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若非零复数

x, y

满足

x^{2}+x y+y^{2}=0

，则 $\left(\frac{x}{x+y}\right)^{2020}+\left(...

Mar 8, 2024

若非零复数

x, y

满足

x^{2}+x y+y^{2}=0

，则

\left(\frac{x}{x+y}\right)^{2020}+\left(\frac{y}{x+y}\right)^{2020}

的值是

Solution by Steps

step 1

Given the equation

x^2 + xy + y^2 = 0

, solve for

y

in terms of

x

step 2

From the asksia-ll calculation list, the solutions for

y

are

y = -\frac{1}{2} i (\sqrt{3} - i)x

and

y = \frac{1}{2} i (\sqrt{3} + i)x

step 3

Notice that both solutions for

y

are multiples of

x

. This means that

x

and

y

are linearly dependent over the complex numbers

step 4

Since

x

and

y

are non-zero complex numbers and

y

is a multiple of

x

\frac{x}{x+y}

and

\frac{y}{x+y}

are complex conjugates of each other

step 5

The magnitude of a complex number is equal to the magnitude of its complex conjugate. Therefore,

\left|\frac{x}{x+y}\right| = \left|\frac{y}{x+y}\right|

step 6

Raising a complex number and its conjugate to an even power results in two real numbers that are equal. Thus,

\left(\frac{x}{x+y}\right)^{2020} = \left(\frac{y}{x+y}\right)^{2020}

step 7

The sum of two equal numbers is twice one of the numbers, so

\left(\frac{x}{x+y}\right)^{2020} + \left(\frac{y}{x+y}\right)^{2020} = 2\left(\frac{x}{x+y}\right)^{2020}

step 8

Since

x

and

y

are non-zero complex numbers satisfying the given equation,

x+y

cannot be zero. Therefore,

\left(\frac{x}{x+y}\right)^{2020}

is a well-defined non-zero real number

step 9

The value of

\left(\frac{x}{x+y}\right)^{2020}

is a real number whose exact value is not needed to find the sum. We denote it as

k

step 10

The sum is

2k

, where

k

is the real number from the previous step

Answer

2k

, where

k

is the real number

\left(\frac{x}{x+y}\right)^{2020}

Key Concept

Complex Conjugates and Their Properties

Explanation

When a non-zero complex number and its conjugate are raised to an even power, the results are equal real numbers. The sum of these equal numbers is simply twice one of the numbers.