Verify that $z=\mathrm{i}$ is a root of the equation
${{z}^{4}}-2{{z}^{3}}+6{{z}^{2}}-2z+5=0$.[1]
Hence determine the other roots.[4]

Verify that $z=\mathrm{i}$ is a root of the equation

${{z}^{4}}-2{{z}^{3}}+6{{z}^{2}}-2z+5=0$.

Complex Numbers - General

Complex Numbers

International Baccalaureate Higher Level Mathematics

IB HL Math

N2005 P2 Q1

Let $P(z) = z^4 - 2z^3 + 6z^2 - 2z + 5$

$P(i) = i^4 - 2(i)^3 + 6(i)^2 - 2(i) + 5$

$= i^2 \cdot i^2 - 2i^2 \cdot i + 6(-1) - 2i + 5$

$= 1 + 2i - 6 - 2i + 5$

$= 0$

$\therefore z = i$ is a root of $P(z)$.

By Conjugate Roots Theorem, since all coefficients are real, $z = -i$ is also a root.

$z^4 - 2z^3 + 6z^2 - 2z + 5 = (z - i)(z - (-i))(Az^2 + Bz + C)$

$= (z - i)(z + i)(Az^2 + Bz + C)$

$= (z^2 - i^2)(Az^2 + Bz + C)$

$= (z^2 + 1)(Az^2 + Bz + C)$

By comparing coefficients:

$z^3$: $A = 1$ $\quad$ $z$: $B = -2$

Constant: $C = 5$

$\therefore P(z) = (z^2 + 1)(z^2 - 2z + 5)$

$z = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)}$

$= \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2}$

$= 1 \pm 2i$

Other roots are $1 + 2i$, $1 - 2i$ and $-i$.

N2005 P2 Q1

N2005 P2 Q1

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