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Q.  6 i - 3 i 1 2 3 i 4 3 i - 1 = x + iy ,   then

    (1) x = y = 1                                    (2) x = y = 0

    (3) x = 3, y = 1                                (4) x = – 3, y = – 1

Dear student,


We have,6i-3i123i43i-1=x+iy                   ...(1)Solving the determinant=6i3×-1-3i×i--3i2×-1-4×i+12×3i-4×3=6i-3-3i2+3i-2-4i+16i-12=6i-3-3i2-6i-12i2+6i-12using i2=-1=6i-3+3-6i-12-1+6i-12=-6i+12+6i-12=0+0ifrom (1)but 0+0i=x+iy  x=0 and y=0x= y=0thus option (2) is correct

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