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.

$e^{x+iy}=e^x(\cos y+i\sin y).$
If $0<y<\pi$ then $e^x\sin y>0,$ so $\exp$ restricts well to a mapping from the horizontal strip  $A=\{x+iy:x\in\mathbb R,0<y<\pi\}$ to the upper half-plane $B=\{u+iv :u\in\mathbb R,v>0\}.$
To see that this mapping is a bijection, you don't need to treat injectivity and surjectivity separately: it suffices to notice that for any $u+iv\in B$, the system $e^x\cos y=u,e^x\sin y=v$ has a unique solution $x+iy\in A$, given by $x=\ln\sqrt{u^2+v^2},y=\arccos\frac u{\sqrt{u^2+v^2}}.$
For any $x_0\in\mathbb R$, the image of $\{x_0+iy: 0<y<\pi\}$ is $e^{x_0}\{\cos y+i\sin y:0<y<\pi\}$, the upper half of the circle of radius $e^{x_0}$ centered at the origin.
For any $y_0\in(0,\pi)$, the image of $\{x+iy_0: x\in\mathbb R\}$ is $\{e^x:x\in\mathbb R\}(\cos y_0+i\sin y_0)=\mathbb R_+^*(\cos y_0+i\sin y_0)$, an open half line.
