0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Sep 4 '15 at 20:18
  • $\begingroup$ What is a half-line? $\endgroup$ – David House Sep 4 '15 at 20:22
  • $\begingroup$ 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 $\pi/4$. $\endgroup$ – André Nicolas Sep 4 '15 at 20:26
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your input. I will sketch that then. $\endgroup$ – David House Sep 4 '15 at 21:34
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.