Complex numbers domain

  • Thread starter DottZakapa
  • Start date
  • #1
210
13

Homework Statement:

i have to find this domain

Relevant Equations:

complex numbers
i have to find such domain
z=x+iy , y,x∈ℝ , |y-x|<=2, |x|<=2

i'm confused with |y-x|<=2, how should i proceed ? with abs of x i am ok.
 

Answers and Replies

  • #2
13,457
10,516
Have you tried to write ##x,y## is polar coordinates? Or start with reals: What does ##|x|<2## mean for real numbers?
 
  • #3
210
13
##e^{-x}*e^{iy} = e^{-x+iy}## this?
 
  • #4
13,457
10,516
##e^{-x}*e^{iy} = e^{-x+iy}## this?
No. You have to resolve the norm, so you must represent the complex numbers ##x,y## somehow. Do you know how we can write them, in order to calculate ##|x|##?
 
  • #5
210
13
i know that |z| = ##\sqrt {x^2+y^2}##
but i don't know how to handle this |y-x|
 
Last edited:
  • #6
13,457
10,516
i know that |z| = ##\sqrt {x^2+y^2}##
but i don't know how to handle this |y-x|
O.k. Polar coordinates would have been ##z=r\cdot e^{i\varphi}## and ##|z|=r##.
Anyway. In order to write ##|z| = \sqrt {x^2+y^2}## you had to write ##z=x+iy## first. This was a bad choice, as ##x,y## will be needed as complex numbers, so the letters are already occupied. You should have written ##z=a+ib## and ##|z|=\sqrt{a^2+b^2}.##

Now we have two complex variables ##x,y##. We first write them as ##x=a+ib## and ##y=c+id##. Can you calculate ##x-y## and ##|x-y|##? And you still have to answer post #2: Which domain is ##|a|<2## if ##a## is real?
 
  • #7
210
13
so if x = a+ib and y= c+id then
y-x would be (c-a)+i(d-b) then the |y-x|= ##\sqrt {(c-a)^2+(d-b)^2}##

then
##\sqrt {(c-a)^2+(d-b)^2}<=2##=> ##(c-a)^2+(d-b)^2<=4##
 
  • #8
13,457
10,516
so if x = a+ib and y= c+id then
y-x would be (c-a)+i(d-b) then the |y-x|= ##\sqrt {(c-a)^2+(d-b)^2}##

then
##\sqrt {(c-a)^2+(d-b)^2}<=2##=> ##(c-a)^2+(d-b)^2<=4##
Yes, but you must be careful with squaring inequalities. You cannot go back easily. That's why I asked for the meaning of ##|a|<2##.
 
  • #9
210
13
a is comprised between -2 and 2. isn't it?
 
  • #10
13,457
10,516
Yes, and the same is true if ##Z=a+ib## is complex. However, the complex norm is defined to be positive, so here we get for ##z=a+ib## from ##|z|=|a+ib|=\sqrt{a^2+b^2}## that ##0\leq \sqrt{a^2+b^2}<2## must hold. Try to draw the points ##\{(a,b)\in \mathbb{R}\}^2\,|\,0\leq\sqrt{a^2+b^2}<2## in a coordinate system with axis ##a## and ##b##. And then figure out where ##z## has to be, if the coordinates were ##a## and ##ib##.
 
  • #11
pasmith
Homework Helper
1,818
483
This was a bad choice, as ##x,y## will be needed as complex numbers, so the letters are already occupied. You should have written ##z=a+ib## and ##|z|=\sqrt{a^2+b^2}.#
I disagree. [itex]x \in \mathbb{R}[/itex] and [itex]y \in \mathbb{R}[/itex] are defined by the OP's question as being the real and imaginary parts of [itex]z \in \mathbb{C}[/itex], and conditions are then placed upon [itex]x[/itex] and [itex]y[/itex] in order to restrict [itex]z[/itex] to a particular subset of [itex]\mathbb{C}[/itex]. That is the natural way to interpret
i have to find such domain
z=x+iy , y,x∈ℝ , |y-x|<=2, |x|<=2
So the OP is looking for [itex](x,y) \in \mathbb{R}^2[/itex] such that [itex]|y - x| \leq 2[/itex] (and [itex]|x| \leq 2[/itex], but they already know how to deal with that).

So start with something easier. Take [itex]C \in \mathbb{R}[/itex] such that [itex]|C| \leq 2[/itex] and consider [itex]y - x = C[/itex]. What does that look like?
 
  • Like
Likes Delta2 and DottZakapa
  • #12
210
13
I disagree. [itex]x \in \mathbb{R}[/itex] and [itex]y \in \mathbb{R}[/itex] are defined by the OP's question as being the real and imaginary parts of [itex]z \in \mathbb{C}[/itex], and conditions are then placed upon [itex]x[/itex] and [itex]y[/itex] in order to restrict [itex]z[/itex] to a particular subset of [itex]\mathbb{C}[/itex]. That is the natural way to interpret


So the OP is looking for [itex](x,y) \in \mathbb{R}^2[/itex] such that [itex]|y - x| \leq 2[/itex] (and [itex]|x| \leq 2[/itex], but they already know how to deal with that).

So start with something easier. Take [itex]C \in \mathbb{R}[/itex] such that [itex]|C| \leq 2[/itex] and consider [itex]y - x = C[/itex]. What does that look like?
i did try in such way, but the results of the instructor did not coincide. Going forward in the video, he made a mistake and corrected it. So problem sorted. Is like you say. thanks to anyone :thumbup:
 
  • Like
Likes WWGD
  • #13
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,259
5,836
The region |y-x|<2 is clearly bounded by the lines y=x+2 and y=x-2, so sketch those. Superimpose the bounds for |x|<2.
 
  • Like
Likes Delta2 and DottZakapa
  • #14
210
13
thank you, as said above i have sorted the problem. the adopted procedure before posting the problem here was the same as @pasmith and you proposed. A dubdt raised for a discrepancy between my obtained result and the one of the instructor, which later on revealed to be a calculation mistake of the instructor.
 

Related Threads on Complex numbers domain

Replies
3
Views
2K
  • Last Post
Replies
1
Views
879
Replies
1
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
16
Views
4K
Replies
2
Views
911
  • Last Post
Replies
3
Views
926
  • Last Post
Replies
24
Views
6K
Top