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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1$\sqrt{a} \sqrt{b} = \sqrt{ab}$ iff $a$ and $b$ are nonnegative. – Daniel Pietrobon Apr 16 '14 at 07:14
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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 07:17
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@akanksha Can't multiply $\sqrt{-1}\sqrt{-1}$ and get $\sqrt{1}$. Can't multiply radicals of negative numbers of any kind in general. – Cookie Apr 16 '14 at 07:20
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$$i=\sqrt {-1}$$ 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 $[0,\infty)$.
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