Why this is false for complex numbers?

Question

Why this statement is false to $a \in \mathbb{C}$

$(\sqrt[n]{a} * \sqrt[k]{a} ) - (a^{\frac{n+k}{nk}})= 0$

How you can prove it with high school maths ?

Answer 1

The problem arises because of $$(\sqrt{-1})^2=-1$$ and that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is not true if $a,b<0$. If it was true, we would get: $$1=\sqrt{1}=\sqrt{(-1)\cdot (-1)}=\sqrt{-1}\sqrt{-1}=-1$$

Answer 2

$$(\sqrt[n]{a} * \sqrt[k]{a} ) - (a^{\frac{n+k}{nk}})= 0$$

is not always true for complex numbers because $\sqrt[n]{a}$ is not uniquely defined.

In fact in complex numbers you have $n$ different roots of $1$ for each $n$ which makes it more interesting.