Complex numbers

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dionys 101 Oct 14, 2004 at 00:07

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:

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NomadRock 101 Oct 14, 2004 at 04:42

|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

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NeZbiE 101 Oct 14, 2004 at 04:52

NomadRock>> He’s talking about complex numbers =)

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

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NeZbiE 101 Oct 14, 2004 at 04:58

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 =)

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dionys 101 Oct 14, 2004 at 06:41

yes
z=x+jy

thanks a lot :)

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NomadRock 101 Oct 14, 2004 at 15:48

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

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Per_Vognsen 101 Oct 18, 2004 at 01:26

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.

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Mario 101 Nov 05, 2004 at 21:15

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.

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NeZbiE 101 Nov 05, 2004 at 22:30

Ha, I eat quaternions for breakfast.

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SigKILL 101 Jan 22, 2005 at 12:17

@Per Vognsen

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.

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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