Question

What different numbers might be considered conjugates of `1+(root(3)(2))i` and why?

Answer 1

`1-\root[3]2i`

Explanation:

Two conjugate complex numbers have equal real parts & imaginary parts but imaginary parts are opposite in sign.

Each complex number has only one conjugate complex number

Hence the conjugate of given complex number `1+\root[3]2i` is

`1-\root[3]2i`

Answer 2

Some conjugates are:

`1-(root(3)(2))i`

`(1-root(3)(4)+2root(3)(2))+(2-root(3)(2)-2root(3)(4))i`

`(1-root(3)(4))-(root(3)(2))i`

Explanation:

A conjugate of a number `x` is a number `y` such that the product `xy` is a number of simpler form.

Complex conjugate

The standard complex conjugate of `a+bi` (where `a, b in RR`) is `bar(a+bi) = a-bi` since:

`(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2 in RR`

So one conjugate of `1+(root(3)(2))i` is `1-(root(3)(2))i`, with:

`(1+(root(3)(2))i)(1-(root(3)(2))i) = 1^2+(root(3)(2))^2 = 1+root(3)(4)`

This is certainly a simpler form of number in that it is real instead of non-real complex, but it is still a bit messy.

Radical conjugates

In general, if we have an expression of the form `sqrt(a)+sqrt(b)` then we can form a simpler expression by multiplying by `sqrt(a)-sqrt(b)`:

`(sqrt(a)+sqrt(b))(sqrt(a)-sqrt(b)) = (sqrt(a))^2-(sqrt(b))^2 = a-b`

So `sqrt(a)+sqrt(b)` and `sqrt(a)-sqrt(b)` are radical conjugates of one another.

Notice that the complex conjugate above is a special case of this kind of conjugate since `sqrt(-1) = i`

What about cube roots?

The radical conjugate for square roots uses the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

For cube roots we can make use of the sum or difference of cubes identities:

`A^3+B^3 = (A+B)(A^2-AB+B^2)`

`A^3-B^3 = (A-B)(A^2+AB+B^2)`

For example, putting `A=1` and `B=root(3)(4)` we can use the sum of cubes identity to find:

`5 = 1^3+(root(3)(4))^3`

`color(white)(5) = (1+root(3)(4))(1-root(3)(4)+(root(3)(4))^2)`

`color(white)(5) = (1+root(3)(4))(1-root(3)(4)+root(3)(16))`

`color(white)(5) = (1+root(3)(4))(1-root(3)(4)+2root(3)(2))`

`color(white)(5) = (1+root(3)(2)i)(1-root(3)(2)i)(1-root(3)(4)+2root(3)(2))`

`color(white)(5) = (1+root(3)(2)i)((1-root(3)(4)+2root(3)(2))-(1-root(3)(4)+2root(3)(2))(root(3)(2))i)`

`color(white)(5) = (1+root(3)(2)i)((1-root(3)(4)+2root(3)(2))-(root(3)(2)-2+2root(3)(4))i)`

`color(white)(5) = (1+root(3)(2)i)((1-root(3)(4)+2root(3)(2))+(2-root(3)(2)-2root(3)(4))i)`

So `(1+root(3)(2)i)` and `((1-root(3)(4)+2root(3)(2))+(2-root(3)(2)-2root(3)(4))i)` are conjugates of one another, with a rational real product.

One more

How about if we want a simpler number, but not necessarily a real one?

Consider `A=1` and `B=root(3)(2)i`.

Then using the sum of cubes identity, we find:

`1 - 2i = 1^3+(root(3)(2)i)^3`

`color(white)(1-2i) = (1+root(3)(2)i)(1-root(3)(2)i+(root(3)(2)i)^2)`

`color(white)(1-2i) = (1+root(3)(2)i)(1-root(3)(2)i-root(3)(4))`

`color(white)(1-2i) = (1+root(3)(2)i)((1-root(3)(4))-(root(3)(2))i)`

So `(1+root(3)(2)i)` and `((1-root(3)(4))-(root(3)(2))i)` are also conjugates with product a Gaussian integer.