Question

How do you simplify `sqrt(9e^6)`?

Answer 1

It depends what you mean by `e`...

Explanation:

• If `e` is the mathematical constant (`~~2.7182818`), then the answer is definitely `3e^3`.

• If `e` is a Real valued variable, then `3abs(e^3)` covers both positive and negative values of `e`.

• If `e` is a Complex valued variable then `3sqrt(e^6)` is about the best you can do.

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The square root symbol `sqrt` denotes the principal square root.

We have:

`(3e^3)^2 = 3^2(e^3)^2 = 9e^6`

`(-3e^3)^2 = (-3)^2(e^3)^2 = 9e^6`

So regardless of the value of `e` (Real or Complex), both `3e^3` and `-3e^3` are square roots of `9e^6`, but which is the principal one?

If `3e^3` is a non-negative Real number then it is the principal square root. Hence for `e >= 0` we have `sqrt(9e^6) = 3e^3`

If `3e^3` is a negative Real number then `-3e^3` is positive and `sqrt(9e^6) = -3e^3`.

So if `e` is a Real number then `sqrt(9e^6) = abs(3e^3) = 3abs(e^3)`

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How about the Complex case?

Suppose `e=sqrt(i)=cos(pi/4)+i sin(pi/4)`

Then:

`3e^3 = 3(cos((3pi)/4)+i sin((3pi)/4))`

`sqrt(9e^6) = 3sqrt(cos((6pi)/4)+i sin((6pi)/4))`

`=3sqrt(cos(-pi/2)+i sin(-pi/2))`

`=3(cos(-pi/4)+i sin(-pi/4))`

`=-3(cos((3pi)/4)+sin((3pi)/4))`

`=-3e^3`

So in the general Complex case, about the best we can say is:

`sqrt(9e^6) = +-3e^3`

or better:

`sqrt(9e^6) = 3sqrt(e^6)`

It would be possible to pick out individual cases according to their `Arg` value, but it would hardly be simplifying.