Question

$ {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
\[
  \left( A \right)\,\,|{z_0}| > 1 \\
  \left( B \right)\,\,|{z_0}| > \dfrac{1}{2} \\
  \left( C \right)\,\,|{z_0}| > \dfrac{1}{4} \\
  \left( D \right)\,\,|{z_0}| > \dfrac{3}{2} \\
 \]

Answer 1

Hint: For this we first put $ {z_0} $ in given equation as it is given solution of the given equation and then taking mode on both side and simplifying it by taking value of mode of cosine function as one and finding sum of the G.P. series so formed to get a solution to the given problem.

Formulas used: $ |a + b| \leqslant |a| + |b|,\,\,\,\max |\cos \theta | = 1\,\,and\,\,\,{S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right),\,r < 1 $

Complete step by step solution:
Here, it is given that $ {z_0} $ is the root of the equation: $ {z^n}\cos {\theta _0} + {z^{n - 1}}\cos {\theta _1} + .......... + z\cos {\theta _{n - 1}} + \cos {\theta _n} = 2 $
Substituting value of $ {z_0} $ in above equation we have
 $ {z_0}^n\cos {\theta _0} + {z_0}^{n - 1}\cos {\theta _1} + .......... + {z_0}\cos {\theta _{n - 1}} + \cos {\theta _n} = 2 $
Taking mode on both side we have
 $ |{z_0}^n\cos {\theta _0} + {z_0}^{n - 1}\cos {\theta _1} + .......... + {z_0}\cos {\theta _{n - 1}} + \cos {\theta _n}| = |2| $
But, we know that: $ |a + b|\, \leqslant \,|a| + |b| $ .
Above result is also true for a number of terms.
Using it we have,
 $ |{z_0}^n\cos {\theta _0}| + |{z_0}^{n - 1}\cos {\theta _1}| + ......... + |{z_0}\cos {\theta _{n - 1}}| + |\cos {\theta _n}| \geqslant 2 $
Or above equation can be written as:
 $ |{z_0}^n||\cos {\theta _0}| + |{z_0}^{n - 1}||\cos {\theta _1}| + .......|{z_0}||\cos {\theta _{n - 1}}| + |\cos {\theta _n}| \geqslant 2 $
We know that maximum value of each $ |\cos \theta |\, = 1,\,\,\theta \in R $
Hence, above equation becomes:
 $ |{z_0}^n| + |{z_0}^{n - 1}| + ........|{z_0}| + 1 \geqslant 2 $
If $ |{z_0}| > 1 $ then the result of the above series is true.
Therefore, considering $ |{z_0}| \leqslant 1 $
Right hand side of the above equation is a geometric progression.
Sum of the G.P. is given as:
\[
  1\left( {\dfrac{{1 - |{z_0}{|^{n + 1}}}}{{1 - |{z_0}|}}} \right) \geqslant 2 \\
   \Rightarrow 1 - |{z_0}{|^{n + 1}} \geqslant 2 - 2|{z_0}| \\
   \Rightarrow - \left\{ {|{z_0}{|^{n + 1}} - 1} \right\} \geqslant 2 - 2|{z_0}| \\
   \Rightarrow |{z_0}{|^{n + 1}} - 1 \leqslant 2|{z_0}| - 2 \\
   \Rightarrow |{z_0}{|^{n + 1}} \leqslant 2|{z_0}| - 2 + 1 \\
   \Rightarrow |{z_0}{|^{n + 1}} \leqslant 2|{z_0}| - 1 \\
 \]
Since, we have taken $ |{z_0}| < 1 $ . Which implies that as value of $ |{z_0}{|^{n + 1}} $ increasing its value will be decreasing or we can say that for n being infinite value will tend to zero.
Therefore, from the above equation. We can say that
 $
  0 < 2|{z_0}| - 1 \\
   \Rightarrow 1 < 2|{z_0}| \\
   \Rightarrow \dfrac{1}{2} < |{z_0}| \;
  $
Therefore, required value of $ |{z_0}| $ is greater than $ \dfrac{1}{2}. $
So, the correct answer is “Option C”.

Note: In geometric progression, to find sum of the series there are two formulas of sum, out of which one formula is for that G.P. in which value of common ratio less than one and other formula is for that G.P. series in which the value of common ratio is greater than one. So, one should choose the correct formula according to G.P. series so formed to get the correct solution of the problem.