Let z1,z2,…,z20 be the twenty (complex) roots of the equationz20−4z19+9z18−16z17+⋯+441=0.
Calculate cot(∑20k=1arccotzk).Note that the addition formula for cotangent is still valid when working with complex numbers.
Given the roots z1,z2,…,z20 of the polynomial equation
z20−4z19+9z18−16z17+⋯+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+⋯+z201−z1z2⋯z20).
Given that arccotzk=arctan1zk, it follows that:
20∑k=1arccotzk=20∑k=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−(1z1⋅1z2⋯1z20)).
Now, we need to analyze the roots of the given polynomial z20−4z19+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:
z1z2⋯z20=(−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(20∑k=1arccotzk)=cot(arctan(∑20k=11zk1−1z11z2⋯1z20)).
Since ∑20k=1zk=4 and ∏20k=1zk=441,
20∑k=11zk=1z1+1z2+⋯+1z20=20441.
Then:
1−20∏k=11zk=1−1441=440441.
Substituting in,
20441440441=20440=122.
Therefore, we have
arctan(122),
which means
cot(arctan122)=22.
Hence, the final answer is:
22.