# If (3+ i) ( z-bar z )- (2+ i) ( z-bar z) + 14i = 0, then z bar z is equal to (1) 5 (2) 8 (3) 10 (4) 40

Solution:

Given $$(3+i)(z+\bar{z})-(2+i)(z-\bar{z})+14i = 0$$ ..(1)

Let z = x+iy

$$\bar{z}$$ = x-iy

$$z+\bar{z}$$ = 2x

$$z-\bar{z}$$ = 2iy

Substituting above values in (1)

(3+i)2x-(2+i)2iy+14i = 0

6x+2ix-4iy+2y+14i = 0

(6x+2y)+(2x-4y+14)i = 0

Comparing real part

6x+2y = 0

6x = -2y

y =6x/-2

y = -3x

Comparing imaginary part

2x-4y+14 = 0

Put y = -3x in above equation

2x+12x+14 = 0

14x = -14

x = -1

So y = 3

Now z = -1+3i

$$\bar{z}$$ = -1-3i

$$z\bar{z}$$ = (-1+3i)(-1-3i)

= 1+9

= 10

Hence option (3) is the answer.