# Math Help - analytic function

1. ## analytic function

$f:\mathbb{C}\longrightarrow\mathbb{C}$ is an analytic function and

$f(x+iy) = u(x,y) + iv(x,y)$

What geometric relationship exists between the level curves of $u$ and those of $v$?

Draw a rough sketch of these level curves for the case $f(z) = z^{2 }$to support your answer.

I haven't really any idea where to start this.

2. Originally Posted by Tekken
$f:\mathbb{C}\longrightarrow\mathbb{C}$ is an analytic function and

$f(x+iy) = u(x,y) + iv(x,y)$

What geometric relationship exists between the level curves of $u$ and those of $v$?

Draw a rough sketch of these level curves for the case $f(z) = z^{2 }$to support your answer.

I haven't really any idea where to start this.
Hint #1 :
Spoiler:
Cauchy-Riemann conditions

Hint #2 :
Spoiler:
Orthogonality of gradients of $u$ and $v$

Hint #3 :
Spoiler:
The gradient is perpendicular to level curves

Hint #4 : do the example first (it could also be hint #0 if you really don't know where to start).

3. well is this a start? Im clueless at picturing things in my head as regards graphing and plotting stuff

$f(z) = z^{2}$ and as $z = x+iy$

=> $f(x+iy) = (x^{2} - y^{2}) + 2xyi$

let $u = x^{2} - y^{2}$ and $v = 2xy$

therefore $U_{x} = 2x$ and $V_{y} = 2x$

so
$U_{x} = V_{y}$

and $U_{y} = -2y$ and $V_{x} = 2y$

so
$U_{y} = -V_{x}$

Therefore as the Cauchy Riemann equations are satisfied and the function $f(z) = z^{2}$ is analytic...

4. Originally Posted by Tekken
well is this a start? Im clueless at picturing things in my head as regards graphing and plotting stuff

$f(z) = z^{2}$ and as $z = x+iy$

=> $f(x+iy) = (x^{2} - y^{2}) + 2xyi$

let $u = x^{2} - y^{2}$ and $v = 2xy$

therefore $U_{x} = 2x$ and $V_{y} = 2x$

so
$U_{x} = V_{y}$

and $U_{y} = -2y$ and $V_{x} = 2y$

so
$U_{y} = -V_{x}$

Therefore as the Cauchy Riemann equations are satisfied and the function $f(z) = z^{2}$ is analytic...
What you need to do is plot a few level curves, like $u=-3,-2,-1,1,2,3,\ldots$ and same for $v$. It is easier for $v$: we have $y=\frac{v}{2x}$ (hyperbola). For $u$, if you don't know conics (these are also hyperbola), you can plot the level curves on your calculator or computer (equation $y=\pm\sqrt{x^2-u}$ (either + or -)), together with level curves for v and see what relation there is between both (you'll need an orthonormal basis to see it).

5. Originally Posted by Laurent
What you need to do is plot a few level curves, like $u=-3,-2,-1,1,2,3,\ldots$ and same for $v$. It is easier for $v$: we have $y=\frac{v}{2x}$ (hyperbola). For $u$, if you don't know conics (these are also hyperbola), you can plot the level curves on your calculator or computer (equation $y=\pm\sqrt{x^2-u}$ (either + or -)), together with level curves for v and see what relation there is between both (you'll need an orthonormal basis to see it).
Thanks i really appreciate your help. Unfortunately im not at the stage where i really understand your reply... Would you have any link to a website where i could read up on this sort of question, particularly "level curves"?

6. Originally Posted by Tekken
Thanks i really appreciate your help. Unfortunately im not at the stage where i really understand your reply... Would you have any link to a website where i could read up on this sort of question, particularly "level curves"?
I don't know if this help, but you can look here or rather there.

For a function $u(x,y)$, the level curves of $u$ are the sets of values $(x,y)$ that give the same value $u(x,y)$: for any $u_0\in\mathbb{R}$, the curve of level $u_0$ is $\{(x,y)\in\mathbb{R}^2|u(x,y)=u_0\}$.