Sketch the following regions:
- $\operatorname{Arg}(e^z)>\dfrac{π}{4}$
- ${e^z|\operatorname{Im}(z) = 1}$
- $|e^z| > 2$
I am confused of graphing $e^z$ functions.
Let $z = x + iy$. For number 1, should I draw the region from angle $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{4}$ or just a vertical line at $y = \dfrac{\pi}{4}$ and shade the right side?
For number 2, should I draw an angle of $1$ or a graph like $e^(x+1)$?
For number 3, should I draw an opened circle with radius $2$ and shade out side or a vertical line at $x = \ln 2$?
For $1$:
$Arg(e^z) \equiv y$ modulo $2\pi$ but it is not equal to $y$. This is because of the choice of principal value since capitalized $Arg$.
$Arg(e^{x+i*(\pi+\epsilon)})=-\pi+\epsilon \equiv \pi + \epsilon$ modulo $2 \pi$. But that is less than $\frac{\pi}{4}$ now. It keeps wrapping around to $-\pi$. This means you need to shade in above but not including the horizontal line $y=\frac{\pi}{4}$, but then stop shading when you reach $y=\pi$ including that line.
You want to shade in the region when $\pi \geq Arg(e^z)>\frac{\pi}{4}$ where the first inequality is vacuous because of how we chose principal value. Repeat this pattern of shading vertically every $2 \pi$.
For $2$:
You've assumed $y=1$ so you are sketching $e^{x+i}$. The $Arg$ is fixed at $1$, but the radius is $e^x$ for all possible values of $x$. That is all positive reals. So you sketch the half line that starts at the origin and goes out along the phase $1$ in the first quadrant. The part near the origin corresponds to $x \to -\infty$. The part far away to $x \to + \infty$. The part near the unit circle $x \approx 0$.
For $3$:
As with $1$ $e^z=e^x e^{iy}$. So $\mid e^z \mid = e^x$. Requiring this be greater than $2$ is the same as requiring $x > \ln 2$. So you draw the vertical line at $x=\ln 2$ and shade in the right part.