(x+a)n=nC0xn+nC1xn−1a+nC2xn−2a2+nC3xn−3a3+⋯ =T0+T1+T2+T3+⋯
In (1) replacing a by ai (x+ai)n=nC0xn+nC1xn−1ai+nC2xn−2(ai)2+nC3xn−3(ai)3+⋯ =(nC0xn−nC2xn−2a2+nC4xn−4a4−⋯) +i(nC1xna−nC3xn−3a3+nC5xn−5a5−⋯) =(T0−T2+T4−⋯)+i(T1−T3+T5−⋯)
Taking modulus on both sides and squaring ∣x+ai∣2n=∣(T0−T2+T4−⋯)+i(T1−T3+T5−⋯) ⇒(x2+a2)n=(T0−T2+T4−⋯)2+(T1−T3+T5−⋯)2