Question 295781
{{{2(x+iy)+i(x-iy)=2}}}
{{{2x+2yi+xi-i^2y=2}}}
{{{(2x+y)+(x+2y)i=2+0i}}}
So then solving for both the real and imaginary part leads to two equations,
1.{{{2x+y=2}}}
2.{{{x+2y=0}}}
From eq. 2,
{{{x+2y=0}}}
{{{x=-2y}}}
{{{2x=-4y}}}
Substitute into eq. 1,
{{{-4y+y=2}}}
{{{-3y=2}}}
{{{highlight( y=-2/3))}}}
Then from eq. 1,
{{{x=-2y}}}
{{{x=-2(-2/3)}}}
{{{highlight( x=4/3)}}}
{{{z=(4/3)-(2/3)i}}}