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.
 A: 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 $\arg{z}=y$
A: $e^{x+iy}=e^x(\cos y+i\sin y).$
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.
A: Let $w=e^z$
$\implies z=\ln (w)$
$\implies x+iy=\ln(Re^{i\phi})=\ln R+i\phi$
$\implies x+iy=\ln R+i\phi$ 
Picking up imaginary parts, you get $y=\phi$.
Now, for $0<y<\pi\implies 0<\phi<\pi$ which suggests that the image is restricted in upper half plane described by the set $\{u+iv: v>0\}$. This proves that map is onto.
For the second part, you can see that $\ln R=k$ (constant) gives you $R=e^k=c(>0)$ so that the image is circle with radius $c$ in $w-$ plane. Similarly, for $y=constant$ you will get $\phi=c$ (constant) suggesting the image is line through origin with fixed angle $c$ with positive real axis.
