Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $i=\sqrt{-1}$.

share|improve this question
    
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
add comment

2 Answers 2

up vote 2 down vote accepted

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 $-2i$?)

share|improve this answer
    
Ok. Thank @mixedmath I understood now. –  Enve Feb 8 '13 at 8:55
add comment

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.

share|improve this answer
1  
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.