Really Complex?

Algebra Level 3

$\large \iota z^3 + z^2 - z + \iota = 0 \quad \quad ; \quad |z| = ?$

None of the Above Irrational Rational

1 solution

Chew-Seong Cheong
Dec 17, 2015

We note that $z = i$ is a root of the equation.

$\begin{aligned} iz^3 +z^2-z+i & = i(-i)+(-1)-i+i \\ & = 1-1-i+i \\ & = 0 \end{aligned}$

Multiply throughout by $-i$ and factorize, we have:

$\begin{aligned} z^3 -iz^2+iz+1 & = 0 \\ (z-i)(z^2+i) & = 0 \end{aligned}$

$\Rightarrow \begin{cases} z_1 & = i \\ z^2 & = -i = e^{2n-\frac{\pi}{2}} \quad n = 0,1 \\ & \Rightarrow \begin{cases} z_2 = e^{-\frac{\pi}{4}} \\ z_3 = e^{\frac{3\pi}{4}} \end{cases} \end{cases}$

We note that $|z_1| = |z_2|=|z_3|=|z|= 1 \text{ an } \boxed{\text{integer}}$

How do you know that $z = \iota$ is the root of given equation.That's not common.It's better if you show it how it comes.

Akhil Bansal - 5 years, 5 months ago

The first part shows it. Sometimes we solve problems by observation.

$\begin{aligned} iz^3 +z^2-z+i & = i(-i)+(-1)-i+i \\ & = 1-1-i+i \\ & = 0 \end{aligned}$

Chew-Seong Cheong - 5 years, 5 months ago

Here is the proof that $\iota$ is the root of $\iota z^3 + z^2 - z + \iota = 0$
$\Rightarrow \iota z^3 - \iota^2 z^2 - z + \iota = 0$ $\Rightarrow \iota z^2(z-\iota) - 1(z-\iota) = 0$ $\Rightarrow (\iota z^2 - 1)(z - \iota) = 0$ $\Rightarrow z - \iota = 0 \ \text{or} \ \iota z^2 - 1 = 0$ $\Rightarrow z = \iota \ \text{or} \ z^2 = - \iota$

Akhil Bansal - 5 years, 5 months ago

@Akhil Bansal – Of course, it can be shown that it is a root because it is a root. Anyway, thanks.

Chew-Seong Cheong - 5 years, 5 months ago

My solution too relied on observation. Its like what we do for finding the first root of a cubic polynomial. Clearly, we can't 'show' that step.

Pulkit Gupta - 5 years, 5 months ago

@Pulkit Gupta – Yes, we sometimes call it intuition.

Chew-Seong Cheong - 5 years, 5 months ago

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