Answer:
[tex]|\alpha| = 2[/tex]
Step-by-step explanation:
Since [tex]\alpha[/tex] and [tex]\beta[/tex] are complex conjugates, let's define them as follows:
[tex]\alpha = a+bi[/tex]
[tex]\beta = a-bi[/tex]
[tex]\frac{\alpha}{\beta^2}=\frac{a+bi}{a^2-b^2-2abi} =\frac{(a+bi)*(a^2-b^2+2abi)}{(a^2-b^2-2abi)*(a^2-b^2+2abi)} =\frac{a^3-3ab^2+(3a^2b-b^3)i}{a^4+2a^2b^2+b^4}[/tex]
Since [tex]\frac{\alpha}{\beta^2}[/tex] is a real number, complex part of above result must be zero.
[tex]3a^2b-b^3=0[/tex]
From to hold above equality, [tex]b=0[/tex] or [tex]b^2=3a^2[/tex].
However, since [tex]|\alpha-\beta|=2\sqrt 3[/tex], [tex]b\neq 0[/tex]
So, [tex]b =\sqrt 3 a[/tex] or [tex]b =-\sqrt 3 a[/tex]
And since [tex]\alpha[/tex] and [tex]\beta[/tex] are complex conjugates, taking plus or minus sign as found above will not affect the result, so let's write the last version of [tex]\alpha[/tex] and [tex]\beta[/tex] as follows:
[tex]\alpha = a+\sqrt 3 ai[/tex]
[tex]\beta = a-\sqrt 3 ai[/tex]
Since [tex]|\alpha-\beta|=2\sqrt 3[/tex]
[tex]|a+\sqrt 3 ai-a+\sqrt 3 ai|=2\sqrt 3[/tex] ⇒ [tex]a = 1[/tex]
Finally, [tex]\alpha = 1+\sqrt 3 i[/tex] ⇒ [tex]|\alpha| = \sqrt{(1^2+\sqrt 3^2)}=2[/tex]
