Show $e^z$ is 1-1 onto from $\{x+iy: 00\}$ and find images under $\{x+iy: \text{$x$ is constant}\}$ and $y$ constant. Show $e^z$ is 1-1 onto from $\{x+iy: 0<y<\pi\}$ onto $\{u+iv : v>0\}$ and find images under $\{x+iy: \text{$x$ is constant}\}$ and  $\{x+iy: \text{$y$ is constant}\}$.
For $1-1$ do I consider say $z_1=x_1+iy_1, z_2 = x_2+iy_2$ such that either $x_1 \neq x_2$ or $y_1 \neq y_2$? Then compute $e^{z_1}$ and $e^{z_2}$ and show they're not equal? And for onto how do I show every element of the upper half plane gets hit by $e^z$?
For the second part I know $x$ is the radius and $y$ is the angle so is it asking what the image is when the radius is fixed and when the angle is fixed? Any help greatly appreciated.

For completeness here's a solution following my comments.
Suppose $z_1=x_1+iy_1,z_2=x_2+iy_2\in\{x+iy:y\in(0,\pi)\}$. Then $e^{z_1}=e^{z_2}$ gives
$$
e^{x_1}e^{iy_1}=e^{x_2}e^{iy_2}
$$
with both sides in polar form. Since polar form is unique up to argument we deduce $x_1=x_2$ and $y_1=y_2+2k\pi$  for some integer $k$. But $\lvert y_1-y_2\rvert < \pi$ so $y_1=y_2$ and injectivity follows.
Now write $e^z=e^xe^{iy}$ for $z$ in  range, in polar form. As $x$ ranges over $\mathbb{R}$, $e^x$ (our polar radius) ranges over $(0,\infty)$. As $y$ ranges over $(0,\pi)$, our polar angle ranges over $(0,\pi)$. This is enough to deduce that the mapping is onto to the upper-half plane.
It should be clear from this construction that fixing $x$ and varying $y$ gives the upper-half semicircle of radius $e^x$ with centre at the origin, and fixing $y$ and varying $x$ giving the half-line $\arg{z}=y$
