# Complex numbers

## Homework Statement

Lets find the set of points z in the complex plane which satisfy the condition:

a)|z-a|=r , r>0 where a is fixed point from the same plane, and r is positive real number.

b)|z-a|=|z-b|, $a \neq b$

## The Attempt at a Solution

I don't know how to solve this at all. Is z something to do with z=a+bi ? is a=Re(z) ? ?

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malawi_glenn
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z=x+yi, where x is coordinate on real axis and y coordinate on imaginary axis.

Then you have to know how |z'| is defined (z' is an arbitrary complex number)

tiny-tim
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… where a is fixed point from the same plane …

… is a=Re(z) ? ?
Hi Physicsissuef! Nooo … a is another "complex number", exactly like z. (and so is b)

malawi_glenn
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Gib Z
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Take a geometric interpretation of the problem; a is some point on the Argand Plane, |z-a| is the distance between the satisfying points z and the point a, and this distance must be exactly r units. Does this remind you of any shape =] ?

We can use a geometric approach for the second one as well: "Find all points where the distance from point a is equal to the distance from point b". Good luck.

Do you have some diagram which shows |z-a|?
Lets say z=x+yi and a=c+di then we can represent z and a as points.

z(x,y)
a(c,d)

So |z-a|=$$\sqrt{(x-c)^2+(y-d)^2}$$, right?

Gib Z
Homework Helper
Well that definitely is true. By doing that, we change back to Cartesian co-ordinates, and now that weve done that : You know that expression is equal to r. Square both sides. Does this look familiar =] ?

dx
Homework Helper
Gold Member
Yes thats right. Now you have an expression for |z-a|, and you know its supposed to be equal to r. If you equate |z-a| to r, you get an equation that must be satisfied by (x,y). Ofcourse, theres more than one (x,y) that satisfies this equation, so you have to figure out how to obtain the whole set of solutions. A good idea is to look at what this equation means geometrically. |z-a| is the distance from z to a in the complex plane. What is the locus of the set of z that are a distance r from a?

Ohhh.... Circular is $$r^2=(x-p)^2+(y-q)^2$$

I also got the same...
$$|z-a|=r=\sqrt{(x-c)^2+(y-d)^2}$$

$$r^2=(x-c)^2+(y-d)^2$$

So it is equation of circular right?

Gib Z
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Yes, it is a circle. Now just the easy questions of what its radius and center is.

And what about the second one?
b)|z-a|=|z-b|,
$a \neq b$

?

Gib Z
Homework Helper
Come on, you don't need to be baby stepped through both =] Again, think of them as distances from points a and b. Give it a good try yourself.

z(x,y)
a(c,d)
b(e,f)
$$|z-a|=\sqrt{(x-c)^2+(y-d)^2}$$
$$|z-b|=\sqrt{(x-e)^2+(y-f)^2}$$
$$\sqrt{(x-c)^2+(y-d)^2}=\sqrt{(x-e)^2+(y-f)^2}$$

Equal distance from one point?

Gib Z
Homework Helper
Well, yes...But you don't need to set it up like that! From the point of Euclidean geometry, can you tell me the locus of a point equidistant from two points?

tiny-tim
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$$\sqrt{(x-c)^2+(y-d)^2}=\sqrt{(x-e)^2+(y-f)^2}$$
Hi Physicsissuef! Assuming you're just being slow and careful, and wanting to prove things from the axiomatic definition of the complex plane rather than from the analogy with ordinary plane geometry …

When you have an equation with a square root on one side (or both sides, of course), just square the equation!
You'll find lots of things then cancel. Well, yes...But you don't need to set it up like that! From the point of Euclidean geometry, can you tell me the locus of a point equidistant from two points?
$$x=\frac{x_1+x_2}{2}$$

$$y=\frac{y_1+y_2}{2}$$

But isn't mine correect?

Gib Z
Homework Helper
Yes, that is ONE point that satisfies it =] Ok, connect the two points, and draw the perpendicular bisector, which will go through the midpoint ( the point you just mentioned). Draw some triangles, can you see that every point on this line will be equidistant from the two points as well =] ?

EDIT: O yes, as tiny-tim said, yours is correct, but it might be easier to understand it this way..

You mean like a symmetrical line ? It is same like mine...

Gib Z
Homework Helper
It is called the perpendicular bisector. Find the equation of that line, you should be set.

Ok, thank you very much for the help. Thanks for all of you, who helped me...