Complex number

Question

The real values of $(a+ib)^{\dfrac{1}{n}} + (a-ib)^{\dfrac{1}{n}}$ is

Comments

Here , Complex Number z = (a+ib)^1/n + (a-ib)^1/n

Now , Let , a = r cosθ  and b = r sinθ , It implies a² + b² = r²

put 'a' and 'b' in z ,

We get ,  z = (rcosθ + irsinθ)^1/n + (rcosθ - rsinθ)^1/n

Now , Apply De Moivre's theorem ie (cos θ + i sin θ)ⁿ = cos n θ + i sin n θ ,

So ,

z = r^1/n(cos θ/n + isin  θ/n ) + r^1/n (cos θ/n - i sin  θ/n)

z = r^1/n  (cos θ/n + isin  θ/n + cos θ/n - isin  θ/n)

z = r^1/n (2cos θ/n)

So , Real part will be ,Re(z) = 2r^1/n cos θ/n..

Now , put values of r and  θ,

Re(z) = 2 (sqrt(a² + b²))^1/n * cos (cos^-1(a/r))/n

Re(z) = 2 (sqrt(a² + b²))^1/n * cos (cos^-1(a/sqrt(a² + b²)))/n

Please correct me if I am wrong..