How can I calculate $(1+\cos(\theta)+i\sin(\theta))^n$ being $n\in{\mathbb{Z}}$?

Question

How can I calculate $(1+\cos(\theta)+i\sin(\theta))^n$ being $n\in{\mathbb{Z}}$ ?

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I think that I can use the half angle formulas for it, but I don't know how. Can anyone help me solve this problem?

Answer 1

well, $$\cos\theta+i\sin\theta=e^{i\theta}$$ so your question is $$(1+e^{i\theta})^n=\sum_{k=0}^n{n\choose k}e^{ik\theta}$$

Answer 2

Use the half-angle identities $$ 1+\cos(θ)=2\cos^2(θ/2),\\ \sin(θ)=2\cos(θ/2)\sin(θ/2) $$ and the Euler-Moivre formulas as per the comments. This gives an expression where all powers are of real numbers, with a phase factor depending on $n$.

Answer 3

Another trick:

Your expression is \begin{aligned} (1+e^{i\theta})^n & = \big(e^{i\theta / 2}(e^{-i\theta/2}+e^{i\theta/2})\big)^n \\ &= e^{ni\theta/2}(2\cos(\theta/2))^n\\ &= 2^n \big(\cos(n\theta/2) + i\sin(n\theta/2)\big)\cdot \cos^n(\theta/2) \end{aligned}

Answer 4

You have\begin{align}\bigl(1+\cos(\theta)+i\sin(\theta)\bigr)^n &=\left(1+e^{i\theta}\right)^n\\&=\left(2e^{i\theta/2}\frac{e^{-i\theta/2}+e^{i\theta/2}}2\right)^n\\&=\left(2e^{i\theta/2}\cos\left(\frac{\theta}2\right)\right)^n\\&=2^n\cos^n\left(\frac{\theta}2\right)\left(\cos\left(\frac{n\theta}2\right)+i\sin\left(\frac{n\theta}2\right)\right).\end{align}

Answer 5

Start from $$\cos\theta+i\sin\theta=e^{i\theta}$$ Then $$(1+e^{i\theta})^n=e^{in\theta/2}(e^{i\theta/2}+e^{-i\theta/2})^n$$ Now we use $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ Then $$(1+e^{i\theta})^n=2^ne^{in\theta/2}\cos^n\frac\theta 2\\=2^n\left(\cos\frac{n\theta}2+i\sin\frac{n\theta}2\right)\cos^n\frac\theta 2$$

Answer 6

The figure shows the sum of the unit vectors $1+e^{i\theta}=re^{i\phi}$.

As we have an isosceles triangle,

$$\phi=\dfrac\theta2$$ and $$r=2\cos\phi.$$

Hence

$$(1+\cos\theta+i\sin\theta)^n=2^n\cos^n(\phi)\,(\cos(n\phi)+i\sin(n\phi)).$$