A) \[5\]
B) \[8\]
C) \[10\]
D) \[40\]
Correct Answer: C
Solution :
Let \[z=x+iy,\] then \[\bar{z}=x-iy\] \[\therefore \] \[z+\bar{z}=2x\] and \[z-\bar{z}=2iy\] Given, \[(3+i)(z+\bar{z})-(2+i)(z-\bar{z})+14i=0\] \[\Rightarrow \] \[(3+i)2x-(2+i)2iy+14i=0\] \[\Rightarrow \] \[6x+2ix-4yi+2y+14i=0+0i\] On comparing real and imaginary part, we get \[6x+2y=0\] and \[2x-4y+14=0\] On solving, we get \[x=-1,\,y=3\] \[\therefore \] \[z\bar{z}=|z{{|}^{2}}={{(\sqrt{{{(-1)}^{2}}+{{(3)}^{2}}})}^{2}}=10\]
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