Question 838942
{{{(a+bi)(a+bi)=a^2+abi+abi+(bi)^2}}}
{{{(a+bi)(a+bi)=(a^2-b^2)+2abi}}}
1.{{{a^2-b^2=-15}}}
2.{{{2ab=8}}}
From eq. 2,
{{{ab=4}}}
{{{a=4/b}}}
{{{a^2=16/b^2}}}
Substitute into eq. 1,
{{{16/b^2-b^2=-15}}}
{{{16-b^4=-15b^2}}}
{{{b^4-15b^2-16=0}}}
{{{(b^2-16)(b^2+1)=0}}}
Only the real solution.
{{{b^2-16=0}}}
{{{b^2=16}}}
{{{b=4}}}
From eq. 2,
{{{ab=4}}}
{{{a(4)=4}}}
{{{a=1}}}
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{{{sqrt((-15+8i))=1+4i}}}