# Sketching the Points in the Complex Plane

I am asked to sketch the points in the complex plane satisfying the given inequality:

$-\pi < \arg(z) < \pi/2$

If $\arg(z) = \pi/4$, what exactly is there to sketch?

• To (almost) answer your specific question, a certain half-line. If it is not obvious which half-line, find a couple of points on it. – André Nicolas Sep 4 '15 at 20:18
• What is a half-line? – David House Sep 4 '15 at 20:22
• Take a full line, and a point $P$ on it. A half-line is the set of points on the line that are on one side of $P$, possibly including $P$. Big further hint: The point $z=1+i$, which we can think of as $(1,1)$, has arg equal to $\pi/4$. – André Nicolas Sep 4 '15 at 20:26

If you view $\mathbb{C}$ as $\mathbb{R}^2$, then $-pi<arg(z)<\frac{pi}{2}$ is the 2nd quadrant.
$arg(z) = \frac{pi}{4}$ is the ray originating from (0,0) and going towards $(\infty,\infty)$ i.e. following the line y=x in the first quadrant.
for $-\pi<\arg(z)<\frac{\pi}{2}$ you need to shade all the argand diagram except for the second quadrant where $\frac{\pi}{2}<\arg(z)<\pi$. Exclude the positive imaginary axis and the negative real axis.
$\arg(z)=\frac{\pi}{4}$ is a part line equivalent to the line $y=x$ in the first quadrant only and excluding the origin.