# Complex analysis

• September 16th 2012, 08:52 AM
Skruven
Complex analysis
let: $u(x,y) = \sin (x^2-y^2)\cosh (2xy)$

Is there a function $f(x+iy) = u(x,y) + iv(x,y)$ that is analytic in the complex plane.
If so determine all the functions f.

........................................

So i know that if u(x,y) and v(x,y) are harmonic and satisfy the cauchy riemann equations f are analytic.

My standard method for this kind of problems is:

Find ${u}'_{x}$ then with CR(1): ${v}'_{y} = {u}'_{x}$ i integrate so i get v(x,y) = something + g(x).
Then with CR(2): ${u}'_{y} = -{v}'_{x}$ i get g(x).

But with my method it gets really messy! i cant solve the integration, so i wonder if theres another way to do this?

regards
• September 16th 2012, 09:32 AM
Prove It
Re: Complex analysis
Quote:

Originally Posted by Skruven
let: $u(x,y) = \sin (x^2-y^2)\cosh (2xy)$

Is there a function $f(x+iy) = u(x,y) + iv(x,y)$ that is analytic in the complex plane.
If so determine all the functions f.

........................................

So i know that if u(x,y) and v(x,y) are harmonic and satisfy the cauchy riemann equations f are analytic.

My standard method for this kind of problems is:

Find ${u}'_{x}$ then with CR(1): ${v}'_{y} = {u}'_{x}$ i integrate so i get v(x,y) = something + g(x).
Then with CR(2): ${u}'_{y} = -{v}'_{x}$ i get g(x).

But with my method it gets really messy! i cant solve the integration, so i wonder if theres another way to do this?

regards

Why are you integrating? Just evaluate the partial derivatives and show that they are equal (as required by the CR equations).
• September 16th 2012, 09:59 AM
Skruven
Re: Complex analysis
Quote:

Originally Posted by Prove It
Why are you integrating? Just evaluate the partial derivatives and show that they are equal (as required by the CR equations).

cause i dont have v(x,y)? and i want to find it so i can set up f = u + iv.
• September 16th 2012, 12:55 PM
SworD
Re: Complex analysis
This is rather difficult, I'm also interested to see if anyone knows any tricks besides the brute force way which Skruven described.
• September 16th 2012, 01:28 PM
Plato
Re: Complex analysis
Quote:

Originally Posted by SworD
This is rather difficult, I'm also interested to see if anyone knows any tricks besides the brute force way which Skruven described.

There is no simple way. Look at this.
That gives $u_x$. But even mathamatica times out on integrating that with respect to y.
• September 16th 2012, 01:46 PM
SworD
Re: Complex analysis
Maple gives the following result. So according to this, V(x,y) is the following function plus some g(x).

$(\frac{1}{4}\cdot e^{4\cdot x\cdot y}-\frac{1}{4})\cdot e^{-2\cdot x\cdot y}\cdot e^{I\cdot x^2}\cdot e^{-I\cdot y^2}+(\frac{1}{4}\cdot e^{4\cdot x\cdot y}-\frac{1}{4})\cdot e^{-2\cdot x\cdot y}\cdot e^{-I\cdot x^2}\cdot e^{I\cdot y^2}$

Now.. constructing f(z) from that is a problem of a whole new level, lol. But the complexity of this doesn't preclude the possibility of there being a neat trick. Maybe f(z) collapses into something simple.
• September 16th 2012, 02:09 PM
SworD
Re: Complex analysis
This is actually not that complicated. I guessed that there must be some "symmetry" between the hyperbolic/trigonometric sines and cosines, as usually exhibited by the real and imaginary parts of an analytic function, and indeed, if U(x,y) is defined as in the initial post, then

$U(x,y) = \sin (x^2-y^2)\cosh (2xy)$

$V(x,y) = \cos(x^2-y^2)\sinh(2xy)$

The above post is maple expanding that into complex exponentials for some reason. But the C-R equations do equal for these functions, try them.

To find f(z) from these, remember that if two analytic functions coincide on any line, say on the real axis, they coincide everywhere. So if you express u + iv as a function of a real variable, by treating y=0, and x = z, you can get the general equation for f(z). Note that on the real axis, the imaginary part will vanish, because sinh(0) is 0. Also, cosh(0) = 1. So from this you can get:

$f(z) = \sin(z^2) + ci$

Where c is any constant. We need that there because given the real part of an analytic function, the imaginary part can vary up to a constant added/subtracted. Technically I should have put it there when writing out V(x,y).