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若非零复数 x,yx, y 满足 x2+xy+y2=0x^{2}+x y+y^{2}=0 ，则 (xx+y)2020+\left(\frac{x}{x+y}\right)^{2020}+ $\le...
Mar 8, 2024
若非零复数 x,yx, y 满足 x2+xy+y2=0x^{2}+x y+y^{2}=0 ，则 (xx+y)2020+\left(\frac{x}{x+y}\right)^{2020}+ (yx+y)2020\left(\frac{y}{x+y}\right)^{2020} 的值是
Solution by Steps
step 1
Given the equation x2+xy+y2=0x^2 + xy + y^2 = 0, we can factor it as (x+ωy)(x+ω2y)=0(x + \omega y)(x + \omega^2 y) = 0, where ω\omega and ω2\omega^2 are the complex cube roots of unity
step 2
Since xx and yy are non-zero complex numbers, we can deduce that x=−ωyx = -\omega y or x=−ω2yx = -\omega^2 y
step 3
Substituting x=−ωyx = -\omega y into (xx+y)2020+(yx+y)2020\left(\frac{x}{x+y}\right)^{2020} + \left(\frac{y}{x+y}\right)^{2020}, we get (−ωy(1−ω)y)2020+(y(1−ω)y)2020\left(\frac{-\omega y}{(1-\omega)y}\right)^{2020} + \left(\frac{y}{(1-\omega)y}\right)^{2020}
step 4
Simplifying the expression from step 3, we find that both terms are equal to (11−ω)2020\left(\frac{1}{1-\omega}\right)^{2020}
step 5
Since 1+ω+ω2=01 + \omega + \omega^2 = 0, we have 1−ω=−(ω+ω2)=ω21 - \omega = -(\omega + \omega^2) = \omega^2. Thus, (11−ω)2020=ω4040\left(\frac{1}{1-\omega}\right)^{2020} = \omega^{4040}
step 6
Observing that ω3=1\omega^3 = 1, we can simplify ω4040\omega^{4040} to ω(3⋅1346+2)=ω2\omega^{(3 \cdot 1346 + 2)} = \omega^2
step 7
Therefore, the value of (xx+y)2020+(yx+y)2020\left(\frac{x}{x+y}\right)^{2020} + \left(\frac{y}{x+y}\right)^{2020} is 2ω22\omega^2
Answer
2ω22\omega^2
Key Concept
Complex cube roots of unity and their properties
Explanation
The equation x2+xy+y2=0x^2 + xy + y^2 = 0 can be factored using the complex cube roots of unity, leading to a simplification of the given expression when xx and yy are substituted accordingly. The properties of the cube roots of unity, such as 1+ω+ω2=01 + \omega + \omega^2 = 0 and ω3=1\omega^3 = 1, are essential in simplifying the expression to the final answer.
m^2+m+1=0, m^1010+1/m^1010=?
Solution by Steps
step 1
Given the equation m2+m+1=0 m^2 + m + 1 = 0 , we can find the value of m1010+1m1010 m^{1010} + \frac{1}{m^{1010}} by using the properties of complex numbers
step 2
Since m m is a root of m2+m+1=0 m^2 + m + 1 = 0 , we can use the fact that m3=1 m^3 = 1 for any complex root of this equation
step 3
We can express m1010 m^{1010} as (m3)336⋅m2 (m^3)^{336} \cdot m^2 because 1010=3⋅336+2 1010 = 3 \cdot 336 + 2
step 4
Since m3=1 m^3 = 1 , m1010=m2 m^{1010} = m^2 . Similarly, 1m1010=1m2 \frac{1}{m^{1010}} = \frac{1}{m^2}
step 5
We can now add m2 m^2 and 1m2 \frac{1}{m^2} to find m1010+1m1010 m^{1010} + \frac{1}{m^{1010}}
step 6
Using the original equation m2+m+1=0 m^2 + m + 1 = 0 , we can express m2 m^2 as −m−1 -m - 1
step 7
Substituting m2 m^2 with −m−1 -m - 1 in m1010+1m1010 m^{1010} + \frac{1}{m^{1010}} , we get −m−1+1−m−1 -m - 1 + \frac{1}{-m - 1}
step 8
To simplify −m−1+1−m−1 -m - 1 + \frac{1}{-m - 1} , we need a common denominator, which is −m−1 -m - 1
step 9
The expression becomes (−m−1)2+1−m−1 \frac{(-m - 1)^2 + 1}{-m - 1}
step 10
Expanding (−m−1)2 (-m - 1)^2 gives m2+2m+1 m^2 + 2m + 1
step 11
Substituting m2 m^2 with −m−1 -m - 1 in m2+2m+1 m^2 + 2m + 1 , we get −m−1+2m+1 -m - 1 + 2m + 1
step 12
Simplifying −m−1+2m+1 -m - 1 + 2m + 1 results in m m
step 13
Now we have m+1−m−1 \frac{m + 1}{-m - 1}
step 14
Simplifying the fraction m+1−m−1 \frac{m + 1}{-m - 1} gives us −1 -1 since m+1 m + 1 is the negative of the denominator
Answer
m1010+1m1010=−1 m^{1010} + \frac{1}{m^{1010}} = -1
Key Concept
Complex Roots of Unity and Symmetry in Powers
Explanation
The key concept is using the properties of complex roots of unity, where m3=1 m^3 = 1 for roots of m2+m+1=0 m^2 + m + 1 = 0 , to simplify high powers of m m and find the sum of a power and its reciprocal.
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