Expert: Josh - 3/3/2009

Question
(x+iy)/(x-iy)=i
|1-(x+iy)|=x+iy
FIND X n Y

Answer
Using Euler's notation, let x+iy=R*exp(iq).
Here, q represents the argument, R*exp(iq)=R[cos(q)+i*sin(q)].
Per usual, i=sqrt(-1).

(x+iy)/(x-iy)=i becomes
[R*exp(iq)]/[R*exp(-iq)]=exp(i((pi/2)+2*k*pi)) for any integer k
...since phasor is 2*pi periodic.
...with the R's canceling on the LHS,
exp(i*2q)=exp(i((pi/2)+2*k*pi))
Equating the angles,
2q = (pi/2)+2*k*pi
q = (pi/4)+k*pi = pi*(1/4+k) for any integer k.

Using the result z(z*)=|z|^2, where z* is the conjugate of z,
squaring both sides of |1-(x+iy)|=x+iy gives
[1-R*exp(iq)][1-R*esp(-iq)]=R^2*exp(i2q)
1-R(2cos(q))+R^2=R^2[cos(2q)+i*sin(2q)] ...[#1]

The imaginary term on the RHS has to be zero, since LHS is real.
Therefore, 2q=pi*k, q=pi(0.5*k).

Rewriting [#1],
with cos(q)=cos(k(pi/2))=0 for odd k
                       =1 for even k=0,4,...
                       =-1 for even k=2,6,...
and cos(2q)=cos(pi*k)=-1 for odd k
                    =1  for even k
case 1: for odd k
1-2R*(0)+R^2=-R^2
R^2=-1/2 has no real solution

case 2: for k=0,4,...
1-2R*(1)+R^2=R^2
R=1/2

case 3: for k=2,6,...
1-2R*(-1)+R^2=R^2
R=-1/2 (not feasible solution, as we require real value R >= 0)
Alternately, rotate solution vector by pi radian, which gets us back the proper solution obtained from case 2.

quick check: R=0.5; q=pi*0;
let n=x+iy=R*(cos(q)+i*sin(q))
n=0.5
LHS: |1-n|^2=(1-n)*conj(1-n)=0.25
RHS: R^2=0.25