Question 1073485
<pre>

let z = x + yi, where x and y are real

Substitute in

{{{z - abs(z) }}}{{{""=""}}}{{{ 8 + 4i}}}

{{{x + yi - abs(x + iy) }}}{{{""=""}}}{{{ 8 + 4i

{{{x + yi - sqrt(x^2 + y^2) }}}{{{""=""}}}{{{ 8 + 4i}}}

Set real parts equal, and imaginary parts equal:

{{{x-sqrt(x^2+y^2)}}}{{{""=""}}}{{{8}}},  {{{yi }}}{{{""=""}}}{{{ 4i}}}
                   {{{y }}}{{{""=""}}}{{{ 4}}}

Substitute y=4

{{{x-sqrt(x^2+4^2)}}}{{{""=""}}}{{{8}}}

{{{x-sqrt(x^2+16)}}}{{{""=""}}}{{{8}}}

{{{x-8}}}{{{""=""}}}{{{sqrt(x^2+16)}}}

Square both sides:

{{{x^2-16x+64}}}{{{""=""}}}{{{x^2+16}}}

Cancel the x<sup>2</sup>'s

{{{-16x+64}}}{{{""=""}}}{{{16}}}

{{{-16x}}}{{{""=""}}}{{{-48}}}

{{{x}}}{{{""=""}}}{{{3}}}

But we must check to see if 3 is 
a solution or only an extraneous answer.
Substitute in:

{{{x-sqrt(x^2+16)}}}{{{""=""}}}{{{8}}}
{{{3-sqrt(3^2+16)}}}{{{""=""}}}{{{8}}}
{{{3-sqrt(9+16)}}}{{{""=""}}}{{{8}}}
{{{3-sqrt(25)}}}{{{""=""}}}{{{8}}}
{{{3-5}}}{{{""=""}}}{{{8}}}
{{{-2}}}{{{""=""}}}{{{8}}}

That's false so 3 is extraneous and there
is no solution.

Edwin</pre>