+0  
 
+1
2
2
avatar+58 

Let \(z_1, z_2, \dots, z_{20} \) be the twenty (complex) roots of the equation\( z^{20} - 4z^{19} + 9z^{18} - 16z^{17} + \dots + 441 = 0.\)
Calculate  \(\cot \left( \sum_{k = 1}^{20} \operatorname{arccot} z_k \right).\)Note that the addition formula for cotangent is still valid when working with complex numbers.

 Jul 2, 2024
 #1
avatar+867 
0

Given the roots \( z_1, z_2, \ldots, z_{20} \) of the polynomial equation


\[
z^{20} - 4z^{19} + 9z^{18} - 16z^{17} + \dots + 441 = 0,
\]


we need to determine the value of \( \cot \left( \sum_{k = 1}^{20} \operatorname{arccot} z_k \right) \).

 

First, recall the properties of the \(\arccot\) function and how it relates to complex roots and trigonometric identities. Specifically, for any complex number \( z \),


\[
\operatorname{arccot} z = \arctan \frac{1}{z}.
\]

 

Next, using the identity for the sum of \(\arctan\) values, we have:


\[
\arctan z_1 + \arctan z_2 + \cdots + \arctan z_{20} = \arctan \left( \frac{z_1 + z_2 + \cdots + z_{20}}{1 - z_1 z_2 \cdots z_{20}} \right).
\]


Given that \(\operatorname{arccot} z_k = \arctan \frac{1}{z_k}\), it follows that:


\[
\sum_{k=1}^{20} \operatorname{arccot} z_k = \sum_{k=1}^{20} \arctan \frac{1}{z_k}.
\]

 

We use the identity for the sum of arctangents. For simplicity, let's assume that the complex numbers \( z_1, z_2, \ldots, z_{20} \) are such that we can use:


\[
\arctan \frac{1}{z_1} + \arctan \frac{1}{z_2} + \cdots + \arctan \frac{1}{z_{20}} = \arctan \left( \frac{\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_{20}}}{1 - \left( \frac{1}{z_1} \cdot \frac{1}{z_2} \cdots \frac{1}{z_{20}} \right)} \right).
\]

 

Now, we need to analyze the roots of the given polynomial \( z^{20} - 4z^{19} + 9z^{18} - \cdots + 441 = 0 \).

 

By Vieta's formulas, the sum of the roots (considering the coefficients of the polynomial) is equal to the coefficient of \( z^{19} \) divided by the leading coefficient (for \( z^{20} \)), which is \( -\frac{-4}{1} = 4 \).

 

However, for the product of the roots, by Vieta's formulas, the constant term (considering all terms up to the last one) will give us:


\[
z_1 z_2 \cdots z_{20} = (-1)^{20} \times \text{constant term}/\text{leading coefficient} = 441.
\]

 

Thus, for the cotangent sum, we need the real parts of the roots in terms of trigonometric identities:


\[
\cot \left( \sum_{k=1}^{20} \operatorname{arccot} z_k \right) = \cot \left( \arctan \left( \frac{\sum_{k=1}^{20} \frac{1}{z_k}}{1 - \frac{1}{z_1} \frac{1}{z_2} \cdots \frac{1}{z_{20}}} \right) \right).
\]

 

Since \(\sum_{k=1}^{20} z_k = 4\) and \(\prod_{k=1}^{20} z_k = 441\),
\[
\sum_{k=1}^{20} \frac{1}{z_k} = \frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_{20}} = \frac{20}{441}.
\]

 

Then:
\[
1 - \prod_{k=1}^{20} \frac{1}{z_k} = 1 - \frac{1}{441} = \frac{440}{441}.
\]

 

Substituting in,


\[
\frac{\frac{20}{441}}{\frac{440}{441}} = \frac{20}{440} = \frac{1}{22}.
\]

 

Therefore, we have


\[
\arctan \left( \frac{1}{22} \right),
\]


which means


\[
\cot \left( \arctan \frac{1}{22} \right) = 22.
\]

 

Hence, the final answer is:
\[
\boxed{22}.
\]

 Jul 2, 2024

3 Online Users

avatar
avatar
avatar