Question

What is the square root of -49?

Answer 1

`sqrt(-49) = 7i`

Explanation:

A square root of a number `n` is a number `x` such that `x^2 = n`

Note that if `x` is a Real number then `x^2 >= 0`.

So any square root of `-49` is not a Real number.

In order to be able to take square roots of negative numbers, we need Complex numbers.

That's where the mysterious number `i` comes into play. This is called the imaginary unit and has the property:

`i^2 = -1`

So `i` is a square root of `-1`. Note that `-i` is also a square root of `-1`, since:

`(-i)^2 = (-1*i)^2 = (-1)^2*i^2 = 1*(-1) = -1`

Then we find:

`(7i)^2 = 7^2*i^2 = 49*(-1) = -49`

So `7i` is a square root of `-49`. Note that `-7i` is also a square root of `-49`.

What do we mean by the square root of `-49`

For positive values of `n`, the square root is usually taken to mean the principal square root `sqrt(n)`, which is the positive one.

For negative values of `n`, the square roots are both multiples of `i`, so neither positive nor negative, but we can define:

`sqrt(n) = i sqrt(-n)`

With this definition, the principal square root of `-49` is:

`sqrt(-49) = i sqrt(49) = 7i`

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Footnote

The question remains: Where does `i` come from?

It is possible to define Complex numbers formally, as pairs of Real numbers with rules for arithmetic like this:

`(a, b) + (c, d) = (a+c, b+d)`

`(a, b) * (c, d) = (ac-bd, ab+cd)`

These rules for addition and multiplication work as expected with commutativity, distributivity, etc.

Then Real numbers are just Complex numbers of the form `(a, 0)` and we find:

`(0, 1)*(0, 1) = (-1, 0)`

That is `(0, 1)` is a square root of `(-1, 0)`

Then we can define `i = (0, 1)` and:

`(a, b) = a(1, 0) + b(0, 1) = a+bi`

Answer 2

I'm not altogether happy with the `7i` being described as the square root of `-49` or even the principal value of the multi-valued function, though it seems to be common practice.

Explanation:

The notation doesn't really extend to `n`th roots, nor roots of complex numbers. What is the fourth root of `1`? (`1`, `i`, `-1`, `-i`). In the special case of square roots of negative numbers it seems to be common practice to take the root to be the complex number with a positive imaginary part: here, `[0,7]` (or `7i` in the more common notation. You also get problems with sudden jumps in the `n` root of `z` if `z` moves, say, along the unit circle, if you use the concept of principal root.

Decades ago at school we learnt a lot about the `n`th complex roots of unity (and hence of any complex number including the pure real negative ones), but as far as I recall we never used the notation `sqrt(x)` (or other roots) if `x` might be negative or complex.

However, we did use the notion of primitive roots of a complex number `z`, from which all others can be derived by raising the primitive root to integral powers. So `i` is a primitive fourth root of `1` but `-1` is not. I am sure we did not use the concept of principal root, analogous to principal values of inverse trig functions.

I prefer the notion that the `n`th root is a function which maps a complex number on to a set of `n` complex numbers, where each such number raised to the `n`th power is the original number. I don't consider "multi-valued function" to be an oxymoron!