0
$\begingroup$

How is $f(z)=e^{5z}$ ENTIRE?

This is a part of the solution in between a bigger question. But i don't seem to understand on how to check the above $f(z)$ holomorphicity?

I know that when $f(z)$ is entire means that it is holomorphic in the entire complex plane.

  • 1
    I'm tempted to say that the answer is entirely obvious.2012-10-22

2 Answers 2

2

Putting $\,z=x+iy\,\,,\,\,x,y\in\Bbb R\,$:

$e^{5z}=e^{5x}e^{5yi}=e^{5x}\cos 5y+i\,e^{5x}\sin 5y$

Now put $\,u(x,y):=e^{5x}\cos 5y\;\;,\;\;v(x,y)=e^{5x}\sin 5y$ , and let us check the Cauchy-Riemann Equations:

$u_x=5e^{5x}\cos 5y=v_y$

$u_y=-5e^{5x}\sin 5y=-v_x$

And since the above is true for any $\,(x,y)\in\Bbb R^2\cong \Bbb C\,$ we're thus done.

  • 2
    Being differentiable is not equivalent to being $C^1$ (but $C^1$ implies differentiable, so the implication goes the right way).2012-10-23
0

The sums, products and compositions of entire functions are entire.