Fun with sets

1. Mar 20, 2006

stunner5000pt

For the following open sets
a) |Arg z| <pi/4
b) -1 < Im z <= 1
c) (Re z)^2 > 1

Sketch the sets
Which are domains
Describe the boundary

a) |Arg z |< pi/4

well ok that means
$$|\theta_{0} + 2 \pi k| < \pi/4$$
so the theta must alwasy be less than pi/4
im not sure how to sketch this... i mthinking its the attached diagram, but only the triangle in the first quadrant
yes it is a domain
umm is there a way to do with without diagram??

b) -1 < Im z <= 1
yes it is a domain
the region bounded by y = -1 from the bottom and unbounded above

c) (Re z)^2 > 1
like 3?
not a domain
unbounded

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2. Mar 20, 2006

d_leet

Is this the principal argument: Arg as opposed to arg? If so then where is the branch cut for Arg? And what does that absolute value sign do?

3. Mar 20, 2006

stunner5000pt

it is Arg
what do u mean branch cut??

4. Mar 20, 2006

d_leet

How is Arg z (the principal argument of z ) defined?

5. Mar 20, 2006

stunner5000pt

the branct cut is at pi/2 for the principle argument yes??

6. Mar 21, 2006

d_leet

No... And if it were the question posed to you would be somewhat impossible.

7. Mar 21, 2006

HallsofIvy

Staff Emeritus
$$|Arg z|< \frac{\pi}{4}$$

It really doesn't matter whether or not "Arg" means the principal argument: since it's between $-\frac{/pi}{4}$ and $\frac{\pi}{4}$, it IS, except for the negative values, the principal argument!
Stunner5000pt, it's not just in the first quadrant- since this is absolute value, you can have negative values of the argument and so include the fourth quadrant.

"b) -1 < Im z <= 1