+893 On the assumption that alpha, beta, a and b are real,
$$\alpha -\imath \beta=\frac{1}{a-\imath b}=\frac{a+\imath b}{a^{2}+b^{2}}.$$
Taking the complex conjugate of both sides,
$$\alpha + \imath\beta=\frac{a-\imath b}{a^{2}+b^{2}},$$
and multiplying the equations together,
$$\alpha^{2}+\beta^{2}=\frac{a^{2}+b^{2}}{(a^{2}+b^{2})^{2}},$$
from which the result follows.
+2353 I think you either forgot to mention some of the details, or the variables (other than i off course) have an implicit meaning which I'm unaware of.
Anyway, I can make it a little easier for you by doing the following.
$$\begin{array}{lcl}
\alpha - i\beta &= \frac{1}{a-ib}\\
\alpha - i\beta &= \frac{1}{a-ib}\frac{a+ib}{a+ib}\\
\alpha - i\beta &= \frac{a+ib}{a^2+ aib - aib -i^2b^2}\\
\alpha - i\beta &= \frac{a+ib}{a^2+b^2}\\
(\alpha - i\beta)(\alpha + i\beta) &= \frac{(a+ib)(\alpha + i\beta)}{a^2+b^2}\\
\alpha^2 +i\alpha \beta - i\alpha \beta - i^2 \beta^2 &= \frac{(a+ib)(\alpha + i\beta)}{a^2+b^2}\\
\alpha^2 + \beta^2 &= \frac{(a+ib)(\alpha + i\beta)}{a^2+b^2}\\
(\alpha^2 + \beta^2)(a^2+b^2)&= (a+ib)(\alpha + i\beta)
\end{array}$$
Now if you can prove that
$$(a+ib)(\alpha + i\beta) = 1$$
You're there.
Reinout
+893 On the assumption that alpha, beta, a and b are real,
$$\alpha -\imath \beta=\frac{1}{a-\imath b}=\frac{a+\imath b}{a^{2}+b^{2}}.$$
Taking the complex conjugate of both sides,
$$\alpha + \imath\beta=\frac{a-\imath b}{a^{2}+b^{2}},$$
and multiplying the equations together,
$$\alpha^{2}+\beta^{2}=\frac{a^{2}+b^{2}}{(a^{2}+b^{2})^{2}},$$
from which the result follows.
