Sketch the sets Which are domains Describe the boundary

[stunner5000pt]

For the following open sets
a) |Arg z| <pi/4
b) -1 < I am 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 < I am z <= 1
shaded part of 2
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

 

[d_leet]

For the following open sets
a) |Arg z| <pi/4

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??

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?

 

[stunner5000pt]

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?

it is Arg
what do u mean branch cut??

 

[d_leet]

it is Arg
what do u mean branch cut??

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

 

[stunner5000pt]

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

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

 

[d_leet]

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

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

 

[HallsofIvy]

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

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 < I am z <= 1
shaded part of 2
yes it is a domain
the region bounded by y = -1 from the bottom and unbounded above"

Why unbounded above? It say "Im z<= 1".

"c) (Re z)^2 > 1"

In other words, Re z> 1 or Re z< -1.