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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)

```
3 years ago Saurabh Koranglekar
10341 Points
```							Dear student(a-ib)m/n is expected to be(a-ib)^m/nKindly verify and repost the question in standard notationRegards
```
9 months ago
```							LHS=( a + ib)m/n + ( a - ib)m/n Let, a = r cosΦ       b = r sinΦThen, 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
```
2 months ago
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