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

In summary, the conversation was about sketching a line described by the equation |z − u| = |z| and finding the equation of the line. The resulting equation was (x+2)/√3 = y. Later, the conversation discussed another equation and concluded that it represents a line y = -1-x.
  • #1
kiwi101
26
0

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
 
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  • #2
kiwi101 said:

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.
 
  • #3
oh yeah
and um did you mean |z|^2=(x+y)^2?
 
  • #4
kiwi101 said:
oh yeah
and um did you mean |z|^2=(x+y)^2?

Definitely not! |z|=sqrt(x^2+y^2). Look it up.
 
  • #5
Dick said:
|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 said:
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.
 
  • #6
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?
 
  • #7
kiwi101 said:
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.
 
  • #8
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?
 
  • #9
Wait do I solve for y and then rationalize like

y = (-1 -x)/j
 
  • #10
kiwi101 said:
Wait do I solve for y and then rationalize like

y = (-1 -x)/j

Im(z)=y, not jy. Check the definition again.
 
  • #11
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
 
  • #12
kiwi101 said:
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!
 
  • #13
Thanks once again! :)
 

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

1. What are complex numbers?

Complex numbers are numbers that are written in the form a + bi, where a and b are real numbers and i is the imaginary unit equal to the square root of -1.

2. How do you sketch the line |z − u| = |z|?

To sketch the line |z − u| = |z|, first rewrite the equation as |z − u| = |z − 0|. This means that the distance from the complex number z to the complex number u is equal to the distance from z to 0. This creates a circle centered at u with radius equal to the distance from u to 0. The line |z − u| = |z| is the boundary of this circle.

3. What does the line |z − u| = |z| represent?

The line |z − u| = |z| represents all the complex numbers that are equidistant from the complex number u and the origin (0,0). It is a circle with center at u and radius equal to the distance from u to 0.

4. How many solutions does the equation |z − u| = |z| have?

This equation has infinitely many solutions. Any complex number z that is equidistant from u and 0 satisfies the equation. This means that there is a continuum of solutions along the circle, rather than a finite number of solutions.

5. Can you use the Pythagorean Theorem to solve this equation?

Yes, the Pythagorean Theorem can be used to solve this equation. Since the equation represents a circle, the distance from z to u and the distance from z to 0 can be represented as the legs of a right triangle, with the distance between u and 0 as the hypotenuse. Using the Pythagorean Theorem, you can find the value of z that satisfies the equation.

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