# Math complex numbers.

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 $i=\sqrt{-1}$.

-
possible duplicate – mez Feb 8 '13 at 8:45
Where is the duplicated question please? – Enve Feb 8 '13 at 8:45
I do not recall the exact one, but this is similar question that has all the answer you might seek. math.stackexchange.com/questions/49169/… – mez Feb 8 '13 at 8:46
$\sqrt{-4}$ are the complex numbers $x$ that satisfy $x^2=-4$. Thus $x=2i$ or $x=-2i$. – Stefan Hansen Feb 8 '13 at 8:47
@StefanHansen also x = -2i – mez Feb 8 '13 at 8:48

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 $-2i$?)

-
Ok. Thank @mixedmath I understood now. – Enve Feb 8 '13 at 8:55

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.

-
Thanks. This answer is correct. But I accepted @mixedmath answer. He was first. – Enve Feb 8 '13 at 9:01
You don't need to explain your decision as one chooses whatever answer seems better, but I thank you for doing it. – DonAntonio Feb 8 '13 at 9:10
You're being a bit careless with the equivalence arrows here, since also $x=-\sqrt{a}$ satisfies $x^2=a$... – Hans Lundmark Feb 8 '13 at 10:15