Originally Posted by

**linalg123** $\displaystyle f(z) = exp[(z)^2] $

$\displaystyle = exp[(x+iy)^2] $

$\displaystyle = exp[x^2-y^2]\cdot exp[(2xy)i] $

$\displaystyle = exp[x^2-y^2]\cdot (cos(2xy)+isin(2xy)) $

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

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

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

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

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.