Proving $ z^n + \frac{1}{z^n} = 2\cos(n\theta) $ for $z = \cos (\theta) + i\sin(\theta)$

Question

Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then

<p><span class="math-container">$$ z^n +  {1\over z^n} = 2\cos(n\theta) $$</span></p>

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What I have attempted

If $$ z = \cos (\theta) + i\sin(\theta) $$

then $$ z^n = \cos (n\theta) + i\sin(n\theta) $$

$$ z^n + {1\over z^n} $$

$$ \cos (n\theta) + i\sin(n\theta) + {1\over \cos (n\theta) + i\sin(n\theta)} $$

$$ (\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1 $$

$$ \left[ {(\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1\over \cos (n\theta) + i\sin(n\theta)} \right] $$

$$ \left[ {\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1\over \cos (n\theta) + i\sin(n\theta)} \right] $$

Now this is where I am stuck.. I tried to use a double angle identity but I can't eliminate the imaginary part..

Answer 1

setting the value of $z=\cos\theta+i\sin\theta$, $$z^n+\frac{1}{z^n}$$$$=z^n+z^{-n}$$ $$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$ Using d-Moivre's theorem, $$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$ $$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$ $$=2\cos(n\theta)$$

Answer 2

Hint. If $$z=\cos (\theta) + i\sin(\theta)= e^{i\theta}$$ then $$ \frac1{z^n}=z^{-n}=e^{-in\theta}=\cos (n\theta) - i\sin(n\theta). $$

Answer 3

Using geometry of complex:

By construction, point $C$ corresponds to $z^n$ an $C^\prime$ to $z^{-n}$. i.e. $\beta=-n\theta$ (the symmetry of the image is due to the fact that $|z|=1$, and thus $z\overline{z}=1$). Then $z^n+z^{-n}$ is, interpreting as a sum of vectors, the vector $\vec{OD}$. Then, by means of Cosine Law's, it is easy to conclude.

Answer 4

If you want to continue in the way you started, you can simply rewrite $1=\cos^2(n\theta)+\sin^2(n\theta)$ and then simplify: $$ \frac{\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1}{ \cos (n\theta) + i\sin(n\theta)}= \frac{\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + \cos^2(n\theta)+\sin^2(n\theta)}{ \cos (n\theta) + i\sin(n\theta)}= \frac{2\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta)}{ \cos (n\theta) + i\sin(n\theta)}= \frac{2\cos(n\theta) (\cos(n\theta)+i\sin(n\theta))}{ \cos (n\theta) + i\sin(n\theta)}= \underline{\underline{2\cos(n\theta)}} $$

Which shows that you were almost there. (But it is still useful that you posted your question here - you have seen other approaches.)

Answer 5

Hint: To simplify $\dfrac{1}{\cos(n\theta)+i\sin(n\theta)}$, multiply top and bottom by $\cos(n\theta)-i\sin(n\theta)$.

Answer 6

Without recalling the De Moivre's formula... Olivier Oloa used it but didn't recall the Euler's Formula: $e^{ix}=\cos x+ i\sin x$. From this, we have $e^{-ix}=\cos x- i\sin x$ and hence $2\cos x=e^{ix}+e^{-ix}$ and $2i\sin x=e^{ix}-e^{-ix}$ for any $x\in\Bbb R$.

Now, we have a quick computation for OP's question: $$\large z^n+z^{-n}=(e^{i\theta})^n+(e^{i\theta})^{-n}=e^{in\theta}+e^{-in\theta}=2\cos n\theta.$$

Answer 7

We know that any complex number can be written in polar form as:

z = r(cosθ + i sinθ)

where r is the modulus of z and θ is its argument.

Now, we can write:

z^-1 = 1/z = 1/r(cos(-θ) + i sin(-θ)) = cos(-θ)/r - i sin(-θ)/r = cosθ/r + i sinθ/r

Using this, we have:

z + z^-1 = r(cosθ + i sinθ) + r(cosθ/r + i sinθ/r) = r(cosθ + cosθ/r) + i(sinθ + sinθ/r) = r(cosθ + cos^2θ/r + sin^2θ/r) + i(sinθ + sin^2θ/r + cosθsinθ/r) = r(1 + cosθcos(x) + sinθsin(x)) + i(sinθ + cosθsin(x)) = 2r cos(x + θ)

where we have used the identity cos(a+b) = cos(a)cos(b) - sin(a)sin(b) with a = θ and b = x.

Now, since z is a non-zero complex number, its modulus r is also non-zero. Therefore, we can divide both sides by 2r to obtain:

z + z^-1 = 2 cos(x + θ)/2 = 2 cos(x)

and the proof is complete.