# Value of $i^2$ in complex numbers [duplicate]

Please solve this doubt : we know that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ and $i^2 = -1$. But $i= \sqrt{-1}$ which implies that $i^2 = i \cdot i = \sqrt{-1}\sqrt{-1} = \sqrt{1} = 1$ that is $i^2 = 1$. So what is the correct value of $i^2$? Is it $1$ or is it $-1$? Please explain.

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## marked as duplicate by mrf, Hans Lundmark, Sami Ben Romdhane, John, J. W. PerryApr 16 '14 at 8:55

$\sqrt{a} \sqrt{b} = \sqrt{ab}$ iff $a$ and $b$ are nonnegative. –  01000100 Apr 16 '14 at 7:14
How are you defining $i$, akanksha? If, like the rest of us, you are defining it by $i^2=-1$, then of course $i^2=-1$. –  Gerry Myerson Apr 16 '14 at 7:17
@akanksha Can't multiply $\sqrt{-1}\sqrt{-1}$ and get $\sqrt{1}$. Can't multiply radicals of negative numbers of any kind in general. –  dragon Apr 16 '14 at 7:20
$$i=\sqrt {-1}$$ is not entirely true. Since there are two numbers, both $-i$ and $i$ that satisfy $x^2=-1$, making the function $f(x)=\sqrt{x}$ more complicated to define for numbers outside $[0,\infty)$.