What's wrong with this proof that $i^5=i$ implies $\sqrt{i}=0$?

Question

I just learn the complex number at school. When I was playing with it, a confused result appeared. $$\because i^4=1$$ $$\therefore i^5=i$$ Now we have $$\sqrt {i^5}=\sqrt i$$ $$i^\frac{5}{2}=i^2\cdot\sqrt i=-\sqrt i=\sqrt i$$ Which lead to $$2\sqrt i=0$$ $$\therefore \sqrt i=0$$ Where I do something wrong? Are there any rules doesn't work in the complex field?

Answer 1

The step where you went wrong was in taking the square root of both sides of $i = i^5$.

The fact is, square roots are a multi-valued function (and this is true for the reals too). Every complex number (except $0$) has exactly two square roots.

In the real case, we can get by by defining the positive square root function, which will be single-valued. In the complex case, it's not so easy to distinguish. It might seem "obvious" in some sense that, say $\sqrt{-1} = i$ and not $\sqrt{-1} = -i$ because that's often how $i$ is described to people at first. But, $i$ and $-i$ are both equally valid square roots of $-1$, and neither is more "positive" than the other. In fact, the distinction between $i$ and $-i$ is completely arbitrary. You can replace $i$ with $-i$, and complex numbers will work basically the same way.

If that still doesn't convince you, note that $(-1 + i)^2 = (1 - i)^2 = -2i$, so $-1 + i$ and $1 - i$ are both square roots. Which one is the value of $\sqrt{-2i}$? By what rule do we choose which is "the" square root?

So, the problem is, you took the same number, expressed in two ways ($i$ and $i^5$), and took "the" square root of this number, assuming that it is possible to do this in a well-defined way.

People will say that rules like $\sqrt{ab} = \sqrt{a}\sqrt{b}$ fail to hold in complex space, which is true in a sense, but I think it's more accurate to say that $\sqrt{\;}$ doesn't exist; only $\pm \sqrt{\;}$ exists.

Answer 2

You made the standard assumption that

$$x^{a+ b} = x^a\cdot x^b$$

is always true, when in fact, it is only always true for positive real numbers.

So, your mistake is in writing $$i^\frac52=i^2\cdot\sqrt{i}.$$ This equality does not in fact hold.