Can someone please verify that my diagram plot based on the calculations below is correct -Thanks.
$$w=e^z=e^x\cos(y)+ie^x\sin(y)$$
$$\implies u=e^x\cos(y)\space and \space v=e^x\sin(y) \tag{1}$$
$$for \space -1 <\Re(z) \le 1 \space \space i.e. \space \space 1 < x \le 1$$
$$\dfrac{u^2}{e^{2x}}+\dfrac{v^2}{e^{2x}}=1$$
$$=\dfrac{u^2}{e^2}+\dfrac{v^2}{e^2}=1 \space \space and \space \space \dfrac{u^2}{e^{-2}}+\dfrac{v^2}{e^{-2}}=1 \space for \space x=1 \space and \space x=-1 \space respectively$$
$$for \space -\frac{\pi}{2} <\Im(z) \le \frac{\pi}{2}\space \space i.e. \space \space -\frac{\pi}{2} < y \le \frac{\pi}{2}$$
$$\color{green}{0<cos(y)<1 \space \space and \space \space -1<sin(y)\le1}$$ from $(1)$ i deduce the following:
$$\therefore \space 0< u \le e \space \space and \space \space -e < v \le e$$
It doesn't seem right.
As you noted: $$w(z) = e^z = e^x\cdot (\color{red}{\cos y+i\sin y})$$
Note that the last part is the equation of a unit-circle since $\color{red}{\cos y+i\sin y} = e^{iy}$.
Since $y\in {\left]\frac{-\pi}{2},\frac{\pi}{2} \right]}$ you know the map goes to the right part of a circle.
How about the radius?
$$\text{the radius = } e^x \in {\left]\frac{1}{e}, e\right[}$$
You get the following: