Given that x, `y in R`. Solve: `x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)`

Given that x,y in R, solve (2+3i)x^(2)-(3-2i)y=2x-3y+5i

Given that x,yinR solve x^2-y^2-i(2x+y)=4i

Solve (1−i)x+(1+i)y=1−3i,

1/(6i)+1/(7i)=x/(8i)

Solve for x,y in R: (x^4+2x i) -(3x^2+yi)=(1+2yi)+34/(3+5i) .

Solve for x: (1-i)x + (1+i)y=1 - 3i .

Find the values of x and y if : ((1+i)x-2i)/(3+i)+((2-3i)y +i)/(3-i) =i , where i=sqrt(-1)

Find the value of x and y which satisfy the following equation (x, y in R): If x(1+3i)+y(2-i)-5+ i^3 = 0, find (x+y)

Given that (1 + i) (1 + 2i) (1 + 3i)…(1 + ni)= x+ iy , show that 2.5.10…(1 + n^(2))= x^(2) + y^(2)