# complex equation

• Nov 21st 2012, 01:27 PM
jakobjakob
complex equation
Im: (z+i)/(1-z) = 1
|z|=1

What i did so far:

im: (a+bi+i)(1-a+bi)/(1-a-bi)(1-a+bi)=1
im: (a-b-(a^2)-(b^2)-ai+bi+i)/((a^2)+(b^2)-2a+1)=1

-a+b+1=(a^2)+(b^2)-2a+1
(a^2)+(b^2)-a-b=0
From here on forward i'm not sure how to proceed because i don't know how to implement the |z|=1 in the equation.What i did is this but i'm fairly certain it's not correct:
(a-(1/2))^2+((b-(1/2))^2 - 1/2 = 0
(a-(1/2))^2+((b-(1/2))^2 = (sqr.root 2/2)^2
My assumption for the solution is: all points on radius of (sqr.root 2/2)^2 with center at (1/2,1/2)

Would appreciate your help. Thank you
• Nov 21st 2012, 02:23 PM
Plato
Re: complex equation
Quote:

Originally Posted by jakobjakob
Im: (z+i)/(1-z) = 1
|z|=1
What i did so far:
im: (a+bi+i)(1-a+bi)/(1-a-bi)(1-a+bi)=1
im: (a-b-(a^2)-(b^2)-ai+bi+i)/((a^2)+(b^2)-2a+1)=1

The numerator is $(a-a^2-b^2-b)+i(b-a+1)$

The denominator is $(1-a)^2+b^2$
• Nov 21st 2012, 04:43 PM
Deveno
Re: complex equation
here is my solution, which avoids "a's and b's".

Given:

$\text{Im}\left(\frac{z+i}{1-z}\right) = 1, |z| = 1$.

the second equation tells us $zz^* = 1$.

thus: $z^* = \frac{1}{z}$.

now:

$\text{Im}\left(\frac{z+i}{1-z}\right) = \frac{1}{2i}\left[\frac{z+i}{1-z} - \left(\frac{z+i}{1-z}\right)^*\right]$

$= \frac{1}{2i}\left[\frac{z+i}{1-z} - \frac{z^*-i}{1-z^*}\right]$

$= \frac{1}{2i}\left[\frac{z+i}{1-z} - \frac{\frac{1}{z}-i}{1-\frac{1}{z}}\right]$ (since z* = 1/z)

the fraction we are subtracting is a bit ugly, so let's simplify it, first:

$\frac{\frac{1}{z}-i}{1-\frac{1}{z}} = \frac{\frac{1-iz}{z}}{\frac{z-1}{z}} = \frac{1-iz}{z-1} = \frac{iz - 1}{1-z}$

oh look, we have a common denominator, now. so:

$\frac{1}{2i}\left[\frac{z+i}{1-z} - \frac{\frac{1}{z}-i}{1-\frac{1}{z}}\right] = \frac{1}{2i}\left[\frac{z+i-iz+1}{1-z}\right]$

now 1/i = -i. so we can replace 1/(2i) with -i/2. so we have:

$\frac{1}{2i}\left[\frac{z+i-1+iz}{1-z}\right] = \frac{(-i)(z+i-iz+1)}{2-2z} = \frac{z(-1-i) + (1-i)}{2-2z}$

remember, this whole mess is the imaginary part of (z+i)/(1-z), and is equal to 1. thus:

$z(-1-i) + (1-i) = 2 - 2z$. this equation is linear in z, so we most certainly can solve it (over C). put all the z terms on one side, and the non-z terms on the other:

$z(-1-i+2) = 2-1+i$. this easily simplifies to:

$z = \frac{1+i}{1-i}$.

but....we're not done, yet. let's simplify the fraction on the right:

$\frac{1}{1-i} = \frac{1+i}{2}$, so:

$\frac{1+i}{1-i} = \frac{(1+i)^2}{2}$.

but:

$(1+i)^2 = 1 + 2i + i^2 = 2i$.

therefore:

$z = \frac{2i}{2} = i$.

verify that this indeed is a solution.

(with all due respect for Plato, his method leads to one "false" solution, which must be eliminated: z = 1. this is due to "squaring introducing extraneous solutions").
• Nov 21st 2012, 04:43 PM
jakobjakob
Re: complex equation
I think i figured it out, but i would appreciate some feedback regarding correctness.

(b-a+1)/((1-a)^2 + b^2) = 1 for all (a,b) =/= (1,0) (denominator can't be 0)

b-a+1=(1-a)^2 + b^2
a^2 + b^2 - a - b = 0

|z|=1 = sqr.root (a^2 + b^2 )
1=sqr.root (a^2 + b^2 ) /^2
1=a^2 + b^2

a^2 + b^2 - a - b = 0
1 - a - b = 0
b = 1 - a
b = 1 - sqr.root(1 - b^2)
sqr.root(1 - b^2)= 1 - b /^2
1 - b^2 = 1 -2b + b^2
2b^2 - 2b = 0
b^2 - b = 0
b(b - 1) = 0, b can't be 0 so the only solution is
b = 1
1 - a - b = 0
1 - a - 1 = 0
a = 0

Solution:
z=i
Could this be correct?
Edit: Ignore this post. I haven't seen Deveno's explanation until after i posted. Thank you Deveno.
• Nov 21st 2012, 05:36 PM
Deveno
Re: complex equation
your solution is fine: you correctly identified the "extraneous solution" by realizing that b = 0 implies z = 1, which would make (z+i)/(1-z) undefined (and 1 certainly ISN'T undefined).

the point is, we only have ONE possible z that works. i prefer to show such things by working with a single variable (z alone) rather than two (a and b), to avoid such difficulties (although i will admit, the algebra can get hairy this way).