Complex Circle Solutions within Given Constraints

[Physicsissuef]

Homework Statement

What will be solutions of http://i26.tinypic.com/jj360y.jpg" complex circles.

Homework Equations

z=a+bi

The Attempt at a Solution

a)
1 \leq Re(z) \leq 2

1 \leq Im(z) \leq 2

What about z?

b)
-2 \leq Re(z) \leq 2

-2 \leq Im(z) \leq 2

What about z?

c) I don't know. I think |z|=\sqrt{2}

 

[DavidWhitbeck]

You are using rectangles or circles for all of them, one is a rectangle, one is an annulus and one is a disk. Let me help you with the non-rectangular ones.

An annulus centered at z_0=x_0+iy_0 with inner and outer radii r_1,r_2 should be described by r_1 \leq |z-z_0| \leq r_2. That should help you. A disk is just an annulus with r_1=0.

 

[rock.freak667]

c) I don't know. I think |z|=\sqrt{2}

remember that the equation for a circle in complex form is |z-a|=r where a is a fixed complex number in the form x_0 +iy_0 and r is the radius. You are correct that r=\sqrt{2}. Then centre of the circle is given by (x_0,y_0). You should now be able to get the proper equation for the circle.

Now to deal with the shaded region. If |z-a| \geq r then that means for the circle |z-a|=r, you would shade everything around the circle.

 

[Physicsissuef]

So |(x,y)-(1,1)| \geq \sqrt{2}.

But what are (x,y)?

z(x,y)

 

[rock.freak667]

You just put |z-(-1-i)| \geq \sqrt{2}

 

[Physicsissuef]

1) Why (-1-i)?

2)Shouldn't it be |z-a|=\sqrt{2}, since as we can see on the picture, it can't be neither lower nor bigger than r...

 

[rock.freak667]

1) Why (-1-i)?

Because in the form |z-a|=r, a is a fixed complex number in the form, a=x_0+iy_0 where the centre of the circle is (x_0,y_0). So to write the equation correctly, you must write a in that form.

2)Shouldn't it be |z-a|=\sqrt{2}, since as we can see on the picture, it can't be neither lower nor bigger than r...

|z-a|=r is just the circle alone with nothing shaded. Everything inside is the circle is shaded. So I think it would have to be such that |z-a| \leq r

 

[Physicsissuef]

I understand. But it should be |z-a| \geq r, right? So we will shade everything inside the circle... But in this case we can't define Re(z) and Im(z), since there are some negative values for Re(z) and Im(z)

 

[rock.freak667]

I understand. But it should be |z-a| \geq r, right? So we will shade everything inside the circle...

I think it would be \leq \sqrt{2} since the distance from the centre to any point on the circumference is \sqrt{2} if it was greater than or equal to \sqrt{2} then you would shade where the distance from the centre to any point is greater than or equal to \sqrt{2}

The inside of the circle is shaded. For all the points within that circle, the maximum distance is \sqrt{2} i.e. the point is on the circumference. Therefore for the shaded part, I think it would be |z-a| \leq r

 

[Physicsissuef]

But if we say like that, what about Re(z) and Im(z), they can receive negative values if we don't define them..

 

[rock.freak667]

But if we say like that, what about Re(z) and Im(z), they can receive negative values if we don't define them..

Well I guess you could restrict Re(z) and Im(z) if you wanted, but I think that leaving it in the circle form would still be correct.

 

[Physicsissuef]

but what about x=-1 and y=-1, you think it will be correct? I don't think so...

 

[Tedjn]

Isn't the center of the circle at 1+i?

 

[DavidWhitbeck]

Isn't the center of the circle at 1+i?

It's worse than that, way long ago they agreed that the region was the complement of a disk, when the picture clearly shows otherwise!

 

[Physicsissuef]

David can you please tell me your opinion?

 

[HallsofIvy]

I don't know if rock.freak667 didn't understand the the question or was just giving you a hard time but the answer he gave is completely wrong.
The last picture is the interior, together with the circle itself, of a disk with center at (1,1) and radius \sqrt{2}. Every point in the figure has distance from (1,1) less than or equal to \sqrt{2}. Written as a complex number, (1, 1) is 1+ i (NOT -1-i) and the distance from z to 1+ i is |z- (1+ i)|.

 

[rock.freak667]

I don't know if rock.freak667 didn't understand the the question or was just giving you a hard time but the answer he gave is completely wrong.
The last picture is the interior, together with the circle itself, of a disk with center at (1,1) and radius \sqrt{2}. Every point in the figure has distance from (1,1) less than or equal to \sqrt{2}. Written as a complex number, (1, 1) is 1+ i (NOT -1-i) and the distance from z to 1+ i is |z- (1+ i)|.

Seems today I keep confusing my + and - signs... but I was hoping that the general method was correct. Seems it was wrong, sorry OP.

 

[Physicsissuef]

Look if |z|=r
r=\sqrt{2}[/tex]<br /> <br /> |z|=r \leq \sqrt{2}<br /> <br /> out from r \leq \sqrt{2}<br /> <br /> and |z-a|=r<br /> <br /> |z-a| \leq \sqrt{2}<br /> <br /> The interior part is r \leq \sqrt{2}<br /> <br /> Is this correct?

 

[Tedjn]

Yes, that looks right. So what would you have for c)?

 

[Physicsissuef]

z=(x,y)

a=(1,1)

|(x,y)-(1,1)| \leq \sqrt{2}

|(x-1,y-1)| \leq \sqrt{2}

(x-1)^2+(y-1)^2 \leq 2

But what about x and y? Should we define them? For ex. x=-1 and y=-1 is not the part that we look for...

 

[HallsofIvy]

Looking back at your original question, I see that I have the direction of the inequality wrong (and rock.freak667 was right): you want |z-(1+i)|\ge \sqrt{2} not less than. x and y are variables any point (x,y) satisfying that inequality will be in the set shown.

 

[Physicsissuef]

HallsofIvy you were right. It is r \leq \sqrt{2}. Just look at the picture. The place which we should find (the circles) have radius less than \sqrt{2}...

 

[Physicsissuef]

All radius which are less then \sqrt{2} are the radius of the shadowed place..

Sp the final answer is:
<br /> (x-1)^2+(y-1)^2 \leq 2<br />

right?

 

[Physicsissuef]

HallsofIvy, somebody?

 

[DavidWhitbeck]

There is a subtle point here that HallsofIvy mentioned but you overlooked-- you are not in R^2. You are in the complex plane. You should write your sets of points in terms of one complex number instead of two real numbers. Just z instead of x and y.

 

[Physicsissuef]

I can write the complex number a=c + di as a point (c,d). And also |z-a|=r is circle with a (circle origin). I am asking just about r \leq \sqrt{2}, since the shadowed place is all places with radius less then \sqrt{2}

I have another question about |i|.

Is |i|=\sqrt{i^2}=i?

 

[Tedjn]

I believe the absolute value of a complex number is defined as its distance from the origin in the complex plane. If you have the complex number z = a + bi, then |z| = \sqrt{a^2+b^2}. Apply that to z = i. What is |z|?

Regarding your first question, I still think the radius should be \leq \sqrt{2}, but I could be wrong.

 

[Physicsissuef]

It will be |z|=|i|=\sqrt{0^2+1^2}=1, right?

 

[HallsofIvy]

"absolute value" is always a non-negative number. The absolute value of a+ ib is \sqrt{(a+ bi)(a- bi)}= \sqrt{a^2+ b^2}. Yes, |i|= 1.