I need to find the real and imaginary part of this $Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}$

Question

I have a test tomorrow and i have some troubles understanding this kind of problems, would really appreciate some help with this $$ Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n} $$ $$ Z_{n}\epsilon \mathbb{C} $$

Answer 1

Note that $z=\left (\frac{ \sqrt{3} + i }{2}\right )^{n}=(a+bi)^n$ we have $$|z|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_z=\arctan(\tfrac{b}{a})=\arctan(\tfrac{1}{\sqrt{3}})=\tfrac{\pi}{6}$$ and for $w=\left (\frac{ \sqrt{3} - i }{2}\right )^{n}=(a-bi)^n$ we have $$|w|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_w=\arctan(-\tfrac{b}{a})=\arctan(-\tfrac{1}{\sqrt{3}})=-\tfrac{\pi}{6}$$ Then, $$z=|z|^n(\cos\theta_z+i\sin\theta_z)^n=\left[\cos\left(\tfrac{\pi}{6}\right)+i\sin\left(\tfrac{\pi}{6}\right)\right]^n$$ $$w=|w|^n(\cos\theta_w+i\sin\theta_w)^n=\left[\cos\left(-\tfrac{\pi}{6}\right)+i\sin\left(-\tfrac{\pi}{6}\right)\right]^n$$ Using de Moivre's formula $$z^n=|z|^n(\cos\theta+i\sin\theta)^n=|z|^n(\cos(n\theta)+i\sin(n\theta))^n$$ we get $$z=\left[\cos\left(\tfrac{\pi}{6}\right)+i\sin\left(\tfrac{\pi}{6}\right)\right]^n=\left[\cos\left(\tfrac{n\pi}{6}\right)+i\sin\left(\tfrac{n\pi}{6}\right)\right]$$ $$w=\left[\cos\left(-\tfrac{\pi}{6}\right)+i\sin\left(-\tfrac{\pi}{6}\right)\right]^n=\left[\cos\left(-\tfrac{n\pi}{6}\right)+i\sin\left(-\tfrac{n\pi}{6}\right)\right]=\left[\cos\left(\tfrac{n\pi}{6}\right)-i\sin\left(\tfrac{n\pi}{6}\right)\right]$$ Finally, $$Z_n=z+w=2\cos\left(\tfrac{n\pi}{6}\right)$$

Answer 2

Let $\omega=\frac{\sqrt{3}+i}{2}$. Then $Z_n$ is twice the real part of $\omega^n$. Now, $\omega$ is a primitive $12$-th root of unity and so $\omega = \exp(\frac{2\pi i}{12})$. Therefore, $\omega^n=\exp(n\frac{2\pi i}{12})$, whose real part is $\cos(n\frac{2\pi}{12})=\cos(\frac{n\pi}{6})$.

Answer 3

The above two can be written using binomial theorem

$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{i}{2}\right)^k$$

And

$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{-i}{2}\right)^k$$

We rewrite the second as

$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{i}{2}\right)^k(-1)^k$$

Such that even k terms are the same and odd k terms are opposite signs

Combining the sums we have

$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^\frac{n-2k}{2}\left(\frac{i}{2}\right)^{2k}$$

$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^{\frac{n}{2}-k}\left(\frac{-1}{4}\right)^{k}$$

$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^{\frac{n}{2}}\left(\frac{-1}{3}\right)^{k}$$

Which calculating the sum from wolfram alpha gives, $$Re(Z_n)=2cos(\frac{n\pi}{6})$$

And

$$Im(Z_n)=0$$

The phase would flip between $\pi$ and $0$ and the function is always real. (Which I should have realized by seeing that this clearly resembles the cos function)

Answer 4

Obviously $\dfrac{\sqrt 3+i}2=\mathrm e^{\tfrac{i\pi}6}$ and $\dfrac{\sqrt 3-i}2$ is its conjugate $\mathrm e^{-\tfrac{i\pi}6}$. Thus $ Z_n= \mathrm e^{\tfrac{in\pi}6}+\mathrm e^{-\tfrac{in\pi}6}$ is the sum ot two conjugate numbers, so that $$\operatorname{Im}(Z_n)=0\quad \text{and}\quad\operatorname{Re}(Z_n)=2\cos \frac{n\pi}6.$$

Let's give explicit formulæ depending on the values of $n$: $$\operatorname{Re}(Z_n)=\begin{cases} 2&\text{if }n\equiv 0 \mod 12, \\\sqrt 3&\text{if }n\equiv 1\;\text{ or }\;11,\\1&\text{if }n\equiv 2\;\text{ or }\;10, \\ 0&\text{if }n\equiv 3\;\text{ or }\;9, \\-1&\text{if }n\equiv 4\;\text{ or }\;8, \\ -\sqrt3 &\text{if }n\equiv 5\;\text{ or }\;7, \\ -2 &\text{if }n\equiv 6. \end{cases}$$