Complex number with a remainder... :O
Algebra
Level
4
Let z be a complex number with |z| = 2014. Let P be the polygon in the complex plane whose vertices are z and every w such that 1/(z+w) = 1/z + 1/w. Then the area enclosed by P can be written in the form n*sqrt(3), where n is an integer. Find the remainder when n is divided by 1000.
The answer is 147.
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1 solution
Solution:
|z| = 2014 => z lies on a circle with radius 2014
1/(z+w) = 1/z + 1/w = (z+w)/(zw)
=> zw = (z+w)² = z² + 2zw +w² => w² + zw + z² = 0
Now any solution w0 found for w for given real z0 = 2014 + 0i will yield equivalent solutions w by rotating w0 by an angle A such that z = 2014 e^(iA). So using z0 in our equation, we have
w² + 2014w + 2014² = 0
=> w = [-2014 ± √(2014²-4*2014²)]/2
=> conjugate roots w, w' = 2014 (-1/2 ± i√3/2)
So |w'| = |w| = 2014 √[(-1/2)²+(3/2)²) = 2014
=> Roots w and w' lie on the same circle.
arg(w) = arg([-1/2,√3/2]) = 120
arg(w') = -arg(w) = -120
So (z0,w,w') are the vertices of an equilateral triangle T inscribed in a circle of radius 2014.
Hence area(T) = 3 (2014² sin 30 cos 30) = 3 (2014² sin 60)/2 = 2014²/4 3√3 = 3*1007² √3 = n √3
Therefore n = 3 * 1007² ≡ 3*7² ≡ 147 [1000] So the remainder is 147.