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



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

Is z something to do with z=a+bi ? Yes, this is the definition of a complex number.is a=Re(z) ? ?In the quote above, yes. But you are mixing variables. Rather, write z = x + yi and note that your questioner says that "a is in the same (i.e. complex) plane" That is, you may assume that a takes the same form as z (of course, you may have 1 or more zeros in there!). I think you may also assume that b also takes the same form.

Are you sure you know what the notation |...| means in this context?

I am not sure if "z" in this task is same with z=a+bi

As far as I know, |z|=r and not |z-a|=r

As I said earlier; you are confusing yourself by using the variable "a" in different contexts. you are given, say, z = x + yi. You are also given that a = d + ci. Note this last equality well, this where you are going astray!

Your task (an easy one) is to show that |z| is real and |a| is real and that |z| - |a| =|z - a| and also (positive) real.

Now do the same for b = p + qi where b not equal a. Fiddle around a bit, and see......

So "a" is just real number, right? Why I need to prove |z| - |a| =|z - a| ?

a is a complex number. first you try to do it in two dimensional plane: Think of z,a, and b as points on a plane, and r is some positive real number, |x-y| is the distance between the points x and y on the plane.

So "a" is just real number, right? Why I need to prove |z| - |a| =|z - a| ?No, no and no again! Read your own opening post, dammit!

Let me remind you.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, ?So what do you think the "a" might be? Real or complex? What plane is "a" in? Hint: you just said it was in the complex plane

Or maybe you don't understand what the complex plane is? If not, I strongly advise some more reading. Of course, someone here may be able to help, if you are confused about this.

Hi StMartin,
This is hard to explain without diagrams.
Here's a resource that might help:
Intuitive arithmetic with complex numbers (http://betterexplained.com/articles/intuitive-arithmetic-with-complex-numbers/)


The key concept in this problem is understanding what |z-a| means (where z and a are complex numbers). Look at the "Magnitude" and "Complex Addition" sections in that link, and see if you can figure it out from there. When you understand it, finding the answer will be trivial.

I think this is a problem you really need to figure out on your own... when you grasp the concepts involved, the problem will be easy. If you don't grasp those concepts on your own, going any further will complex numbers will be very hard.

Ohh... I understand now, both z and a are complex numbers... Why I need to show that |z| is real and |a| is real and that |z| - |a| =|z - a| and also (positive) real? How do you know that I must show that?

Ohh... I understand now, both z and a are complex numbers...
Right!
Why I need to show that |z| is real and |a| is real and that |z| - |a| =|z - a| and also (positive) real? How do you know that I must show that?

No. |z - a| is not |z| - |a|.
What is z - a?
What is the magnitude of a complex number?

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?

So in this case I got equation of circular right?

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?

Well done!
So,
a) Describes a circle with centre at point a, and radius r.

What about b)?

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?

For example. AB=AD