# Sketching variations of $Arg(z)$

I'm asked to describe and sketch the following sets in $\mathbb{C}$:

A) $\{ z \in \mathbb{C}: |Arg(z+1)| < \frac{\pi}{4}$

B)$\{ z \in \mathbb{C}: |z| \le arg \ z \ \ \ \ and\ \ 0 \le arg \ z \le \pi \}$

C) $\{ z \in \mathbb{C}: 1 \le |z| < 2 \ \ \ \ and\ \ \frac{\pi}{4} \le arg \ z \le \pi \}$

D) $|z|^2=(arg \ z)^2 \ and \ \ 0 \le arg \ z < 2 \pi$

Through Wolfram and other online sources, I get different graphs (from spirals to lines). I'm confused as to how I can show these are spirals or lines.

I know $arg \ z=tan^{-1}(y/x)$. I don't know how to solve for these using polar coordinates. Is that even the best way to solve? Any help is appreciated. Thank you.

• Before you start out, regarding that argument, please observe the quadrants. The way how you wrote $arg(z)$, it's true for Q1 and Q4 – imranfat May 5 '14 at 21:16
• for which letter? For all you mean @imranfat – User69127 May 5 '14 at 21:47
• What I meant is that $arg(z)=arctan(y/x)$ is true if the complex number is in the first quadrant or the fourth. If the point is in the third and fourth, you need to "add" that 180 – imranfat May 6 '14 at 14:14
• @imranfat Oh I see. I guess what I'm doing here is trying to solve for these by coverting to cartesian coordinates. Not sure if that's the best route. For example, Part B I have $\sqrt{x^2+y^2} \le tan^{-1}(y/x)$ I'm not sure where that's getting me. – User69127 May 7 '14 at 21:53

A) since $|Arg(z)| < \frac{\pi}{4}$ are the complex numbers for which the argument is under $\pi/4$ excluded that is 1/8th of the plan, then $0<|Arg(z+1)| < \frac{\pi}{4}$ is the same 1/8th of the plan but slided one to the left B) $\{ z \in \mathbb{C}: |z| \le arg \ z \ \ \ \ and\ \ 0 \le arg \ z \le \pi \}$ is the complex for which the modulus is smaller than the argument : that is under the spiral described by $r=\theta$ for $\theta$ from $0$ to $\pi$ included. C) $\{ z \in \mathbb{C}: 1 \le |z| < 2 \ \ \ \ and\ \ \frac{\pi}{4} \le arg \ z \le \pi \}$ is $3/8th$ of an annulus for which the inner radius is 1 (included) and the outside radius is 2 (excluded) going from an angle of $\pi/4$ included to $\pi$ included D) are all points of the spiral $r=\theta$ since all points have an argument into the bound $[0; 2\pi[$ and that r is always positive. • For part $D$, can't we just take the square root of both sides and get $r=\theta$? – User69127 May 9 '14 at 0:02
• yes you are right. – Tétef May 9 '14 at 0:04
• Part D is the set of $z\in\mathbb{C}$ for which $|z|^2=(arg \ z)^2 \ and \ \ 0 \le arg \ z < 2 \pi$. These $z$-values all have $|z|<2\pi$, so your graph should not spiral outward forever. – Steve Kass May 9 '14 at 0:13
• @SteveKass agreed – User69127 May 9 '14 at 0:16