Solving Complex Numbers: Sketching the Line |z − u| = |z|

[kiwi101]

Homework Statement

Sketch the line described by the equation:
|z − u| = |z|

z = x+jy
u = −1 + j√3

The Attempt at a Solution

(x+1)^2 + j(y-√3)^2 = (x+jy)^2



I just don't quite get where to go with this
please give me a headstart

 

[Dick]

Homework Statement

Sketch the line described by the equation:
|z − u| = |z|

z = x+jy
u = −1 + j√3

The Attempt at a Solution

(x+1)^2 + j(y-√3)^2 = (x+jy)^2



I just don't quite get where to go with this
please give me a headstart

|z|^2=(x^2+y^2). There is no j in there. There shouldn't be any j in the left hand side either. It's an absolute value.

 

[kiwi101]

oh yeah
and um did you mean |z|^2=(x+y)^2?

 

[Dick]

oh yeah
and um did you mean |z|^2=(x+y)^2?

Definitely not! |z|=sqrt(x^2+y^2). Look it up.

 

[Mark44]

|z|^2=(x^2+y^2). There is no j in there. There shouldn't be any j in the left hand side either. It's an absolute value.

oh yeah
and um did you mean |z|^2=(x+y)^2?

If z = x + iy, then |z|² = x² + y², which is what Dick wrote. kiwi101, it looks like you need to review the definition of the absolute value or magnitude of a complex number.

 

[kiwi101]

I was just about to write a long argument about how I was right and then I realized you're right. I misinterpreted something.

So this is what I have done, I feel its right.

(x+1)^2 + (y-√3)^2 = x^2 + y^2

x^2 + 2x + 1 + y^2 -2√3y + 3 = x^2 + y^2

2x + 1 -2√3y + 3 = 0

(x+2)/√3 = y

and then rationalize it and this is the equation of the line?

 

[Dick]

I was just about to write a long argument about how I was right and then I realized you're right. I misinterpreted something.

So this is what I have done, I feel its right.

(x+1)^2 + (y-√3)^2 = x^2 + y^2

x^2 + 2x + 1 + y^2 -2√3y + 3 = x^2 + y^2

2x + 1 -2√3y + 3 = 0

(x+2)/√3 = y

and then rationalize it and this is the equation of the line?

That looks ok to me.

 

[kiwi101]

Thanks!
Out of curiosity this question:

Sketch the line or curve described by the equation
ℜe{z} + ℑm{z} = ℜe{u}

would be x + jy = -1 ?

so is this a line or what?

 

[kiwi101]

Wait do I solve for y and then rationalize like

y = (-1 -x)/j

 

[Dick]

Wait do I solve for y and then rationalize like

y = (-1 -x)/j

Im(z)=y, not jy. Check the definition again.

 

[kiwi101]

I just assumed that since it says Im(z) it meant to include the imaginary iota.

So then I guess it is just a line y = -1-x

 

[Dick]

I just assumed that since it says Im(z) it meant to include the imaginary iota.

So then I guess it is just a line y = -1-x

Yep!

 

[kiwi101]

Thanks once again! :)