I have heard that $i=\sqrt{-1}$ and I have also read about it here http://www.mathsisfun.com/numbers/imaginary-numbers.html.
Now I want to ask why in example $\sqrt{-4} = 2i$ as $i=\sqrt{-1}$.
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possible duplicate – mez Feb 08 '13 at 08:45
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Where is the duplicated question please? – Idrizi.A Feb 08 '13 at 08:45
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I do not recall the exact one, but this is similar question that has all the answer you might seek. http://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1 – mez Feb 08 '13 at 08:46
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$\sqrt{-4}$ are the complex numbers $x$ that satisfy $x^2=-4$. Thus $x=2i$ or $x=-2i$. – Stefan Hansen Feb 08 '13 at 08:47
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@StefanHansen also x = -2i – mez Feb 08 '13 at 08:48
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It is because $(2i)^2 = 2^2 i^2 = 4 \cdot -1 = -4$. Thus $2i$ is a possible answer to $\sqrt{-4}$ (though perhaps not the only one. What about $-2i$?)
davidlowryduda
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If you remember the basic definition
$$x=\sqrt a\Longleftrightarrow x^2=a $$
then
$$(2i)^2=2^2i^2=-4\Longleftrightarrow\sqrt{-4}=2i$$
Of course, also $\,(-2i)\,$ makes the job.
DonAntonio
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1Thanks. This answer is correct. But I accepted @mixedmath answer. He was first. – Idrizi.A Feb 08 '13 at 09:01
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You don't need to explain your decision as one chooses whatever answer seems better, but I thank you for doing it. – DonAntonio Feb 08 '13 at 09:10
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1You're being a bit careless with the equivalence arrows here, since also $x=-\sqrt{a}$ satisfies $x^2=a$... – Hans Lundmark Feb 08 '13 at 10:15