Question

If ${z_0}$ is a root of the equation ${z^n}\cos {\theta _0} + {z^{n - 1}}\cos {\theta _1} + ... + z\cos {\theta _{n - 1}} + \cos {\theta _n} = 2$ where $\theta \in R$ , then

A.$\left| {{z_0}} \right| > 1$
B.$\left| {{z_0}} \right| > \dfrac{1}{2}$
C.$\left| {{z_0}} \right| > \dfrac{1}{4}$
D.$\left| {{z_0}} \right| > \dfrac{3}{2}$

Answer 1

Here, we will take modulus on both sides of the given equation. Then, we will use some properties of modulus and convert the equation into an inequality. We will find the range of values of $\left| z \right|$ which simplify the inequality and choose the correct option accordingly.

Formulas used: We will use the following formulas:
$\left| {a + b} \right| \le \left| a \right| + \left| b \right|$
$\left| a \right|\left| b \right| = \left| {ab} \right|$

Complete step-by-step answer:
We have the equation,
${z^n}\cos {\theta _0} + {z^{n - 1}}\cos {\theta _1} + ... + z\cos {\theta _{n - 1}} + \cos {\theta _n} = 2$
We need to find $\left| {{z^n}} \right|$ , so we will take modulus on both sides. Therefore, we get
$\Rightarrow \left| {{z^n}\cos {\theta _0} + {z^{n - 1}}\cos {\theta _1} + ... + z\cos {\theta _{n - 1}} + \cos {\theta _n}} \right| = \left| 2 \right|$
We will use the property $\left| {a + b} \right| \le \left| a \right| + \left| b \right|$ of modulus and simplify the above equation. So, we have
$\Rightarrow \left| {{z^n}\cos {\theta _0}} \right| + \left| {{z^{n - 1}}\cos {\theta _1}} \right| + ... + \left| {z\cos {\theta _{n - 1}}} \right| + \left| {\cos {\theta _n}} \right| \ge \left| {{z^n}\cos {\theta _0} + {z^{n - 1}}\cos {\theta _1} + ... + z\cos {\theta _{n - 1}} + \cos {\theta _n}} \right| = 2$
$\Rightarrow \left| {{z^n}\cos {\theta _0}} \right| + \left| {{z^{n - 1}}\cos {\theta _1}} \right| + ... + \left| {z\cos {\theta _{n - 1}}} \right| + \left| {\cos {\theta _n}} \right| \ge 2$
Now using property $\left| a \right|\left| b \right| = \left| {ab} \right|$ of modulus and simplifying the above equation, we get
$\Rightarrow \left| {{z^n}} \right|\left| {\cos {\theta _0}} \right| + \left| {{z^{n - 1}}} \right|\left| {\cos {\theta _1}} \right| + ... + \left| z \right|\left| {\cos {\theta _{n - 1}}} \right| + \left| {\cos {\theta _n}} \right| \ge 2$…………………$\left( 1 \right)$
We are aware that $\cos \theta$ has 1 as its maximum value and $- 1$ as its minimum value.
$- 1 \le \cos \theta \le 1$
So, modulus of $\cos \theta$ will always be less than 1:
$\Rightarrow \left| {\cos \theta } \right| \le 1$
If we substitute the maximum value of $\cos \theta$ in equation (1), the inequality will still hold true.
Substituting 1 for $\cos \theta$, we get
$\Rightarrow \left| {{z^n}} \right| + \left| {{z^{n - 1}}} \right| + ... + \left| z \right| + 1 \ge 2$ ………………………$\left( 2 \right)$
We can see that the above inequality will hold true if $\left| z \right|$ is greater than or equal to 1 but we cannot say the same in the case where $\left| z \right|$ is less than 1 by merely observing. So, we will assume that $\left| z \right| < 1$ and simplify inequality $\left( 2 \right)$.
Simplifying equation $\left( 2 \right)$, we get
$\Rightarrow 1 + \left| z \right| + ... + \left| {{z^{n - 1}}} \right| + \left| {{z^n}} \right| > 2$
We can see that the terms on the left-hand side are forming a geometric progression with the common ratio as $\left| z \right|$ .
Substituting 1 for $a$ and $\left| z \right|$ for $r$ in the formula of sum of terms of a G.P , $\dfrac{a}{{1 - r}}$, we get
$\begin{array}{l}\dfrac{1}{{1 - \left| z \right|}} > 2\\\ \Rightarrow 1 > 2\left( {1 - \left| z \right|} \right)\\\ \Rightarrow 1 > 2 - 2\left| z \right|\end{array}$
Simplifying the inequality, we get
$\begin{array}{l} \Rightarrow 2\left| z \right| > 1\\\ \Rightarrow \left| z \right| > \dfrac{1}{2}\end{array}$
We have found out that inequality $\left( 2 \right)$ will be true for all $\left| z \right| > \dfrac{1}{2}$ .
$\therefore$ Option B is the correct option.

Note: If $\left| z \right| > \dfrac{1}{2}$ , it will also be greater than 1. So, option B covers the range of option A as well but option A does not cover the cases when $\dfrac{1}{2} < \left| z \right| \le 1$ so, option A is incorrect and cannot be chosen.
We know that the sum of terms of a G.P when the common ratio is less than 1 is $\dfrac{a}{{1 - r}}$ where $a$ is the first term of the G.P. and $r$ is the common ratio. We might make a mistake by taking the common ratio greater than 1 and using the formula of the sum of terms of a G.P as $\dfrac{a}{{r - 1}}$. This will give us the wrong sum and hence we will get incorrect answers.