# Thread: Simple complex number question

1. ## Simple complex number question

Why does sqrt(-6) * sqrt(-6) = sqrt(36) = 6 not work? I know the answer is = sqrt(6)i * sqrt(6)i = -6, but why does the first one not work?

2. ## Re: Simple complex number question

Probably because sqrt(x) * sqrt(y) =sqrt(xy) only works when x,y>0

3. ## Re: Simple complex number question

Originally Posted by pikachu975
Why does sqrt(-6) * sqrt(-6) = sqrt(36) = 6 not work? I know the answer is = sqrt(6)i * sqrt(6)i = -6, but why does the first one not work?
Once we enter the complex world, the square root function is multi-valued (in fact, double-valued) (except when z = 0, in which case sqrt(z) has only one value). (Actually also for real positive numbers, it'S multivalued). We need to specify something called a "branch" to make the square root single-valued. When we use the "√" symbol, we are referring to what is known as the principal square root. For a non-zero complex number z, the principal square root √(z) is defined as √(z) = √(|z|) * cis((1/2)*Arg(z)), where |z| > 0 is the modulus of z and Arg(z) is the principal argument of z (the unique argument of z in (-pi, pi]). Here √(|z|) is the positive square root of the modulus.

Equivalently, the principal square root of z is the square root of it that has positive real part (except when z is pure imaginary, when it refers to the square root with argument pi/2).

Using this definition of principal square roots, it is not generally true that √(wz) = √(w) * √(z).

What is true is that given any two values of sqrt(wz), sqrt(w) and sqrt(z), there is a value for the third square root so that sqrt(zw) = sqrt(z)*sqrt(w).

4. ## Re: Simple complex number question

Originally Posted by trecex1
Probably because sqrt(x) * sqrt(y) =sqrt(xy) only works when x,y>0
Expanding on that idea, we can prove in a different way why $\sqrt{-6 } \sqrt{-6} = 6$ is not correct.
let f(x) be our extended square root function, principal square root, where $(f(-1))^{2} = -1$, the definition of i,. we want to prove any function with that condition can not satisfy $f(x)f(y) = f(xy), \ \ \forall x \in \mathbb{C}.$
suppose
$f(-x)f(-y) =f(xy)$
letting x= y implies
$(f(-x))^{2} (f(-1)f(x))^{2} = (f(-1))^{2}(f(x))^{2} = f(x^{2}) \ \ \ \ (1)$
but we also know
$f(x)f(x) = (f(x))^{2} = f(x^{2}) \ \ \ \ (2)$
from (1) and (2) we get
$(f(-1))^{2} = 1$ which is a contradiction, assuming
$f(x) \neq 0, \forall x \in \mathbb{C}$

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