Show $5z^n=e^z$ has a finite number of zero in $\{a<\Im z < v\}$ and $\{a < \Re z < b \}$ Show that the number of roots $N$ of the function
$$
h(z)= f(z)-g(z)=5z^n - e^z, n\ge 1,
$$
is at most finite in any horizontal strip
$$
\alpha=\{z:a<\Im z<b\},
$$
and any vertical strip,
$$
\beta=\{z:a<\Re z<b\}.
$$
I'm not clear on the meaning of the question.
Should I prove that $N$ is finite on $\alpha$ and $\beta$ separately, or on $\alpha \cap \beta$.  
In any case, I would use Rouche theorem, but has none of the contour is a circle
I don't see how to proceed.   
Also, $|g|$ is not always less than $|f|$ on circle of arbitrary radius $a$ or $b$.
 A: There are two different problems here: 


*

*Show that $h$ has finitely many zeroes on $\alpha$

*Show that $h$ has finitely many zeroes on $\beta$


On $\alpha\cap \beta$ the problem would be too easy: an entire function can have only finitely many zeroes on any bounded region, since the zeroes cannot accumulate.
The right approach is to rewrite $h(z)=0$  as the equation $$5z^n=e^z\tag1$$ and consider the size (absolute value) of both sides of (1). Namely, if (1) holds then we certainly have $$5|z|^n=e^{\mathrm{Re}\,z}\tag2$$
Problem 2 is a bit easier. Since $\mathrm{Re}\,z$ is bounded within the strip, the right side of (2) is bounded (by some constant $M$). Therefore, (2) can only hold when $5|z|^n\le M$. But this constrains $z$ to lie in some bounded region, and $h$ can have only finitely many zeroes on a bounded region (as mentioned before).
In problem 1, you should consider the left and right "tails" of the strip separately. If you go far enough to the left, you  will find $5|z|^n>e^{\mathrm{Re}\,z} $.  If you go far enough to the right, you  will find $5|z|^n<e^{\mathrm{Re}\,z} $.  This leaves only the middle part, where, again, $h$ can have only finitely many zeroes.
