Finding Domain for Complex Numbers: |y-x|<=2, |x|<=2

[DottZakapa]

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.

 

[fresh_42]

Have you tried to write $x,y$ is polar coordinates? Or start with reals: What does $|x|<2$ mean for real numbers?

 

[DottZakapa]

$e^{-x}*e^{iy} = e^{-x+iy}$ this?

 

[fresh_42]

$e^{-x}*e^{iy} = e^{-x+iy}$ 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 $|x|$?

 

[DottZakapa]

i know that |z| = $\sqrt {x^2+y^2}$
but i don't know how to handle this |y-x|

 

[fresh_42]

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?

 

[DottZakapa]

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
$\sqrt {(c-a)^2+(d-b)^2}<=2$=> $(c-a)^2+(d-b)^2<=4$

 

[fresh_42]

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 $|a|<2$.

 

[DottZakapa]

a is comprised between -2 and 2. isn't it?

 

[fresh_42]

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 $ib$.

 

[pasmith]

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. $x \in \mathbb{R}$ and $y \in \mathbb{R}$ are defined by the OP's question as being the real and imaginary parts of $z \in \mathbb{C}$, and conditions are then placed upon $x$ and $y$ in order to restrict $z$ to a particular subset of $\mathbb{C}$. 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 $(x,y) \in \mathbb{R}^2$ such that $|y - x| \leq 2$ (and $|x| \leq 2$, but they already know how to deal with that).

So start with something easier. Take $C \in \mathbb{R}$ such that $|C| \leq 2$ and consider $y - x = C$. What does that look like?

 

[DottZakapa]

I disagree. $x \in \mathbb{R}$ and $y \in \mathbb{R}$ are defined by the OP's question as being the real and imaginary parts of $z \in \mathbb{C}$, and conditions are then placed upon $x$ and $y$ in order to restrict $z$ to a particular subset of $\mathbb{C}$. That is the natural way to interpretSo the OP is looking for $(x,y) \in \mathbb{R}^2$ such that $|y - x| \leq 2$ (and $|x| \leq 2$, but they already know how to deal with that).

So start with something easier. Take $C \in \mathbb{R}$ such that $|C| \leq 2$ and consider $y - x = C$. 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

 

[haruspex]

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.

 

[DottZakapa]

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.