Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
You should say what is your $O$ –  user9413 Dec 18 '11 at 8:48
    
does it matter? in this case $O$ is ($Z+1-2i$) and $\overline{O}$ is ($\overline{Z}+1+2i$). –  Some1 Dec 18 '11 at 9:00
12  
I would like to point out that using the letter O as a variable could be very confusing. –  Potato Dec 18 '11 at 20:45

3 Answers 3

up vote 18 down vote accepted

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 $\bar z + \bar w = \overline{z + w}$.

share|improve this answer

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.

share|improve this answer

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 $\mu_1=-p$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.