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# Let be the twenty (complex) roots of the equation Calculate Note that the addition formula for cotangent is still valid when working

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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

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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

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

$\boxed{22}.$