Prove an equation ($n \in N$, $z \in C$, $a \in R$):
$$ z^n + \frac1{z^n} = 2 \cos\alpha n$$
if $$ z + \frac1{z} = 2 \cos\alpha$$
I tried math induction, but did not solve.
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Do you know Euler's formulae for the trigonometric functions in terms of the exponential function? – Daniel Fischer Oct 02 '13 at 20:41
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Only De Moivre's formula and basic complex numbers properties. – Vladislav Brylinskiy Oct 02 '13 at 20:43
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$e^{ix} = \cos x + i\sin x$, $\cos x = \frac12(e^{ix} + e^{-ix})$, $\sin x = \frac{1}{2i}(e^{ix} - e^{-ix})$. The one for $\cos$ is the one you most want here. – Daniel Fischer Oct 02 '13 at 20:45
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In book, where i faced this equation, "Euler's formulae for the trigonometric functions in terms of the exponential function" is not listed before this exercise. As i said, only De Moivre's formula. So i have to use exponential function to solve this? – Vladislav Brylinskiy Oct 02 '13 at 20:48
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@VladislavBrylinskiy: Have you learned that, for every complex number $z$, we can write $z = re^{i\theta}$ for some real numbers $r$ and $\theta$? – BaronVT Oct 02 '13 at 20:49
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Unfortunately, no. Only De Moivre's formula. – Vladislav Brylinskiy Oct 02 '13 at 20:53
3 Answers
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Using de Moivre's formula $(\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n\alpha$, we see that setting $z = r(\cos \alpha + i\sin \alpha)$, we have $z^{-1} = r^{-1}(\cos \alpha - i \sin \alpha)$; thus
$z + z^{-1} = r(\cos \alpha + i \sin \alpha) + r^{-1}(\cos \alpha - i \sin \alpha) = 2 \cos \alpha. \tag{1}$
This implies
$ r\cos \alpha + r^{-1}\cos \alpha = 2 \cos \alpha, \tag{2}$
as well as
$i(r\sin \alpha -r^{-1}\sin \alpha) = 0, \tag{3}$
this latter equation (3) since $2 \cos \alpha$ is real.
If $\cos \alpha \ne 0$, (2) yields
$r + r^{-1} = 2, \tag{4}$
or
$(r - 1)^2 = r^2 - 2r + 1 = 0; \tag{5}$
$r = 1$ and $z = \cos \alpha + \sin \alpha$; then $z^n = \cos (n\alpha) + i\sin (n\alpha)$, $z^{-n} = \cos (-n\alpha) + i\sin (-n\alpha) = \cos (n\alpha) - i\sin (n\alpha)$, so
$z^n + z^{-n} = 2 \cos n \alpha \tag{6}$
in this case.
In the event that $\cos \alpha = 0$, we have $\sin \alpha \ne 0$, and from (3) we infer
$r \sin \alpha = r^{-1} \sin \alpha, \tag{7}$
whence
$r^2 = 1, \tag{8}$
implying $r = 1$ since $r > 0$. Thus we still have $z = \cos \alpha + i\sin \alpha$ and (6) still follows as in the case $\cos \alpha \ne 0$. I credit Daniel Littlewood's comment for making clear to me the utility of equation (3) in this context.
Well, I hope that helps. Cheers,
and as always,
Fiat Lux!!!
Robert Lewis
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Your hypothesis means that $$ z^2 - 2z\cos\alpha + 1 = 0 $$ Solve the quadratic equation: $$ z=\cos\alpha\pm\sqrt{\cos^2\alpha-1} $$ so $$ z=\cos\alpha+i\sin\alpha \quad\text{or}\quad z=\cos\alpha-i\sin\alpha=\cos(-\alpha)+i\sin(-\alpha). $$ Apply de Moivre's formula.
egreg
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Assuming you are familiar with the formula $z^{n}=(\cos(\alpha)+i\sin(\alpha))^{n}=\cos(n\alpha)+i\sin(n\alpha)$ at least for integers $n$, then you can see that$$z^{n}+z^{-n}=\cos(n\alpha)+i\sin(n\alpha)+\cos(-n\alpha)+i\sin(n\alpha)$$ Since $\cos$ is even and $\sin$ is odd, this simplifies to your result.
preferred_anon
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1As i understand, $$z^n + z^{-n} = r^n (\cos (na) + i\sin (na)) + r^{-n}(\cos (-na) + i\sin (-na))$$ – Vladislav Brylinskiy Oct 02 '13 at 20:59
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Let $z=re^{i\theta}$, so that $z+1/z=r\cos\theta+ir\sin\theta+1/r\cos\theta-i1/r\sin\theta=2\cos\alpha$ is purely real, meaning $r\sin\theta=1/r \sin\theta$. Thus, either $\sin\theta=0$, so $z$ is a real number, or $r^{2}=1$ so $r=1$ (note $r>0$). Is this sufficient? – preferred_anon Oct 02 '13 at 22:00