# How to make the nth root of a product act the same as simple multiplication in regard to parity?

I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors.

I've done some initial research, and reviews these Stack questions: Square roots — positive and negative and The Product Rule of Square Roots with Negative Numbers but I couldn't find the information I was seeking (or am not fully understanding the answers.)

Are the following expressions true? If not, how can I produce the those results?

$$\sqrt[2]{1*-1} = -1$$

$$\sqrt[3]{1*1*-1} = -1$$

$$\sqrt[3]{1*-1*-1} = 1$$

[update] This is what the function does:

$$\sqrt[n]{\overline{\Delta_1}*\overline{\Delta_2} *...*\overline{\Delta_n}} \text{ }*\text{ } \frac{\overline{\Delta_1}*\overline{\Delta_2} *...*\overline{\Delta_n}}{\Delta_1*\Delta_2*...*\Delta_n}$$

such that if there are an odd number of negative factors, the product is negative, otherwise positive.

• Is there a more compact way to express this?

also, any tips on notation are appreciated.

• $\sqrt{-1} =i$, so the first statement is wrong. – Larry Nov 26 '18 at 18:04

We have that

• $$\sqrt[2]{1\times(-1)} = \sqrt{-1} \neq -1\quad \color{red}\checkmark$$ indeed $$-1\times -1=1$$
• $$\sqrt[3]{1\times1\times(-1)} = \sqrt[3]{-1} = -1\quad \color{green}\checkmark$$ indeed $$-1\times -1\times -1=-1$$
• $$\sqrt[3]{1\times(-1)\times(-1)} = \sqrt[3]{1}=1\quad \color{green}\checkmark$$ indeed $$1\times 1\times 1=1$$

As a general rule

• for $$n\in \mathbb{N}$$ even and $$a\ge 0$$ we have

$$\sqrt[n] a=b \iff b\ge 0 \quad b^n=a$$

• for $$n\in \mathbb{N}$$ odd and we have

$$\sqrt[n] a=b \iff b^n=a$$

I am not sure if this is the method you want.

We have Euler's formula $$e^{i\theta} = i\sin\theta+\cos\theta$$ We can take $$n^{th}$$ root of both sides to obtain $$\sqrt[n]{e^{i\theta}}=\sqrt[n]{i\sin\theta+\cos\theta}\tag{1}$$

It seems like you are only asking for the cases for $$1$$ and $$-1$$, so let's do the following.

It is clear that $$\sqrt[n]{1}$$ is $$1$$ or $$-1$$ regardless of whether $$n$$ is odd or even. The problem is how to figure out $$\sqrt[n]{-1}$$. For $$n$$ is odd, $$\sqrt[n]{-1}=-1$$.

For $$n$$ is even, let's suppose $$\theta = \pi$$. Then, from $$(1)$$ we have $$\sqrt[n]{e^{i\pi}}= \sqrt[n]{-1}$$ $$e^{i\frac{\pi}{n}} = \sqrt[n]{-1}$$ $$\sqrt[n]{-1}= i\sin\frac{\pi}{n}+\cos\frac{\pi}{n}$$ I apologize if this is not what you are looking for.