Question

Let
`\alpha` and
`\beta` be conjugate complex numbers such that
`(\alpha)/(\beta^2)$ is a real number and $|\alpha - \beta| = 2 √(3)`. Find
`|\alpha|`.

Answer 1

`|\alpha| = 2`

Explanation:

Since
`\alpha` and
`\beta` are complex conjugates, let's define them as follows:

`\alpha = a+bi`

`\beta = a-bi`

`(\alpha)/(\beta^2)=(a+bi)/(a^2-b^2-2abi) =((a+bi)*(a^2-b^2+2abi))/((a^2-b^2-2abi)*(a^2-b^2+2abi)) =(a^3-3ab^2+(3a^2b-b^3)i)/(a^4+2a^2b^2+b^4)`

Since
`(\alpha)/(\beta^2)` is a real number, complex part of above result must be zero.

`3a^2b-b^3=0`

From to hold above equality,
`b=0` or
`b^2=3a^2`.

However, since
`|\alpha-\beta|=2\sqrt 3`,
`b\\eq 0`

So,
`b =\sqrt 3 a` or
`b =-\sqrt 3 a`

And since
`\alpha` and
`\beta` are complex conjugates, taking plus or minus sign as found above will not affect the result, so let's write the last version of
`\alpha` and
`\beta` as follows:

`\alpha = a+\sqrt 3 ai`

`\beta = a-\sqrt 3 ai`

Since
`|\alpha-\beta|=2\sqrt 3`

`|a+\sqrt 3 ai-a+\sqrt 3 ai|=2\sqrt 3` ⇒
`a = 1`

Finally,
`\alpha = 1+\sqrt 3 i` ⇒
`|\alpha| = √((1^2+\sqrt 3^2))=2`