show that f(z) is an entire function

**linalg123** · May 31st 2011, 08:20 PM

$f(z) = exp[(z)^2]$
$= exp[(x+iy)^2]$
$= exp[x^2-y^2]\cdot exp[(2xy)i]$
$= exp[x^2-y^2]\cdot (cos(2xy)+isin(2xy))$

$u= exp[x^2-y^2]\cdot cos(2xy)$
$v= exp[x^2-y^2]\cdot sin(2xy)$

$u_x= 2exp[x^2-y^2]\cdot (xcos(2xy)-ysin(2xy)) = v_y$

$u_y= -2exp[x^2-y^2]\cdot (xsin(2xy)+ycos(2xy))= -v_x$

The partial derivatives of v and u exist, are continuous and satisfy the cauchy-reimann equations for the whole complex plane, therefore, f(z) is an entire function.

is this answer sufficient? or is there a way to show that the equations hold for the entire complex plane.

**TheEmptySet** · May 31st 2011, 08:35 PM

Originally Posted by linalg123

$f(z) = exp[(z)^2]$
$= exp[(x+iy)^2]$
$= exp[x^2-y^2]\cdot exp[(2xy)i]$
$= exp[x^2-y^2]\cdot (cos(2xy)+isin(2xy))$

$u= exp[x^2-y^2]\cdot cos(2xy)$
$v= exp[x^2-y^2]\cdot sin(2xy)$

$u_x= 2exp[x^2-y^2]\cdot (xcos(2xy)-ysin(2xy)) = v_y$

$u_y= -2exp[x^2-y^2]\cdot (xsin(2xy)+ycos(2xy))= -v_x$

The partial derivatives of v and u exist, are continuous and satisfy the cauchy-reimann equations for the whole complex plane, therefore, f(z) is an entire function.

is this answer sufficient? or is there a way to show that the equations hold for the entire complex plane.

Yes this is sufficient. Depending on your definitions another way is to show that the function has power series that has an infinite radius of convergence.

$f(z)=e^{z^2}=\sum_{n=0}^{\infty}\frac{z^{2n}}{n!}$
This power series converges for all z by the ratio test.
So the function is analytic, or holomorphic, or entire.

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