(x+iy)(2-3i)=(4+i)

Express the result in the form x+iy, where x,y are real number i=sqrt(-1) : (i) (2-3i)/(4-i) (ii) (2+3i)/(-5-4i) (iii) (1+i)/(3+i) (iv) (3+2i)/(4-3i)

((x+iy)/(2+3i)+(2+i)/(2-3i)=(9)/(13)(1+i)

If (1+2i) (2+3i)(3+4i)=x+iy,x,y in R then x^(2)+y^(2) is

Find the values of x and y : (i) x+3i=6-9iy (ii) (x+iy)-(3-2i)=5+i (iii) i=x+2yi (iv) 4x+i(x-y)=2-5i (v) (x+2iy)(2-i)^(2)=10(1-i)

Find the values of x and y : (i) x+3i=6-9iy (ii) (x+iy)-(3-2i)=5+i (iii) i=x+2yi (iv) 4x+i(x-y)=2-5i (v) (x+2iy)(2-i)^(2)=10(1-i)

If (x+iy)(3-4i) = 5+12i , then sqrt(x^2+y^2) =

Write the following in the form x+iy: (i) (3+2i)(2-i) (ii) 2i^(2)+6i^(3)+3i^(16)-6i^(19)+4i^(25) . (iii) ((3-2i)(2+3i))/((1+2i)(2-i)) .

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):}