Complex circle

[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=[itex]\sqrt{2}[/tex]

$|z|=r \leq \sqrt{2}$

out from $r \leq \sqrt{2}$

and |z-a|=r

$|z-a| \leq \sqrt{2}$

The interior part is $r \leq \sqrt{2}$

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:



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

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.