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Often complex numbers are descirbed as $$\mathbb{C} = {\{a + bi|a,b\in\mathbb{R},i^2=-1\}}$$

Since $i^2 = -1$, I would assume that $i = \pm \sqrt{-1}$.

My math said that $\sqrt{-1}$ doesn't exist because of $$-1 = \sqrt{-1} \times\sqrt{-1} = \sqrt{(-1)^2} = \sqrt{1} = 1$$

But of course $-1 \neq 1$. Equivalently you could do $$-1 = (-\sqrt{-1}) \times (-\sqrt{-1}) = \sqrt{(-1)^2} = \sqrt{1} = 1$$

which gives the same result.

Could someone explain what I'm missing? Thanks in advance.

Kevin
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  • $i=\pm \sqrt{-1}$ is incorrect. The correct is $i=\sqrt{-1}$ – msm Sep 13 '16 at 10:26
  • Okay thanks. However if $i = \sqrt{-1}$ to my knowledge $-1 = \sqrt{-1} \times\sqrt{-1} = \sqrt{(-1)^2} = \sqrt{1} = 1$ still holds. – Kevin Sep 13 '16 at 10:27
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    $\sqrt a \sqrt b = \sqrt{ab}$ is only valid if $a$ and $b$ are nonnegative. Otherwise, you must write $\sqrt{-1}\sqrt{-1} = (\sqrt{-1})^2 = -1$ in this case. – Decaf-Math Sep 13 '16 at 10:27
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    That definition of complex numbers is not sufficient. Besides $i^2=-1$ is enough, no need to express $i$ like that. Did you check out the popular question http://math.stackexchange.com/questions/438/why-sqrt-1-times-1-neq-sqrt-12? – StubbornAtom Sep 13 '16 at 10:28
  • The point is that $-1$ is in fact a square root of $1$ as $(-1)^2$. However, when applied to positive reals the square-root symbol will typically only be taken to denote the positve one. Further details in the proposed duplicate. – quid Sep 13 '16 at 10:47
  • You just rediscovered that $(\sqrt z)^2=\sqrt{z^2}$ doesn't necessarily hold in the complex numbers. –  Sep 13 '16 at 11:02

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$$\sqrt{-1} \times\sqrt{-1} = \sqrt{(-1)^2}$$

This holds only for non-negative real numbers, and not for complex numbers in general.

Actually, complex multiplication is defined by: $$(a,b)\times(c,d)=(ac-bd,ad+bc)$$ for any complex number $(a,b)$ and $(c,d)$.

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