Suppose z is any root of 11z^8+21iz^7+10iz-22=0 where i=√(-1). Then, S=|z|^2+|z|+1 satisfies

Correct answer: (B) 3<S<7. We have, 11z^8+20iz^7+10iz-22=0 ⇒z^8+20/11iz^7+10i/11z-2=0 Root of equation are z_1,z_2,z_3,z_4,z_5,…,z_8 ∑i=18zi=20/11i z_1z_2z_3z_4z_5z_6z_7z_8=2 ⇒z_1z_2z_3z_4z_5z_6z_7z_8=2 ⇒z_1z_2z_3z_4z_4z_5z_6z_7z_8=2 ∵1<z_i<2 ∴S=|z|^2+|z|+1 ⇒1+1+1<|z|^2+|z|+1<2^2+2+1 ⇒3<|z|^2+|z|+1<7 ∴3<S<7

Q60 / 91 · KVPY 2018 (SB/SX) · Mathematics

Suppose $z$ is any root of $11 z^{8} + 21 i z^{7} + 10 i z - 22 = 0$ where $i = \sqrt{- 1} .$ Then, $S = | z |^{2} + | z | + 1$ satisfies

A
$S \leq 3$
Correct answer
B
$3 < S < 7$
C
$7 \leq S < 13$
D
$S \geq 13$

Solution

We have,

$11 z^{8} + 20 i z^{7} + 10 i z - 22 = 0$

$\Rightarrow z^{8} + \frac{20}{11} i z^{7} + \frac{10 i}{11} z - 2 = 0$

Root of equation are $z_{1}, z_{2}, z_{3}, z_{4}, z_{5}, \dots, z_{8}$

$\sum_{i = 1}^{8} z i = \frac{20}{11} i$

$z_{1} z_{2} z_{3} z_{4} z_{5} z_{6} z_{7} z_{8} = 2$

$\Rightarrow |z_{1} z_{2} z_{3} z_{4} z_{5} z_{6} z_{7} z_{8}| = 2$

$\Rightarrow |z_{1}| |z_{2}| |z_{3}| |z_{4}| |z_{4}| |z_{5}| |z_{6}| |z_{7}| |z_{8}| = 2$

$∵ 1 < |z_{i}| < 2$

$∴ S = | z |^{2} + | z | + 1$

$\Rightarrow 1 + 1 + 1 < | z |^{2} + | z | + 1 < 2^{2} + 2 + 1$

$\Rightarrow 3 < | z |^{2} + | z | + 1 < 7$

$∴ 3 < S < 7$

Hence, the correct answer is option: 'B'.

Suppose z is any root of 11z8+21iz7+10iz-22=0 where i=-1. Then, S=|z|2+|z|+1 satisfies

Solution

More Complex Number Questions

Suppose $z$ is any root of $11 z^{8} + 21 i z^{7} + 10 i z - 22 = 0$ where $i = \sqrt{- 1} .$ Then, $S = | z |^{2} + | z | + 1$ satisfies