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\begin{document}


{\bf Question}

Solve the equation $$(x+ia)6{2n} + (x-ia)^{2n} = (x^2 + a^2)^n$$

\vspace{.25in}

{\bf Answer}

Assumption $x$ and $a$ are real $x+ia = re^{i\theta}$

$\ds(r^2e^{2i\theta})^n + (r^2e^{-2i\theta})^n = (r^2)^n$

${}$

If $r\not= 0$

$\ds e^{2ni\theta} + e^{-2ni\theta} = 1$

i.e. $\cos 2n\theta = \frac{1}{2}$

$\ds\theta = \left(\frac{2k}{n} + \frac{1}{6n}\right)\pi
\hspace{.2in} k = 0, 1, 2, .., 2n-1$

${ }$

${ }$

If $r=0$

then $a=0$ so the equation becomes $\ds x^{2n} + x^{2n} = x^{2n}
\Rightarrow x=0$

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