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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?

rschwieb
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Yuanming Luo
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  • Did you learn as well roots of complex numbers? Beware with roots and all power functions with non integers arguments, the devil is always in the details. – Martigan Sep 28 '18 at 13:19
  • Your line four (of equations) makes no sense. Also, taking square roots of complex numbers is not the same as taking square roots of real numbers. – AnyAD Sep 28 '18 at 13:21
  • Thanks, but which property of complex number leads that? – Yuanming Luo Sep 28 '18 at 13:25
  • @YuanmingLuo Some unsolicited advice: don't use $\therefore$ or $\because$ in mathematical writing. There are probably some mathemeticians who use them well, but I think exceedingly few mathematicians like or use that notation. Just use words. Also, please think about using titles that are more useful. I took care of it for you this time. – rschwieb Sep 28 '18 at 13:28
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    @rschwieb Thanks, I will notice that. – Yuanming Luo Sep 28 '18 at 13:30

2 Answers2

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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.

Theo Bendit
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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.

5xum
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