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

Solve for x(i)|x^(3)-1|>=1-x

Solve (1−i)x+(1+i)y=1−3i,

Find the real values of x and y for which : {:((i),(1-i)x+(1+i)y=1 - 3i,(ii),(x+iy)(3-2i)=(12+5i)),((iii),x+4yi =ix + y + 3,(iv),(1+i)y^(2)+(6+i)=(2+i)x),((v),((x+3i))/((2+iy))=(1-i),(vi),((1+i)x-2i)/((3+i))+((2-3i)y+i)/((3-i))=i):}

Solve 1/z + 1/(2+i) = 1/(1+3i)

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

Solve (1−i)x+(1+i)y=1−3i, find x and y

If ((1 + i) / (1 - i))^3 - ((1-i) / (1 + i))^3 = x + iy then find x and y.

Finid the vlues of x and y if: (i) (x+iy)(1+i)=1-i (ii) ((1+i)x-2i)/(3+i)+((2-3i)y+i)/(3-i)=i , where i=sqrt(-1) .

Solve ( sin x + "i" cos x )/( 1 +i),i =sqrt(-1) when it is purely imaginary .