^2+2expleft(-fracz2right)coshleft(fracz2right)-1$$
$$f(z)=z^2+exp(-z)$$
But with this function, $f(0)=1$. Are you sure about $f(0)=0$?
Thanks to Davide Morgante, you can find out more about this way in the second paragraph of this paper or in this question.
answered Jul 15 at 11:11
Botond
3,8882632
1
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
1
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
1
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
1
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
1
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
 |Â
show 30 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If the real part is $u(x,y)$ we have that:
$$f(z)=2uleft(fracz2, -fraciz2right)-u(0,0)$$
$$f(z)=2left(fracz^24-frac(iz)^24+expleft(-fracz2right)cosleft(-fraciz2right)right)-1$$
$$f(z)=z^2+2expleft(-fracz2right)coshleft(fracz2right)-1$$
$$f(z)=z^2+exp(-z)$$
But with this function, $f(0)=1$. Are you sure about $f(0)=0$?
Thanks to Davide Morgante, you can find out more about this way in the second paragraph of this paper or in this question.
answered Jul 15 at 11:11
Botond
3,8882632
If the real part is $u(x,y)$ we have that:
$$f(z)=2uleft(fracz2, -fraciz2right)-u(0,0)$$
$$f(z)=2left(fracz^24-frac(iz)^24+expleft(-fracz2right)cosleft(-fraciz2right)right)-1$$
$$f(z)=z^2+2expleft(-fracz2right)coshleft(fracz2right)-1$$
$$f(z)=z^2+exp(-z)$$
But with this function, $f(0)=1$. Are you sure about $f(0)=0$?
Thanks to Davide Morgante, you can find out more about this way in the second paragraph of this paper or in this question.
answered Jul 15 at 11:11
Botond
3,8882632
edited Jul 18 at 12:32
answered Jul 15 at 11:11
Botond
3,8882632
answered Jul 15 at 11:11
Botond
3,8882632
answered Jul 15 at 11:11
Botond
3,8882632
3,8882632
1
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
1
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
1
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
1
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
1
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
 |Â
show 30 more comments
1
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
1
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
1
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
1
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
1
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
1
1
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
@Did I think I understand: $$2u(x,y) = f(x+iy)+overlinef(x+iy) = f(x+iy)+overlinef(overlinex+iy) = f(x+iy)+overlinef(x-iy)$$ choose $$z=x+iy;;w=x-iy$$. Solving for $x$ and $y$ we get $x=1over 2(z+w);;y=1over2i(z-w)$ and so $$2u(1over 2(z+w), 1over2i(z-w)) = f(z)+overlinef(w)= f(z)+overlinef(overlinew);;forall z,w$$ Let's choose now $w=z_0 = 0+i0$ for our problem, then we get $$f(z)=2u(1over 2(z+z_0), 1over2i(z-z_0))-overlinef(z_0) Rightarrow f(z)=2u(1over 2z, 1over2iz)-u(0,0)$$
– Davide Morgante
Jul 16 at 22:03
1
1
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
I stand to be corrected! Even I, like @Botond, have had only a course on complex analysis in my mathematical methods for physics class, so I'm clearly not an expert at it. I have to say that never have I ever seen this method for determining analytic functions from one of their parts as the approach that makes more sense to me is to use the Cauchy-Riemann conditions, but still, I found it interesting. Probably the method proposed by Botond has something to do with the analytic continuation of $u(x,y)$
– Davide Morgante
Jul 16 at 22:33
1
1
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
@Did I don't find my answer inadequate. I provided a great tool, and it's working well. So I won't delete this answer, but please feel free to downvote and/or flag it if you think that it should not be here.
– Botond
Jul 17 at 7:22
1
1
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
The given function $u(x,y)$ is analytical on $Bbb R^2$ and can thus be continued to a function on $Bbb C^2$ having the same power series coefficients. If one wants to stay simple but less universal one can compute $Re(f(x))=u(x,0)=x^2+e^-x$ and then compute $Re(f(z)-z^2-e^-z)=0$. In general one will lose terms with this approach, that is the last formula will still have terms on the right side that are zero for $y=0$.
– LutzL
Jul 17 at 7:49
1
1
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
I've started an MSE forum to have a definite clarification math.stackexchange.com/questions/2854399/… and a very helpful user answered pretty well! Hope you find it useful
– Davide Morgante
Jul 17 at 15:20
 |Â
show 30 more comments
up vote
2
down vote
We know that $Re(f(z)) = u(x,y)$, to be the real part of an analytic function this has to be harmonic, so: $$nabla^2 u(x,y) = 0$$ which is easy to prove.
