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# vishal's doubt made me to write a note

$\sqrt{-1}\times\sqrt{-1}$ $case (1)=>\sqrt{-1}\times\sqrt{-1}=\sqrt{-1\times-1}=\sqrt{1}=1$ $case(2)=>\sqrt{-1}\times\sqrt{-1}=(\sqrt{-1})^2=-1$ which one is correct either first case or the second case can anyone say please calvin lin can you say it

Note by Sudoku Subbu
2 years, 11 months ago

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$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$$ if and only if $$a,b> 0$$

- 2 years, 11 months ago

Case 2 is correct...

Since $$\sqrt{-1} = i$$, then, the equation above is simply equal to $$i^{2} = -1$$.. :)

- 2 years, 11 months ago

Case 1 is incorrect since you cannot split the square-root if both radicands are negative.

- 2 years, 11 months ago

mr.subbu case 2 is correct

- 2 years, 2 months ago