Let Pz - z3 - az2 - bz- c where a- b and c are real numbers- There exists a complex number - such that the three roots of Pz are - -3i- - -9i and 2- -4 where i2 -1- The value of -a-b-c- is

Z^3 + az^2 + bz+ c=0 Let w=p+iq Then roots are p+(q+3)i, p+(q+9)i, nbsp;2p 4+2qi Sum of roots = a=real ⇒ 4p 4+ (4q +12)i =real ⇒ q= 3 So, roots are p, p+6i, 2 ...

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Byju's Answer

Standard XI

Mathematics

Complex Numbers

Let Pz = z3 +...

Question

Let $P (z) = z^{3} + a z^{2} + b z + c$ where $a, b$ and $c$ are real numbers. There exists a complex number $ω$ such that the three roots of $P (z)$ are $ω + 3 i, ω + 9 i$ and $2 ω - 4$ where $i^{2} = - 1.$ The value of $| a + b + c |$ is

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Solution

$z^{3} + a z^{2} + b z + c = 0$
Let $w = p + i q$
Then roots are $p + (q + 3) i, p + (q + 9) i, 2 p - 4 + 2 q i$
Sum of roots $= - a = real$
$\Rightarrow 4 p - 4 + (4 q + 12) i = real$
$\Rightarrow q = - 3$
So, roots are $p, p + 6 i, 2 p - 4 - 6 i$
As the coefficients of the equation are real, there must be one real and two complex conjugate roots.
$p + 6 i, 2 p - 4 - 6 i$ are conjugates.
$\Rightarrow p = 2 p - 4$
$\Rightarrow p = 4$
Roots are $4, 4 + 6 i, 4 - 6 i$

$z^{3} + a z^{2} + b z + c = (z - 4) (z - (4 + 6 i)) (z - (4 - 6 i))$
Putting $z = 1,$
$1 + a + b + c = - 135$
$a + b + c = - 136$

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Let P(z)=z3+az2+bz+c where a,b and c are real numbers. There exists a complex number ω such that the three roots of P(z) are ω+3i,ω+9i and 2ω−4 where i2=−1. The value of |a+b+c| is

Let $P (z) = z^{3} + a z^{2} + b z + c$ where $a, b$ and $c$ are real numbers. There exists a complex number $ω$ such that the three roots of $P (z)$ are $ω + 3 i, ω + 9 i$ and $2 ω - 4$ where $i^{2} = - 1.$ The value of $| a + b + c |$ is