To find a complex part such that $f(z)$ will be analytic we integrate the differential form: $$dv = fracpartial vpartial xdx + fracpartial vpartial ydy = -fracpartial upartial ydx + fracpartial upartial xdy$$ for the Cauchy-Riemann conditions. To integrate the differential form we choose the spline connecting the points $(0,0)rightarrow (x,0)rightarrow (x,y)$, then $$v(x,y)-v(0,0) = int_gamma-fracpartial upartial ydx + fracpartial upartial xdy\
= -undersety'=0int_0^xfracpartial u(x',y')partial y'dx'+undersetx'=xint_0^yfracpartial u(x',y')partial x'dy'\
= undersety'=0int_0^x(2y'+e^-x'sin(y'))dx'+undersetx'=xint_0^y(2x'+e^-x'cos(y'))dy' \
= left.(2x'y'-e^-x'sin(y'))right|_0,;y'=0^x + left.(2x'y'-e^-x'sin(y'))right|_0, ;x'=x^y \
= 0 + 2xy-e^-xsin(y) = 2xy-e^-xsin(y)$$
If this is a good answer, we could see if $$nabla^2v(x,y)=0$$ which is true.
For to be $f(0) = f(0+i0)=1$ we have $v(0,0)=0$ because $u(0,0) = 1$. So in the end $$f(z) = (x^2-y^2+e^-xcos(y))+i(2xy-e^-xsin(y))$$ which can be rewritten knowing that $$(x+iy)^2 = x^2-y^2+2ixy= z^2;;;e^-x(cos(y)-isin(y)) = e^-xe^-iy = e^-z$$
$$f(z) = z^2+e^-z$$
Edit
I've edited the answer to take count of the correction $f(0)=1$ but if the initial condition was $f(0)=0$ I think that you could cheat using $v(0,0)=i$, this would give you the right answer, but an integral in $mathbbR$ cannot give you a complex result
answered Jul 15 at 11:44
Davide Morgante
1,904220
add a comment |Â
up vote
2
down vote
We know that $Re(f(z)) = u(x,y)$, to be the real part of an analytic function this has to be harmonic, so: $$nabla^2 u(x,y) = 0$$ which is easy to prove.
To find a complex part such that $f(z)$ will be analytic we integrate the differential form: $$dv = fracpartial vpartial xdx + fracpartial vpartial ydy = -fracpartial upartial ydx + fracpartial upartial xdy$$ for the Cauchy-Riemann conditions. To integrate the differential form we choose the spline connecting the points $(0,0)rightarrow (x,0)rightarrow (x,y)$, then $$v(x,y)-v(0,0) = int_gamma-fracpartial upartial ydx + fracpartial upartial xdy\
= -undersety'=0int_0^xfracpartial u(x',y')partial y'dx'+undersetx'=xint_0^yfracpartial u(x',y')partial x'dy'\
= undersety'=0int_0^x(2y'+e^-x'sin(y'))dx'+undersetx'=xint_0^y(2x'+e^-x'cos(y'))dy' \
= left.(2x'y'-e^-x'sin(y'))right|_0,;y'=0^x + left.(2x'y'-e^-x'sin(y'))right|_0, ;x'=x^y \
= 0 + 2xy-e^-xsin(y) = 2xy-e^-xsin(y)$$
If this is a good answer, we could see if $$nabla^2v(x,y)=0$$ which is true.
For to be $f(0) = f(0+i0)=1$ we have $v(0,0)=0$ because $u(0,0) = 1$. So in the end $$f(z) = (x^2-y^2+e^-xcos(y))+i(2xy-e^-xsin(y))$$ which can be rewritten knowing that $$(x+iy)^2 = x^2-y^2+2ixy= z^2;;;e^-x(cos(y)-isin(y)) = e^-xe^-iy = e^-z$$
$$f(z) = z^2+e^-z$$
Edit
I've edited the answer to take count of the correction $f(0)=1$ but if the initial condition was $f(0)=0$ I think that you could cheat using $v(0,0)=i$, this would give you the right answer, but an integral in $mathbbR$ cannot give you a complex result
answered Jul 15 at 11:44
Davide Morgante
1,904220
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We know that $Re(f(z)) = u(x,y)$, to be the real part of an analytic function this has to be harmonic, so: $$nabla^2 u(x,y) = 0$$ which is easy to prove.
