If $f(z)=e^z$ then $f(A)=?$...

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Let A denote the set of all complex numbers lying within the square having vertices $0,1,\pi i,\pi i +1$. let $f(z)=e^z$ then $f(A)=$

(1) $ \{ \omega : 1\leq \omega \le e \}$

(2) $\{\omega : |\omega| \le e \}$

(3) $\{\omega : 1\leq |ω| \le e, 0\leq \arg \omega \le \pi \}$

(4) $\{\omega : 1\leq |ω| \le e, 0\leq \arg \omega \le \frac{\pi}{2} \}$

I am not getting what to use here. The region they mentioned is a rectangle in first quadrant.so can I say $0 \le \arg \omega \le \frac{\pi}{2}$ The given function is entire, so is analytic in the given region. and $f(z)=e^z$ is continuous also, so can I say by maximum modulus principle, the maximum and minimium lies in the boundary of the region. i.e. $e^0< |f(z)| <e^{\pi i+1}=e$, but still little doubtful here.

So option (3) should be correct. am I correct?

Thanks in advance!!