Let L = $\{(x,y): y = 2x\}$. Find the image of L under $e^z$. Also sketch the image.
I am really confused about all of this but here is my attempt. Say we are mapping from the z-plane to the w-plane.
In the z-plane, we are taking all points on the line y = 2x. We want to put each point through the transformation $e^z = e^{x+iy}$ (if we let z = x + iy) that maps to the w-plane.
Since y = 2x, we get $e^z = e^{x+iy} = e^{x + i{x \over 2}} = e^xe^{i{x \over 2}}$. This is equivalent to
$e^x(cos({x \over 2})+isin({x \over 2}))$.
$=e^xcos({x \over 2})+ie^xsin({x \over 2})$
$= u + iv$ (so we have the u-coordinate and v-coordinate).
I have no clue if this is the image, and I also don't know how to draw this. Do I just start plugging in points?
Thank you.
Best Answer
Yes, you're right. The image is the set $\{ e^x (\cos(x/2) + i \sin(x/2)) \colon x \in \mathbf{R} \}$ in the $w$ plane. To draw it, note that without the factor $e^x$ you would have the point $w=\cos(x/2) + i \sin(x/2)$ which moves counterclockwise around the unit circle as $x$ increases. Multiplying this by the real number $e^x$ has the effect of increasing the radius (i.e., the distance from the origin to $w$) as $x$ increases. This means that you get a spiral instead of a circle. (As $x \to -\infty$, the spiral approaches the origin in the $w$ plane, and as $x \to +\infty$, it grows without bound.)