Theory of Intelligence
11TH•Mathematics•Complex Numbers•Exercise 1.3

1.3 Complex Polynomials as a Product of Linear Factors

Practice set 1.3 for Complex Numbers (conjugate & division).

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Q2

Question 2

Factorize into linear factors.

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(i)

z3+8z^3 + 8

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Step-by-Step Solution

Question 2 • Part (i)

verifiedVerified Answer

(i) z3+8z^3 + 8

Solution

We know the sum of cubes formula:a3+b3=(a+b)(a2−ab+b2)So, z3+8=z3+(2)3a=z,b=2Applying the formula:z3+8=(z+2)(z2−(2)(z)+4)z3+8=(z+2)(z2−2z+4)Now factor z2−2z+4:a=1,  b=−2,  c=4Δ=b2−4ac=(−2)2−4(1)(4)=4−16=−12Δ=−12=2i3Using the quadratic formula:z=−b±Δ2a=−(−2)±2i32=2±2i32=1±i3z1=1+i3,z2=1−i3Therefore, the factors are:z2−2z+4=(z−z1)(z−z2)==(z−(1+i3))(z−(1−i3))Therefore,z3+8=(z+2)(z−1−i3)(z−1+i3)\begin{aligned} & \boxed{\text{We know the sum of cubes formula:}} \\ \\ & a^3 + b^3 = (a + b)(a^2 - ab + b^2) \\ \\ & \text{So, } \\ & z^3 + 8 = z^3 + (2)^3 \\ & a = z,\quad b = 2 \\ & \\ &\boxed{\text{Applying the formula:}}\\ & z^3 + 8 = (z + 2)(z^2 - (2)(z) + 4) \\ & z^3 + 8 = (z + 2)(z^2 - 2z + 4) \\ & \\ &\boxed{\text{Now factor } z^2 - 2z + 4:} \\ & a = 1,\; b = -2,\; c = 4 \\ & \Delta= b^2 - 4ac = (-2)^2 - 4(1)(4) = 4 - 16 = -12 \\ & \sqrt{\Delta} = \sqrt{-12} = 2i\sqrt{3} \\ \\ & \boxed{\text{Using the quadratic formula:}} \\ & z = \frac{-b \pm \sqrt{\Delta}}{2a} =\frac{-(-2) \pm 2i\sqrt{3}}{2} = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3} \\ & \\ & z_1 = 1 + i\sqrt{3}, \quad z_2 = 1 - i\sqrt{3} \\ \\ & \boxed{\text{Therefore, the factors are:}} \\ & z^2 - 2z + 4 = (z - z_1)(z - z_2) = \\ & \quad = \big(z - (1 + i\sqrt{3})\big)\big(z - (1 - i\sqrt{3})\big) \\ & \\ & \boxed{\text{Therefore,}} \\ & z^3 + 8 = (z + 2)\left(z - 1 - i\sqrt{3}\right)\left(z - 1 + i\sqrt{3}\right) \end{aligned}
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