# Thread: Does âˆš-1 evaluate to i or -i?

1. ## Does âˆš-1 evaluate to i or -i?

When we use take the square root using the radical sign âˆš in the reals we take only the positive root.

So $\sqrt{64} = 8$ and not $-8$.

But what does the radical sign âˆš actually mean when used on complex numbers?

What does $\sqrt{-1}$ evaluate to? $i$ or $-i$?

2. ## Re: Does âˆš-1 evaluate to i or -i?

The former, so sqrt(-1) = i, and 1 = -i

3. ## Re: Does âˆš-1 evaluate to i or -i?

Why is it $i$ and not $-i$?

4. ## Re: Does âˆš-1 evaluate to i or -i?

Originally Posted by fan96
Why is it $i$ and not $-i$?
I'm not familiar at all with the complex mathematical background of it, but the general idea is that it's impossible to square root a negative number i.e. with the definition of a square root where the result squared itself is the number inside of the square root (a bit confusing, but it leads), you therefore need imaginary numbers, which we use $i$ to describe. Don't think too much into it, there is not always a logic to why it happens, rather it is a convenient definition used.

5. ## Re: Does âˆš-1 evaluate to i or -i?

There is no purpose to make sqrt(1) a definition as it is clearly already = 1. It's more convenient to use $i$ and makes more sense to do that. Since sqrt(-1) doesn't exist without the use of imaginary numbers, $i$ can be used to define and introduce an impossible number as a variable and therefore allow expansion across equations that use imaginary numbers.

6. ## Re: Does âˆš-1 evaluate to i or -i?

Originally Posted by fan96
When we use take the square root using the radical sign âˆš in the reals we take only the positive root.

So $\sqrt{64} = 8$ and not $-8$.

But what does the radical sign âˆš actually mean when used on complex numbers?

What does $\sqrt{-1}$ evaluate to? $i$ or $-i$?
A lot of textbooks seem to define $i=\sqrt{-1}$. However, the actual definition is just $i^2=-1$ (that 'i' is this mysterious thing that squares to -1). Radical signs should ideally be avoided when denoting 'i' as square roots of negative numbers do not follow the standard properties of radicals/surds (treating 'i' like a pronumeral in arithmetic operations typically avoids this).

7. ## Re: Does âˆš-1 evaluate to i or -i?

Originally Posted by fan96
When we use take the square root using the radical sign âˆš in the reals we take only the positive root.

So $\sqrt{64} = 8$ and not $-8$.

But what does the radical sign âˆš actually mean when used on complex numbers?

What does $\sqrt{-1}$ evaluate to? $i$ or $-i$?
Basically the convention with the "âˆš" symbol for complex numbers is to take the root that has its principal argument in the range (-pi/2, pi/2]. (For any non-zero complex number z, there is a unique square root of z with principal argument in this range. This square root is called the principal square root of z.)

I don't know if the HSC adheres to this convention though, and you don't really need to know it for the HSC either I think.

8. ## Re: Does âˆš-1 evaluate to i or -i?

The radical symbol (âˆš) is just square root i.e. to the power of 1/2. In the complex field, âˆš-1, its just i, otherwise i.e. in the real number field, it is impossible to square root negative numbers, hence we write no solution. As for -âˆš1, its -i, much like 1 and -1. In reality, we use i was invented because in the past, imaginary numbers were taboo and people didn't like the thought of square rooting a negative number, pioneers of mathematics and complex numbers decided to make i denote âˆš-1. Nowadays its just to follow convention plus its easier and much neater to write -8i than âˆš-64. In any case, we only really convert from âˆš-1 to i when simplifying the square root of a number like after using the quadratic equation and to put numbers in a+bi form.

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