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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.

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    $f'(z) = 5 e^{5z}$ et voila.2012-10-22
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    Or you can just say it is the composition of two entire functions.2012-10-22
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    i think i am thinking too much since the question involves complex planes.2012-10-22
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    I'm tempted to say that the answer is entirely obvious.2012-10-22

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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.

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    Not entirely: you still need to check differentiability (which is easy in this case)2012-10-22
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    I don't get it: differentiability *of what*? What still must be "checked" is that the partial derivatives are *continuous*, which is boringly easy in this case2012-10-22
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    That exactly what I meant. Differentiablity of $u,v$ as functions of two variables, which is equivalent to their partial derivatives being continuous. This is trivial and boring, but still should be mentioned - since there are function for which the CR equations do hold but they are not analytic.2012-10-22
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    Being differentiable is not equivalent to being $C^1$ (but $C^1$ implies differentiable, so the implication goes the right way).2012-10-23
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The sums, products and compositions of entire functions are entire.