Question 1113675
{{{ 3x + yi = (1+i)(x^2+y^2) }}}

Let {{{ k=x^2+y^2 }}} for now:

{{{  3x+yi = (1+i)k }}}
{{{  3x+yi = k+ki }}}<br>

Using Re{LHS} = Re{RHS},  Im{LHS}=Im{RHS} :
<pre>
3x = k  —>  x = k/3  (1)
y = k                (2) <br>
</pre>

But   {{{ x^2+y^2 = k }}} so, using (1) and (2):  
{{{ k^2/9 + k^2  = k }}}
{{{  10k^2 - 9k = 0 }}}
k=0 and/or k=9/10

k=0 is a trivial solution (x=y=k=0)
k=9/10 —> <b> {{{ highlight( x = 3/10) }}} , {{{ highlight(  y = 9/10) }}}