First slide

Algebra of complex numbers

Question

If $z^{3} + (3 + 2 i) z + (- 1 + ia) = 0$ has one real root, then the value of a lies in the interval $(a \in R)$

Moderate

A

(-2, 1)
B

(-1, 0)
C

(0, 1)
D

(-2, 3)

Solution

Let $z = α$ be a real root. Then,
$\begin{array}{l} α^{3} + (3 + 2 i) α + (- 1 + ia) = 0 \\ \Rightarrow (α^{3} + 3 α - 1) + i (a + 2 α) = 0 \\ \Rightarrow α^{3} + 3 α - 1 = 0 and α = - a / 2 \\ \Rightarrow - \frac{a^{3}}{8} - \frac{3 a}{2} - 1 = 0 \\ \Rightarrow a^{3} + 12 a + 8 = 0 \end{array}$
Let $f (a) = a^{3} + 12 a + 8$ .
$f' (a) = 3 a^{2} + 12 > 0$ .
So, f(a) cuts x-axis only once.
Now, f(-1) < 0, f(0) > 0,
f(-2) < 0, f(1) > 0 and f(3) > 0
Hence, $a \in (- 1, 0) or a \in (- 2, 1) or a \in (- 2, 3)$ .

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