# Math Help - Analytic Functions?

1. ## Analytic Functions?

Hello again,

I am trying but it is really hard learning this stuff on your own. Never gonna take a calc course online again :\.

This it he question, I am currently stuck on.

Where is the function $f(z)=e^(^x^^^2^-^y^^^2^) * [cos(2xy) + isin(2xy)]$ analytic?

For now could someone just help me with the derivative? I am in the process of figuring out Couchy-Rieman equations.

Is the derivative only present if $\partial$u/ $\partial$x = $\partial$v/ $\partial$y?

just for my understanding in the equation $1+iy$ there is no derivative because $\partial$u/ $\partial$x = 0 and $\partial$v/ $\partial$y = 1?

Thanks
Rioch

2. ## Re: Analytic Functions?

$u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_x(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))$
$v(x,y)=e^{x^2-y^2}\sin(2xy) \implies v'_y(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))$
$u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_y(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))$
$v(x,y)=e^{x^2-y^2}\sin(2xy) \implies -v'_x(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))$

3. ## Re: Analytic Functions?

Originally Posted by Ridley
$u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_x(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))$
$v(x,y)=e^{x^2-y^2}\sin(2xy) \implies v'_y(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))$
$u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_y(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))$
$v(x,y)=e^{x^2-y^2}\sin(2xy) \implies -v'_x(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))$
HI,

Thanks for the quick response. I understand that you have taken the derivatives and in this example the Cauchy-Riemann equations are satisfied.

I am unsure what this means though and where to go from here? No need to carry out the math (although it is helpful), however if possible a short explanation ?

how would I express the final derivative? $f'(x)$ , is it just the range of equations?

4. ## Re: Analytic Functions?

You asked where the function is analytic. The first step is to check where the partial derivatived exist and satisfy the Cauchy-Riemann equations, and as you can see they are satisfied everywhere. If the function is also continuous in that domain, then the function is analytic there (Looman-Menchoff's theorem).

The derivative at that point can be written as $\partial f/\partial z = u'_x+iv'_x = v'_y - iu'_y$

5. ## Re: Analytic Functions?

Another approach:

\begin{aligned} e^{(x^2-y^2)}(\cos(2xy) + i\sin(2xy)) &= e^{(x^2-y^2)}e^{i2xy} \\ &= e^{x^2-y^2+2ixy} = e^{(x+iy)^2} = e^{z^2}. \end{aligned}

Squaring and exponentiating are both analytic functions, so f(z) is analytic everywhere.