Describe the image of the set under a complex function

Question

Question:

Describe the image of the rectangle under exp(z): {z=x+iy (complex) | 0<=x<=1, 0<=y<=pi/3}

My concepts are unclear in complex analysis, but I read online that I'm supposed to substitute the set into the function:

exp(x+iy) = (e^x)(e^iy) = (e^x)(cos(y)+isin(y))

What can I do next to find the image? And more broadly, what's a general procedure I should be following for these types of problems? Thank you in advance!

Answer 1

$e^z=e^xe^{iy}=e^x(\cos y+i\sin y)$, so it looks like a wedge of angle $\frac{\pi}3$ with the $x$-axis that lies in an annulus of inner radius $1$ and outer radius $e$.

As far as general strategy, whenever you have $e^{ix}$ there is Euler's formula. Also, (and closely related to that fact), $z=re^{i\theta}$ where $(r,\theta)$ are "polar coordinates" for $z$.

Answer 2

Try mapping the line segments forming the rectangle under $\exp$, one at a time, to see what curves they produce. Draw a picture. The resulting curves should form a closed loop, separating a bounded region from an unbounded region.

You should check which region is the image of the rectangle (it's theoretically possible that it's the unbounded region) by picking a point in the rectangle, and figuring out whether that point lies in the bounded or unbounded region.

I'll help get you started. One edge of the rectangle is described by $1 + iy$, where $0 \le y \le \frac{\pi}{3}$. If we map this under $\exp$, we get $$\exp(1 + iy) = e \cdot e^{iy} = e(\cos(y) + i \sin(y)), \quad 0 \le y \le \frac{\pi}{3}.$$ This describes an arc of the circle of radius $e$, centred at $0$, particularly of the points whose principle argument is between $0$ and $\frac{\pi}{3}$.

Try mapping the other edges. Two of them remain line segments, and one maps to a single point!