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Let $O \in \mathbb{C}$. How can I prove that $O^{51}+\bar{O}^{51}$ is a real number, or in other words: $\Im(O^{51}+\bar{O}^{51}) = 0$?

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    $\begingroup$ You should say what is your $O$ $\endgroup$ – user9413 Dec 18 '11 at 8:48
  • $\begingroup$ does it matter? in this case $O$ is ($Z+1-2i$) and $\overline{O}$ is ($\overline{Z}+1+2i$). $\endgroup$ – Some1 Dec 18 '11 at 9:00
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    $\begingroup$ I would like to point out that using the letter O as a variable could be very confusing. $\endgroup$ – Potato Dec 18 '11 at 20:45
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The exponent here doesn't really matter as long as you know that $\bar z\bar w = \overline{zw}$, and this is easily verified by writing out $z = x + iy$ with $x, y$ real and carrying out the algebra. This implies (by induction, perhaps) that \[\bar{z}^{51} = \overline{z^{51}}.\] So the question becomes, why is $z + \bar z$ real? We could verify this by writing out real and imaginary parts as before, and we'll find that $z + \bar z = 2\operatorname{Re} z$. Alternatively, note that a complex number is real if and only if it is equal to its conjugate, and use that $\bar{\bar{z}} = z$ and $\bar z + \bar w = \overline{z + w}$.

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Assuming that $O$ is an arbitrary complex number, write it in exponential form as $O=re^{i\theta}$. Then $\overline{O}=re^{-i\theta}$, so $O^{51}+\overline{O}^{51}=r^{51}e^{51i\theta}+r^{51}e^{-51i\theta}$. Now use Euler’s formula to return to rectangular form; you shouldn’t have any trouble seeing that the imaginary part is zero.

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Slightly high-powered: letting $(u-O)(u-\bar{O})=u^2+pu+q$, it is known, by Newton-Girard, that $\mu_n=O^n+\bar{O}^n$ for nonnegative integer $n$ can always be expressed in terms of $O+\bar{O}=-p$ and $O\bar{O}=q$, which are both easily shown to be real. In particular, the $\mu_n$ satisfy the recursion relation $\mu_{n+1}+p \mu_n+q \mu_{n-1}=0$, with the starting values $\mu_0=1$ and $\mu_1=-p$.

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