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How can i find entire functions $f, g, F, G$ with $f(0)=g(0)=F(0)=G(0)=0$ and $$e^{4f(z)}+e^{4g(z)}=2e^z ,\ \ e^{4F(z)}+e^{4G(z)}=2(z+1)$$ I can't get started, i am stack. I will appreciate your help. Thank you.

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up vote 2 down vote accepted

For the first one $$ f(z)=g(z)=\frac{z}{4}. $$

The second one has no solution. If $e^{4F(z)}+e^{4G(z)}=2(z+1)$, then $$ 1+e^{4(G(z)-F(z))}=2(z+1)e^{-4F(z)}. $$ The left hand side is an entire function that does not take the value $1$. Moreover, it has an essential singularity at $\infty$ (since obviously it is not constant). By Picard's Big Theorem, it takes the value $0$ infinitely often. But the right hand side takes the value $0$ only for $z=-1$.

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yep! this one works, but it goes back to what @mike said, right hand side constant. Lets see the second one and thank you. – jm sol Apr 10 '12 at 20:10

That you cannot do the first is an application of Picard's Thm. Divide through by $e^{4f(z)}$. The rhs can never be zero, so it must take on every other value . When it takes on the value 1 $e^{4g-4f} = 0$ which is impossible. I would suppose the 2nd works out along the same lines.

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Thanks Mike, however, you are very brief that i cannot seem to gasp the point, and later on, the conclusion. Would you elaborate more? – jm sol Apr 10 '12 at 18:58
@mike Your argument fails if the right hand side is constant, what happens when $z-4\,f(z)equiv0$. – Julián Aguirre Apr 10 '12 at 19:07

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