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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)

Grade:12th pass

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
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
15 Points
3 years ago
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 = rm/n(cosΦ + isinΦ)m/n
                                   = rm/n{cos[(m/n)Φ] + i sin[(m/n)Φ]}
                                                        {using De-Moivre's Theorem}
& (a - ib)m/n = rm/n {cos[(m/n)Φ] - i sin[(m/n)Φ]}
now, (a + ib)m/n + (a - ib)m/n = rm/n{2 Cos[(m/n)Φ]}
                                               = 2(a² + b²)m/2n Cos[(m/n)tan-1(b/a)]
      Proved

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