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I am struggling to find out why the following equality is wrong, $$\sqrt{-i}=i\sqrt{i}$$ I know the above equality is wrong from the following derivation, $$\text{L.H.S.}=\sqrt{-i}=\sqrt{e^{-i\pi/2}}=e^{-i\pi/4}=\frac{1}{\sqrt{2}}(1-i)$$ Whereas $$\text{R.H.S.}=i\sqrt{i}=i\sqrt{e^{i\pi/2}}=ie^{i\pi/4}=i\frac{1}{\sqrt{2}}(1+i)=-\frac{1}{\sqrt{2}}(1-i)$$

Thus $$\text{L.H.S.}\neq \text{R.H.S.}$$

Which is true, however, if we do the calculation in the following way, $\sqrt{-i}=\sqrt{(-1)\times i}=i\sqrt{i}$, then the identity seems to be correct. What am I missing here? I know I have done some silly mistake, Somebody can help me?

Blue
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Dhiman Bhowmick
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    The square root is multi-valued. – Randall Feb 20 '23 at 02:02
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    Almost all of the identities concerning exponents break down in some way when you stop limiting the bases to positive real numbers. You've used $(ab)^c =a^c b^c$, and kaboom! – JonathanZ Feb 20 '23 at 02:07
  • for the same reason $-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{1} = 1$ doesn't work. Square roots can be defined for positive real numbers, but there isn't really a way of making them work for complex numbers. – Zoe Allen Feb 20 '23 at 02:12
  • Cf. this question – J. W. Tanner Feb 20 '23 at 02:16
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    @ZoeAllen - I'd say that we can make square roots work for complex numbers - it's kind of a lot of what Riemann surfaces do. It's just that they're a lot more finicky that square roots of positive reals, and we have to careful that we don't (implicitly) assume they work the same way. – JonathanZ Feb 20 '23 at 02:16
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    Essentially a duplicate of https://math.stackexchange.com/q/438/42969 – Martin R Feb 20 '23 at 02:44

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The equality $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds for non-negative real numbers, but not beyond. If this were to hold, you would have $$ -1=i^2=\sqrt{-1}\sqrt{-1}\stackrel?=\sqrt{(-1)(-1)}=\sqrt1=1. $$ Part of the problem is that the square root is multi-valued. In the case of positive numbers there is a canonical way of choosing the root, but not with other numbers.

This also shows that it is a fairly bad practice to say that $i=\sqrt{-1}$. It is way better to avoid the square root and just say that $i^2=-1$.

Martin Argerami
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  • Thanks, I understand that the square root of a negative number can be multivalued and that is why the rule $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for negative numbers does not hold. I understand this in the following way, $$\sqrt{-1}=\sqrt{e^{i\pi}}=e^{i\pi/2}=i$$ But We can also write down, $$\sqrt{-1}=\sqrt{e^{i3\pi}}=e^{i3\pi/2}=-i$$ Thus $\sqrt{-1}$ is actually multivalued and it is not correct to say $\sqrt{-1}=i$. But somehow the opposite identity is correct that is $i=\sqrt{-1}$. – Dhiman Bhowmick Feb 21 '23 at 01:33