Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}. Find alpha.

[tex]\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}[/tex]

Imaginary part of the above =0

i.e. [tex]\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1[/tex]

So the value of alpha = [tex]-3+i\sqrt{3} , 1+\sqrt{3}[/tex]

MrRoyal MrRoyal · Answer 2 · 2021-12-28T18:31:40+01:00

The possible values of alpha are: [tex]\alpha =i\sqrt 3[/tex] and [tex]\alpha =\sqrt 3 + i\sqrt 3[/tex]

The given parameter is:

[tex]\alpha - \beta = 2 \sqrt{3}[/tex]

Given that alpha and beta are conjugates, then they can be represented as:

[tex]\alpha = a + bi[/tex]

[tex]\beta = a - bi[/tex]

Subtract beta from alpha

[tex]\alpha - \beta = a + bi -a +bi[/tex]

Collect like terms

[tex]\alpha - \beta = a -a+ bi +bi[/tex]

Evaluate like terms

[tex]\alpha - \beta = 2bi[/tex]

Given that: [tex]\alpha - \beta = 2 i\sqrt{3}[/tex]

The equation becomes

[tex]2bi = 2i\sqrt 3[/tex]

Divide both sides by 2i

[tex]b = \sqrt 3[/tex]

Also, we have:

[tex]\frac{\alpha}{\beta^2}[/tex]

This becomes

[tex]\frac{\alpha}{\beta^2} = \frac{a + bi}{(a - bi)^2}[/tex]

Substitute [tex]b = \sqrt 3[/tex]

[tex]\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{(a - i\sqrt 3)^2}[/tex]

Expand

[tex]\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{a^2 - 2i\sqrt 3 - 3}[/tex]

Rewrite as:

[tex]\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{a^2 - 3- 2i\sqrt 3 }[/tex]

Rationalize

[tex]\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{a^2 - 3- 2i\sqrt 3 } \times \frac{a^2 - 3+ 2i\sqrt 3 }{a^2 - 3+ 2i\sqrt 3 }[/tex]

Expand

[tex]\frac{\alpha}{\beta^2} = \frac{a(a^2 - 3+ 2i\sqrt 3) + i\sqrt 3(a^2 - 3+ 2i\sqrt 3)}{(a^2 - 3- 2i\sqrt 3)(a^2 - 3+ 2i\sqrt 3) }[/tex]

Set to 0

[tex]\frac{a(a^2 - 3+ 2i\sqrt 3) + i\sqrt 3(a^2 - 3+ 2i\sqrt 3)}{(a^2 - 3- 2i\sqrt 3)(a^2 - 3+ 2i\sqrt 3) } = 0[/tex]

[tex]a(a^2 - 3+ 2i\sqrt 3) + i\sqrt 3(a^2 - 3+ 2i\sqrt 3) = 0[/tex]

Remove all imaginary parts

[tex]a(a^2 - 3) = 0[/tex]

Solve for a

[tex]a = 0[/tex] or [tex]a^2 - 3 = 0[/tex]

This gives

[tex]a = 0[/tex] or [tex]a= \sqrt 3[/tex]

Recall that:

[tex]\alpha = a + bi[/tex]

So, the possible values of alpha are:

[tex]\alpha =i\sqrt 3[/tex] and [tex]\alpha =\sqrt 3 + i\sqrt 3[/tex]