#2**+13 **

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.

Bertie
Mar 24, 2015

#1**+10 **

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

reinout-g
Mar 24, 2015

#2**+13 **

Best Answer

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.

Bertie
Mar 24, 2015