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Let z1,z2,,z20 be the twenty (complex) roots of the equationz204z19+9z1816z17++441=0.
Calculate  cot(20k=1arccotzk).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 z1,z2,,z20 of the polynomial equation


z204z19+9z1816z17++441=0,


we need to determine the value of cot(20k=1arccotzk).

 

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


arccotz=arctan1z.

 

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


arctanz1+arctanz2++arctanz20=arctan(z1+z2++z201z1z2z20).


Given that arccotzk=arctan1zk, it follows that:


20k=1arccotzk=20k=1arctan1zk.

 

We use the identity for the sum of arctangents. For simplicity, let's assume that the complex numbers z1,z2,,z20 are such that we can use:


arctan1z1+arctan1z2++arctan1z20=arctan(1z1+1z2++1z201(1z11z21z20)).

 

Now, we need to analyze the roots of the given polynomial z204z19+9z18+441=0.

 

By Vieta's formulas, the sum of the roots (considering the coefficients of the polynomial) is equal to the coefficient of z19 divided by the leading coefficient (for z20), which is 41=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:


z1z2z20=(1)20×constant term/leading coefficient=441.

 

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


cot(20k=1arccotzk)=cot(arctan(20k=11zk11z11z21z20)).

 

Since 20k=1zk=4 and 20k=1zk=441,
20k=11zk=1z1+1z2++1z20=20441.

 

Then:
120k=11zk=11441=440441.

 

Substituting in,


20441440441=20440=122.

 

Therefore, we have


arctan(122),


which means


cot(arctan122)=22.

 

Hence, the final answer is:
22.

 Jul 2, 2024

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