A simple question related to Complex Numbers?

Question

Ok so this was the equation given in my text book $$\implies\sqrt{-a}\sqrt{-a} $$$$= (-1)a $$$$= -a $$

so my question is why can't i solve it this way $$\implies\sqrt{-a}\sqrt{-a}$$$$=\sqrt{(-a)(-a)}$$$$=\sqrt{a^2}$$$$=a$$

so what is wrong with my approach can anyone explain

Thanks

Akash

This should help http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number?lq=1 — ir7, Feb 17 '14 at 03:43
What's wrong is when you say that $\sqrt{a^2}=a$. For real numbers $a$, that equation is only true if $a\ge0$. Instead, this is what's true for any real number $a$: $\sqrt{a^2}=|a|$ — Steve Kass, Feb 17 '14 at 03:58

score 2 · Answer 1 · answered Feb 17 '14 at 04:10

When dealing with complex numbers, for integer values of n, $\sqrt[n]z$ is not a single number, but rather a set of n numbers, each of which has the property that its n-th power is z. For instance, $\sqrt1=\pm1$, $\sqrt[3]1=\left\{1,\dfrac{-1\pm i\sqrt3}2\right\}$, $\sqrt[4]1=\{\pm1,\pm i\}$, etc. In other words, for complex numbers, the n-th root is a binary relation rather than an actual function. Which is why the property that the n-th root of a product is the same as the product of n-th roots, ultimately no longer holds true anymore.

A simple question related to Complex Numbers?

1 Answers1