Show that the equation
${{z}^{4}}-2{{z}^{3}}+6{{z}^{2}}-8z+8=0$
has a root of the form $k\mathrm{i}$, where $k$ is real.[3]
Hence solve the equation
${{z}^{4}}-2{{z}^{3}}+6{{z}^{2}}-8z+8=0$.[3]

has a root of the form $k\mathrm{i}$, where $k$ is real.

${{z}^{4}}-2{{z}^{3}}+6{{z}^{2}}-8z+8=0$.

Complex Numbers - General

Complex Numbers

International Baccalaureate Higher Level Mathematics

IB HL Math

N2006 P1 Q6

Find a root of the form z = ki where k is real:

Substituting z = ki into z⁴ - 2z³ + 6z² - 8z + 8 = 0:

Equating real and imaginary parts to zero:

Imaginary: -2k³ + 8k = 0  →  -2k(k² - 4) = 0  →  k = 0 or k = ±2

So k² = 2 (k = ±√2) or k² = 4 (k = ±2)

The common solution is k = ±2, so z = 2i is a root (shown).

Since coefficients are real, z = -2i is also a root.

Dividing z⁴ - 2z³ + 6z² - 8z + 8 by (z² + 4):

The four roots are: z = 2i, z = -2i, z = 1+i, z = 1-i

N2006 P1 Q6

N2006 P1 Q6

Video Solution: