Question

Let ula"> 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:

$|\alpha| = 2$

Step-by-step explanation:

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

$\alpha = a+bi$

$\beta = a-bi$

$\frac{\alpha}{\beta^2}=\frac{a+bi}{a^2-b^2-2abi} =\frac{(a+bi)*(a^2-b^2+2abi)}{(a^2-b^2-2abi)*(a^2-b^2+2abi)} =\frac{a^3-3ab^2+(3a^2b-b^3)i}{a^4+2a^2b^2+b^4}$

Since $\frac{\alpha}{\beta^2}$ 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 $|\alpha-\beta|=2\sqrt 3$, $b\neq 0$

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, $\alpha = 1+\sqrt 3 i$ ⇒ $|\alpha| = \sqrt{(1^2+\sqrt 3^2)}=2$