If 8 i z3+12 z2-18 z+27 i=0, then

Question

If 8 i z3+12 z2-18 z+27 i=0, then (A) |z|=(3/2) (B) |z|=(2/3) (C) |z|=1 (D) |z|=(3/4)

Tardigrade Admin · Accepted Answer

8 i z3+12 z2-18 z+27 i =0 ⇒ 8 i z3-12 i2 z2-18 z+27 i =0 ⇒ 4 i z2(2 z-3 i)-9(2 z-3 i) =0 ⇒ (4 i z2-9)(2 z-3 i)=0 ⇒ z2=(9/4 i) text or z=(3 i/2) In any case |z|=3 / 2.

Q. If $8 i z^{3} + 12 z^{2} - 18 z + 27 i = 0$ , then

$∣ z ∣ = \frac{3}{2}$

$∣ z ∣ = \frac{2}{3}$

$∣ z ∣ = 1$

$∣ z ∣ = \frac{3}{4}$

$8 i z^{3} + 12 z^{2} - 18 z + 27 i = 0$
$\Rightarrow 8 i z^{3} - 12 i^{2} z^{2} - 18 z + 27 i = 0$
$\Rightarrow 4 i z^{2} (2 z - 3 i) - 9 (2 z - 3 i) = 0$
$\Rightarrow (4 i z^{2} - 9) (2 z - 3 i) = 0$
$\Rightarrow z^{2} = \frac{9}{4 i} or z = \frac{3 i}{2}$
In any case $∣ z ∣ = 3/2$ .

Q. If 8iz3+12z2−18z+27i=0, then

∣z∣=23​

∣z∣=32​

∣z∣=1

∣z∣=43​