Question

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.

Answer 1

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

Answer 2

The possible values of alpha are: $\alpha =i\sqrt 3$ and $\alpha =\sqrt 3 + i\sqrt 3$

The given parameter is:

$\alpha - \beta = 2 \sqrt{3}$

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

$\alpha = a + bi$

$\beta = a - bi$

Subtract beta from alpha

$\alpha - \beta = a + bi -a +bi$

Collect like terms

$\alpha - \beta = a -a+ bi +bi$

Evaluate like terms

$\alpha - \beta = 2bi$

Given that: $\alpha - \beta = 2 i\sqrt{3}$

The equation becomes

$2bi = 2i\sqrt 3$

Divide both sides by 2i

$b = \sqrt 3$

Also, we have:

$\frac{\alpha}{\beta^2}$

This becomes

$\frac{\alpha}{\beta^2} = \frac{a + bi}{(a - bi)^2}$

Substitute $b = \sqrt 3$

$\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{(a - i\sqrt 3)^2}$

Expand

$\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{a^2 - 2i\sqrt 3 - 3}$

Rewrite as:

$\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{a^2 - 3- 2i\sqrt 3 }$

Rationalize

$\frac{\alpha}{\beta^2} = \frac{a + i\sqrt 3}{a^2 - 3- 2i\sqrt 3 } \times \frac{a^2 - 3+ 2i\sqrt 3 }{a^2 - 3+ 2i\sqrt 3 }$

Expand

$\frac{\alpha}{\beta^2} = \frac{a(a^2 - 3+ 2i\sqrt 3) + i\sqrt 3(a^2 - 3+ 2i\sqrt 3)}{(a^2 - 3- 2i\sqrt 3)(a^2 - 3+ 2i\sqrt 3) }$

Set to 0

$\frac{a(a^2 - 3+ 2i\sqrt 3) + i\sqrt 3(a^2 - 3+ 2i\sqrt 3)}{(a^2 - 3- 2i\sqrt 3)(a^2 - 3+ 2i\sqrt 3) } = 0$

$a(a^2 - 3+ 2i\sqrt 3) + i\sqrt 3(a^2 - 3+ 2i\sqrt 3) = 0$

Remove all imaginary parts

$a(a^2 - 3) = 0$

Solve for a

$a = 0$ or $a^2 - 3 = 0$

This gives

$a = 0$ or $a= \sqrt 3$

Recall that:

$\alpha = a + bi$

So, the possible values of alpha are:

$\alpha =i\sqrt 3$ and $\alpha =\sqrt 3 + i\sqrt 3$

Read more about complex numbers at:

https://brainly.com/question/10662770