Let P(z)=z3+az2+bz+c, where a,b,c∈R. If there exists a complex number w such that the three roots of P(z) are w+3i,w+9i and 2w−4, where i2=−1, then the value of a+b+c is (correct answer + 1, wrong answer - 0.25)
A
136
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B
−136
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C
280
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D
−280
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Solution
The correct option is B−136 Let w=x+iy, so the three roots are x1=x+i(y+3)x2=x+i(y+9)x3=(2x−4)+2iy Now, P(z)=z3+az2+bz+c Sum of roots x1+x2+x3=−a As a∈R, so Im(x1+x2+x3)=0⇒4y+12=0⇒y=−3 Now, x1=xx2=x+6ix3=(2x−4)−6i Product of roots x1x2x3=−c As c∈R, so Im(x1x2x3)=0⇒Im(x(x+6i)((2x−4)−6i))=0⇒(2x−4)×6−6x=0⇒x=4 Now, x1=4x2=4+6ix3=4−6i Therefore, a=−12,c=−208,b=84 Hence, a+b+c=−136