Answer to Question #116905 in Algebra for kay

Question

Show that z = i is a root of the equation z^4 + z^3 + z − 1 = 0. Find the three other roots.

Answer 1

Give z = i is a root of the equation z⁴ + z³ + z - 1 = 0.

Hence, z - i = 0 is a factor of the given quartic equation.

Let (z - i)(az³ + bz² + cz + d) = z⁴ + z³ + z - 1,

Where a, b, c and d are coefficients.

(z - i)(az³ + bz² + cz + d) = z⁴ + z³ + z - 1 .... eq. (1)

az⁴ + bz³ + cz² + dz - iaz³ - ibz² - icz - id = z⁴ + z³ + z - 1

az⁴ + z³(b - ia) + z²(c - ib) + z(d - ic) - id = z⁴ + z³ + z - 1

Equating the coefficients of LHS and RHS, we get :

a = 1

b - ia = 1

i.e., b - i = 1

$\therefore$ b = 1 + i

c - ib = 0

c - i(1 + i) = 0

c - i + 1 = 0

$\therefore$ c = i - 1

d - ic = 1

d - i(i - 1) = 1

d + 1 + i = 1

d = 1 - 1 - i

$\therefore$ d = -i

Substituting the above values of a, b, c and d in equation (1) :

(z - i)(z³ + z²(1 + i) + z(i - 1) - i) = z⁴ + z³ + z - 1

The remaining 3 roots can be found from equating (z³ + z²(1 + i) + z(i - 1) - i) = 0.

z³ + z²(1 + i) + z(i - 1) - i = 0

(z³ + z² - z) + i(z² + z - 1) = 0

z (z² + z - 1) + i(z² + z - 1) = 0

(z² + z - 1)(z + i) = 0

i.e., z² + z - 1 = 0 & z + i = 0

z² + z - 1 = 0 & z = - i

z² + z - 1 = 0

Here, a = 1, b = 1, c = -1.

b² - 4ac = 1 - 4(1)(-1) = 1 + 4 = 5

Therefore,

z = (-b + $\sqrt{\smash[b]{b² - 4ac}}$ )/2a or z = (-b - $\sqrt{\smash[b]{b² - 4ac}}$ )/2a

i.e., z = (-1 + $\sqrt{\smash[b]{5}}$ )/2 or z = (-1 - $\sqrt{\smash[b]{5}}$ )/2

$\therefore$ The three other roots of the given equation are :

z = -i

z = (-1 + $\sqrt{\smash[b]{5}}$ )/2

z = (-1 - $\sqrt{\smash[b]{5}}$ )/2