Question

How do you simplify `\sqrt (z^34)`?

Answer 1

`x^17`

Explanation:

Divide the index by the root you are trying to find.

`sqrt(z^34) = z^(34div2)`

`= z^17`

You can see why this is true if you write two identical factors under the root:

`sqrt 49 = sqrt(7xx7) = 7`

`sqrt(x^6) = sqrt(x^3 xx x^3) = x^3`

`x^34 = sqrt(x^17 xx x^17) = x^17`

Answer 2

`sqrt(z^34) = abs(z^17)` which simplifies to `z^17` if `z >= 0`

Explanation:

Note that:

`z^17 * z^17 = z^(17+17) = z^34`

So `z^17` is a square root of `z^34`.

Note also that:

`(-z^17) * (-z^17) = z^34`

So `-z^17` is also a square root of `z^34`.

The expression `sqrt(z^34)` denotes the principal square root, which in the case of real-valued square roots is the non-negative one.

So to choose the non-negative square root we can write `abs(z^17)`.

If we are told that `z >= 0` then this simplifies to `z^17`.

Complex footnote

If `z` is allowed to range over complex numbers then the "correct" choice for the principal square root `sqrt(z^34)` is still between `z^17` and `-z^17`, but is messy to express.

If we are using the usual definition of `Arg(z) in (-pi, pi]`, then essentially we want to pick whichever of `z^17` or `-z^17` has the non-negative real part, or if they are both pure imaginary then whichever has the non-negative imaginary coefficient.

It is possible to express a suitable condition in terms of `Arg(z)`, but it is messy.

Answer 3

`z^17`

Explanation:

A square root can be rewritten as a base to the power of `1/2`, so

`sqrt(z^34) => (z^34)^(1/2)`

Then we use the identity

`(a^b)^c=a^(bc)`

so

`(z^34)^(1/2) => z^(34*1/2)`

`=> z^17`