To find a complex part such that $f(z)$ will be analytic we integrate the differential form: $$dv = fracpartial vpartial xdx + fracpartial vpartial ydy = -fracpartial upartial ydx + fracpartial upartial xdy$$ for the Cauchy-Riemann conditions. To integrate the differential form we choose the spline connecting the points $(0,0)rightarrow (x,0)rightarrow (x,y)$, then $$v(x,y)-v(0,0) = int_gamma-fracpartial upartial ydx + fracpartial upartial xdy\
= -undersety'=0int_0^xfracpartial u(x',y')partial y'dx'+undersetx'=xint_0^yfracpartial u(x',y')partial x'dy'\
= undersety'=0int_0^x(2y'+e^-x'sin(y'))dx'+undersetx'=xint_0^y(2x'+e^-x'cos(y'))dy' \
= left.(2x'y'-e^-x'sin(y'))right|_0,;y'=0^x + left.(2x'y'-e^-x'sin(y'))right|_0, ;x'=x^y \
= 0 + 2xy-e^-xsin(y) = 2xy-e^-xsin(y)$$
If this is a good answer, we could see if $$nabla^2v(x,y)=0$$ which is true.
For to be $f(0) = f(0+i0)=1$ we have $v(0,0)=0$ because $u(0,0) = 1$. So in the end $$f(z) = (x^2-y^2+e^-xcos(y))+i(2xy-e^-xsin(y))$$ which can be rewritten knowing that $$(x+iy)^2 = x^2-y^2+2ixy= z^2;;;e^-x(cos(y)-isin(y)) = e^-xe^-iy = e^-z$$
$$f(z) = z^2+e^-z$$
Edit
I've edited the answer to take count of the correction $f(0)=1$ but if the initial condition was $f(0)=0$ I think that you could cheat using $v(0,0)=i$, this would give you the right answer, but an integral in $mathbbR$ cannot give you a complex result
answered Jul 15 at 11:44
Davide Morgante
1,904220
We know that $Re(f(z)) = u(x,y)$, to be the real part of an analytic function this has to be harmonic, so: $$nabla^2 u(x,y) = 0$$ which is easy to prove.
To find a complex part such that $f(z)$ will be analytic we integrate the differential form: $$dv = fracpartial vpartial xdx + fracpartial vpartial ydy = -fracpartial upartial ydx + fracpartial upartial xdy$$ for the Cauchy-Riemann conditions. To integrate the differential form we choose the spline connecting the points $(0,0)rightarrow (x,0)rightarrow (x,y)$, then $$v(x,y)-v(0,0) = int_gamma-fracpartial upartial ydx + fracpartial upartial xdy\
= -undersety'=0int_0^xfracpartial u(x',y')partial y'dx'+undersetx'=xint_0^yfracpartial u(x',y')partial x'dy'\
= undersety'=0int_0^x(2y'+e^-x'sin(y'))dx'+undersetx'=xint_0^y(2x'+e^-x'cos(y'))dy' \
= left.(2x'y'-e^-x'sin(y'))right|_0,;y'=0^x + left.(2x'y'-e^-x'sin(y'))right|_0, ;x'=x^y \
= 0 + 2xy-e^-xsin(y) = 2xy-e^-xsin(y)$$
If this is a good answer, we could see if $$nabla^2v(x,y)=0$$ which is true.
For to be $f(0) = f(0+i0)=1$ we have $v(0,0)=0$ because $u(0,0) = 1$. So in the end $$f(z) = (x^2-y^2+e^-xcos(y))+i(2xy-e^-xsin(y))$$ which can be rewritten knowing that $$(x+iy)^2 = x^2-y^2+2ixy= z^2;;;e^-x(cos(y)-isin(y)) = e^-xe^-iy = e^-z$$
$$f(z) = z^2+e^-z$$
Edit
I've edited the answer to take count of the correction $f(0)=1$ but if the initial condition was $f(0)=0$ I think that you could cheat using $v(0,0)=i$, this would give you the right answer, but an integral in $mathbbR$ cannot give you a complex result
answered Jul 15 at 11:44
Davide Morgante
1,904220
edited Jul 15 at 12:21
answered Jul 15 at 11:44
Davide Morgante
1,904220
answered Jul 15 at 11:44
Davide Morgante
1,904220
answered Jul 15 at 11:44
Davide Morgante
1,904220
1,904220
add a comment |Â
add a comment |Â
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$Re(f(z))=x^2-y^2+e^-xcos y$ implies $Re(f(0)) = 1$, which contradicts $f(0) = 0$.
– Martin R
Jul 15 at 11:49
maybe it is a mistake in the question, I will change it to $1$
– newhere
Jul 15 at 11:55
@newhere Beside the constant for the value of $f(0)$ your calculations are correct! The only thing is that you have to do another small step to write the last function in term of $z$
– Davide Morgante
Jul 15 at 12:58