Q.If $n$ is a positive integer, show that: $(P + iQ)^{1/n} + (P - iQ)^{1/n} = 2(P^2 + Q^2)^{1/2n} \cdot \cos\left[\dfrac{1}{n}\tan^{-1}\dfrac{Q}{P}\right]$.

Convert $P\pm iQ$ to polar form, take the $n$th root using De Moivre's theorem, then add the two results — the sine parts cancel and the cosine parts double.

Write $P+iQ$ in polar (modulus-argument) form:

$P+iQ = r(\cos\theta+i\sin\theta), \quad \text{where } r=\sqrt{P^2+Q^2},\ \theta=\tan^{-1}\dfrac{Q}{P}$

Then $P-iQ$ is the conjugate, which in polar form is:

$P-iQ = r(\cos\theta-i\sin\theta)$

Taking the principal $n$th root of each (by De Moivre's theorem, $z^{1/n}=r^{1/n}\left(\cos\dfrac{\theta}{n}+i\sin\dfrac{\theta}{n}\right)$):

$(P+iQ)^{1/n} = r^{1/n}\left(\cos\dfrac{\theta}{n}+i\sin\dfrac{\theta}{n}\right)$

$(P-iQ)^{1/n} = r^{1/n}\left(\cos\dfrac{\theta}{n}-i\sin\dfrac{\theta}{n}\right)$

Adding these two: …