prove that (a+ib)^m/n+ (a-ib)m/n= 2(a^2+b^2)^m/2ncos(m/ntan^-1a/b)

Question

Saurabh Koranglekar · Answer

Dear student (a-ib)m/n is expected to be(a-ib)^m/n Kindly verify and repost the question in standard notation Regards

Owa See · Answer

LHS=( a + ib) m/n + ( a - ib) m/n Let, a = r cosΦ b = r sinΦ Then, r = √(a² + b²) & Φ = tan - ¹(b/a) Now, (a + ib) m/n = r m/n (cosΦ + isinΦ) m/n = r m/n {cos[( m / n )Φ] + i sin[( m / n )Φ]} {using De-Moivre's Theorem} & (a - ib) m/n = r m/n {cos[( m / n )Φ] - i sin[( m / n )Φ]} now, (a + ib) m/n + (a - ib) m/n = r m/n {2 Cos[( m / n )Φ]} = 2(a² + b²) m/2n Cos[( m /n)tan -1 ( b / a )] Proved

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prove that (a+ib)^m/n+ (a-ib)m/n= 2(a^2+b^2)^m/2ncos(m/ntan^-1a/b)

prove that (a+ib)^m/n+ (a-ib)m/n= 2(a^2+b^2)^m/2ncos(m/ntan^-1a/b)

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