Question

if 1,a1,a2,a3 are the fourth roots of unity, then the value of (1+a1) (1+a2) (1+a3)

Answer 1

given, $1,a_1,a_2,a_3$ are the fourth roots of unity.

$so,(x^4-1)=(x-1)(x-a_1)(x-a_2)(x-a_3)$

$or,\frac{(x^4-1)}{(x-1)}=(x-a_1)(x-a_2)(x-a_3)$

or, (x⁴ - x³ + x³ - x² + x² - x + x - 1)/(x -1)= $(x-a_1)(x-a_2)(x-a_3)$

or, (x - 1)(x³ + x² + x + 1)/(x - 1) = $(x-a_1)(x-a_2)(x-a_3)$

or, x³ + x² + x + 1 = $(x-a_1)(x-a_2)(x-a_3)$

putting x = -1

(-1)³ + (-1)² + (-1) + 1 = $(-1-a_1)(-1-a_2)(-1-a_3)$

or, -1 + 1 - 1 + 1 = $-(1 + a_1)(1+a_2)(1+a_3)$

or, $(1+a_1)(1+a_2)(1+a_3)=0$

hence, value of $(1+a_1)(1+a_2)(1+a_3)=0$

Answer 2

Answer:

$(1+a_1)(1+a_2)(1+a_3)=0$

Step-by-step explanation:

We know that the fourth roots of unity are 1, i, -1, -i

All these roots are got by solving the equation $x=1^{\frac{1}{4}}$

by using demovire's theorem.

Now,

$(1+a_1)(1+a_2)(1+a_3)$

$=(1+i)(1+(-1))(1+(-i))$

$=(1+i)(0)(1+(-i))$

$=0$