Why the function $w=e^z$ maps the lines $x=c$ The questions asks:
Explain why the function $w=e^x$
a) maps the lines $x=c$, with $c$ a constant, onto the circles $w=e^c$
b)maps the lines $y=c$, with $c$ a constant, onto half rays $\theta=c$ from the origin to infinity excluding the origin
c)maps the strip $T=\{z=x+iy\mid0<y<2\pi\}$ onto $\Bbb C-\{0\}$
I am completely lost. Can someone help explain what this question is even asking me to do?
 A: HINT: It's all due to the Euler's formula. 

Recall that for a complex number $z = x+ iy$ we have
$$
e^{i\theta} =  \cos \theta + i \sin \theta \iff z = x+ i y = |z| e^{\operatorname{arg}z} = \sqrt{x^2 + y^2}\,e^{\,\operatorname{atan}\!\frac{y}{x}}
$$


*

*The line $x=c $  can be written as $  z_1 = c + i\cdot0$. The $z_1$ gets mapped into $z_2$ by $ w(z)$, i.e. $z_2 = w(z_1) = e^{z_1} = e^{c + i\cdot 0} = e^c$. Using Euler's formula, we write $z_2 = e^{z_1} = e^{c + i\cdot 0} = \sqrt{c^2 + 0^2} \cdot e^0 = |c|$, so the domain set of $\left\{z_1 \ \big|\  z_1 =  c+0\cdot i\right\}$ is mapped by $w$ into the image set $\left\{z_2 \ \big| \ |z_2| = c \right\}$, which is a circle.

*Similarly, the line $y=c$ can be written as $z_1 = 0 + i\cdot c$. Then $z_2 = w(z_1) = e^{z_1} = e^{ 0 + c\cdot i} = 1 \cdot e^{c\cdot i} = \cos c + i \sin c $ and then the image of $w$ applied to the domain set of $\left\{z_1 \ \big|\  z_1 =  0+c\cdot i\right\}$ will be the image set $\left\{z_2 \ \big| \ z_2 = \cos c + i \sin c \right\}$, which is a half ray.

*I am sure that using previous two cases, you can figure out what is the strip $T=\big\{z=x+iy\mid0<y<2\pi\big\}$  getting mapped two.
