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Question

# Let A1,A2,⋯,An be the vertices of a regular polygon of n sides in a circle of radius unity and a=|A1A2|2+|A1A3|2+⋯|A1An|2, b=|A1A2||A1A3|⋯|A1An|, then ab=

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Solution

## Let us assume that O is the centre of the regular polygon and z0,z1,⋯,zn−1 represent the affixes of A1,A2,⋯,An, such that z0=1,z1=α,z2=α2,⋯,zn−1=αn−1, where α=ei2π/n (Since all nth roots of unity can be represented as verticies of regular polygon inscribed in a circle of unit radius) Now, |A1Ar|2=|αr−1|2=|1−αr|2 =∣∣∣1−cos2rπn−isin2rπn∣∣∣2 =(1−cos2rπn)2+(sin2rπn)2 =2−2cos2rπn ∴n∑r=2|A1A2|2=n∑r=1(2−2cosrπn) =2(n−1)−2[cos2πn+cos4πn+⋯cos2(n−1)πn] =2(n−1)−2 Real part of (α+α2+⋯+αn−1) =2(n−1)−2(−1)=2n [∵1+α+α2+⋯+αn−1=0] ∴|A1A2|2+|A1A3|2+⋯|A1An|2=2n Also, let E=|A1A2||A1A3|⋯|A1An| =|1−α||1−α2||1−α3|⋯|1−αn−1| =|(1−α)(1−α2)(1−α3)⋯(1−αn−1)| Since, 1,α,α2,⋯,αn−1 are the roots of zn−1=0 ⇒(z−1)(z−α)(z−α2)⋯(z−αn−1)=zn−1 ⇒(z−α)(z−α2)⋯(z−αn−1)=zn−1z−1 =1+z+z2+⋯+zn−1 Substituting z=1, we have =(1−α)(1−α2)(1−α3)⋯(1−αn−1)=n ∴|1−α||1−α2||1−α3|⋯|1−αn−1|=n Hence, the value of ab=2nn=2

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