# Complex numbers

9 replies to this topic

### #1dionys

New Member

• Members
• 6 posts

Posted 14 October 2004 - 12:07 AM

Hi guys .

Can anyone help me to solvethis equation algebraically plz

note : z=x+yj or in cartesian coordinates z=|z|(cosθ+jsinθ)

|z+2|=|z-1| :wink:

Senior Member

• Members
• 785 posts

Posted 14 October 2004 - 04:42 AM

|z+2| = z-1 or -z+1
so
z+2 = z-1 ==> 2 = -1 (clearly not the right solution)
or
-z-2 = z-1 ==> -2z = 1 ==> z = -1/2
or
z+2 =-z+1 ==> 2z = -1 ==> z = -1/2
or
-z-2 = -z+1 ==> -2 = 1 (also clearly not a solution)

Therefore the solution is z = -1/2
Jesse Coyle

### #3NeZbiE

Member

• Members
• 61 posts

Posted 14 October 2004 - 04:52 AM

By using |z| you mean the complex norm, right? I'm used to different notation ;)

### #4NeZbiE

Member

• Members
• 61 posts

Posted 14 October 2004 - 04:58 AM

Alright, assuming you are talking about complex norms, here goes:

Let z = a+b*i

then |z-1| = root((a-1)^2 +b^2)
and |z+2| = root((a+2)^2 + b^2)

That is, we need to find solutions to (a-1)^2 = (a+2)^2 to be able to find valid complex numbers fitting the |z-1| = |z+2| property.

So we have:
(a-1)^2 = (a+2)^2
a^2 - 2*a +1 = a^2 + 4a +4
a = -1/2

Therefore, complex numbers fitting the solution should be all numbers in the form:
-1/2 + b*i, where b is a real number.

Hope this helps =)

### #5dionys

New Member

• Members
• 6 posts

Posted 14 October 2004 - 06:41 AM

yes
z=x+jy

thanks a lot :)

Senior Member

• Members
• 785 posts

Posted 14 October 2004 - 03:48 PM

Oh hell, I completely missed that. That's what I get for posting after many hours of coding.
Jesse Coyle

### #7Per Vognsen

New Member

• Members
• 8 posts

Posted 18 October 2004 - 01:26 AM

Slightly different approach that easily generalizes to the complex vector case:

We want to solve for z such that

|z-1| = |z+2|

Square both sides:

|z-1|^2 = |z+2|^2

Rewrite using conjugates:

(z-1)(conj(z)-1) = (z+2)(conj(z)+2)
z conj(z) - conj(z) - z + 1 = z conj(z) + 2 conj(z) + 2 z + 4

The z conj(z) terms cancel each other, leaving us with

-3(conj(z) + z) = 3
conj(z) - z = -1

Note that conj(z) - z = 2 Re(z) so we want 2 Re(z) = -1. Thus Re(z) = -1/2. There are no restrictions on Im(z) so the general solution is z = -1/2 + r i for any real number r.

### #8Mario

New Member

• Members
• 5 posts

Posted 05 November 2004 - 09:15 PM

Heh... this was simple for complex numbers. Have we any volunteer for solving this equation in quaternion algebra? Did I say quaternion? Octionian algebra would be good enough.

### #9NeZbiE

Member

• Members
• 61 posts

Posted 05 November 2004 - 10:30 PM

Ha, I eat quaternions for breakfast.

### #10SigKILL

Valued Member

• Members
• 200 posts

Posted 22 January 2005 - 12:17 PM

Per Vognsen said:

Slightly different approach that easily generalizes to the complex vector case:

We want to solve for z such that

|z-1| = |z+2|

Square both sides:

|z-1|^2 = |z+2|^2

Rewrite using conjugates:

(z-1)(conj(z)-1) = (z+2)(conj(z)+2)
z conj(z) - conj(z) - z + 1 = z conj(z) + 2 conj(z) + 2 z + 4

The z conj(z) terms cancel each other, leaving us with

-3(conj(z) + z) = 3
conj(z) - z = -1

Note that conj(z) - z = 2 Re(z) so we want 2 Re(z) = -1. Thus Re(z) = -1/2. There are no restrictions on Im(z) so the general solution is z = -1/2 + r i for any real number r.

You're just almost right, -3(conj(z)+z) = 3 => conj(z)+z = -1. But you get the right answer because you state that conj(z)-z = 2 Re(z) which is incorrect (conj(z)-z = -2 Im(z)).

-Si

#### 1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users