Question 913204
{{{a+bi+c+di=4}}}
{{{a+c+(b+d)i=4}}}
{{{a+c=4}}} and {{{b+d=0}}}
{{{highlight_green(b=-d)}}}


{{{(a+bi)(c+di)=13}}}
{{{ac-bd+(bc+ad)i=13}}}
{{{ac-bd=13}}} and {{{bc+ad=0}}}


Substituting for b into these last two equations
{{{ac-(-d)d=13}}}
{{{ac+d^2=13}}}
and
{{{(-d)c+ad=0}}}
{{{-cd+ad=0}}}
and summarizing the THREE equations which use just variables a,c,d is
{{{highlight_green(system(a+c=4,ac+d^2=13,-cd+ad=0))}}}


The last equation in the summary of the three-equation system equivalently gives
{{{d(a-c)=0}}}


Looking at the first equation of that system and the last equation and assuming {{{highlight_green(d<>0)}}}, then
{{{system(a+c=4,(a-c)d=0)}}}
with the assumption about d allows to have the system:
{{{system(a+c=4,a-c=0)}}}


Find through adding first of those to the last of those,
{{{2a=4}}}
{{{highlight(a=2)}}}
and then
{{{a+c=4}}}
{{{2+c=4}}}
{{{highlight(c=2)}}}


Return to the two equation which came from the product 13.
-
{{{ac-bd=13}}}
{{{2*2-bd=13}}}
{{{-bd=13-4}}}
{{{bd=-9}}}----coming back to this after other steps...
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{{{-bc+ad=0}}}, also from the product 13 situation
{{{2*b+2*d=0}}}, because we know a and c
{{{2(b+d)=0}}}
{{{b+d=0}}}
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Again {{{bd=-9}}}
{{{bd=(-d)d=-9}}}
{{{d^2=9}}}
{{{highlight(d=3)}}}----- or negative 3 as well.
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Because we found {{{b=-d}}}
{{{-b=d}}}
{{{-b=3}}}
{{{highlight(b=-3)}}}


The variables a, b, c, and d, have now been solved for.
ANSWER:  <b>The two complex numbers are 2-3i and 2+3i.</b>
{{{b=-d}}}----which we can soon use in substitution.