Question 1152633
<pre>
{{{Conjugate(z) - 12/z = 1-i}}}

Let z = x+yi, where x and y are real numbers

{{{Conjugate(x+yi) - 12/(x+yi) = 1-i}}}

{{{(x-yi) - (12/(x+yi))((x-yi)/(x-yi)) = 1-i}}}

{{{x-yi - (12x-12yi^"")/(x^2-y^2i^2) = 1-i}}}

i² = -1

{{{x-yi - (12x-12yi^"")/(x^2-y^2(-1)) = 1-i}}}

{{{x-yi - (12x-12yi^"")/(x^2+y^2) = 1-i}}}

Put parentheses around the fraction term:

{{{x-yi - ((12x-12yi^"")/(x^2+y^2)) = 1-i}}}

Break the fraction term into two fractions:

{{{x-yi - ((12x^"")/(x^2+y^2)-(12yi^"")/(x^2+y^2)) = 1-i}}}

Remove the parentheses:

{{{x-yi - (12x^"")/(x^2+y^2)+(12yi^"")/(x^2+y^2) = 1-i}}}

In an equation with both real and imaginary terms, the real
and imaginary terms are like oil and water. The real terms
on the left equals the real terms on the right, and the 
imaginary terms on the left equal the imaginary terms on the
right.  So we split it into two equations.

The REAL equation is:

{{{x- (12x^"")/(x^2+y^2) = 1}}}

The IMAGINARY equation is

{{{-yi+(12yi^"")/(x^2+y^2) = -i}}}

which we divide through by -i

{{{y-(12y^"")/(x^2+y^2) = 1}}}

So we have the system of equations:

{{{system(x- (12x^"")/(x^2+y^2) = 1,y-(12y^"")/(x^2+y^2) = 1)}}}

The equations are symmetrical in x and y.  So we can substitute 
x for y.  

{{{x- (12x^"")/(x^2+y^2) = 1}}}

Replace y by x:

{{{x- (12x^"")/(x^2+x^2) = 1}}}

{{{x- (12x^"")/(2x^2) = 1}}}

{{{x- 6/x = 1}}}

{{{x^2-6 = x}}}

{{{x^2-x-6=0}}}

{{{(x-3)(x+2)=0}}}

x - 3 = 0; x + 2 = 0
    x = 3;     x = -2

So there are two solutions: (x,y) = (3,3), (x,y) = (-2,-2)

Expressing in a+bi form:

z = x+yi = 3+3i   and  z = x+yi = -2-2i

Edwin</pre>