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Complex numbers De Moivre's theorem

If C=1+cos⁡θ+...+cos⁡(n−1)θC = 1+\cos\theta+...+\cos(n-1)\theta and S=sin⁡θ+...+sin⁡(n−1)θ.S = \sin\theta+...+\sin(n-1)\theta.

prove that C=sin⁡nθ2sin⁡θ2cos⁡(n−1)θ2C=\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}} \cos\frac{(n-1)\theta}{2} and S=sin⁡nθ2sin⁡θ2sin⁡(n−1)θ2S =\frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\sin\frac{(n-1)\theta}{2} if θ≠2kπ,k∈Z \theta\neq2k\pi, k\in\mathbb{Z}

Attempt

C+iS=1+(cos⁡θ+isin⁡θ)+...+(cos⁡(n−1)θ+isin⁡(n−1)θ)C+iS = 1+(\cos\theta+i\sin\theta)+...+(\cos(n-1)\theta+i\sin(n-1)\theta)

=1+eiθ+...+ei(n−1)θ=1+e^{i\theta}+...+e^{i(n-1)\theta}

=1+z+...+zn−1=1+z+...+z^{n-1} where z=eiθz=e^{i\theta}

=1−zn1−z=\frac{1-z^n}{1-z} if z≠1z\neq1

=1−einθ1−eiθ=einθ2(e−inθ2)−einθ2eiθ2(e−iθ2)−eiθ2=\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2})-e^\frac{in\theta}{2}}{e\frac{i \theta}{2}(e\frac{-i\theta}{2})-e\frac{i\theta}{2}}

Is this correct so far for the derivation? How can I prove that the real parts are equal to the complex parts?

Thanks.
I would guess that you first rationalise the fraction so the denominator is real by multiplying by the complex conjugate, then just equate the real and imaginary parts
Original post by Darth_Narwhale
I would guess that you first rationalise the fraction so the denominator is real by multiplying by the complex conjugate, then just equate the real and imaginary parts


but is it derived correctly?
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math42
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Original post by AishaGirl
but is it derived correctly?


looks fine
Original post by AishaGirl
=1−einθ1−eiθ=einθ2(e−inθ2)−einθ2eiθ2(e−iθ2)−eiθ2=\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2})-e^\frac{in\theta}{2}}{e\frac{i \theta}{2}(e\frac{-i\theta}{2})-e\frac{i\theta}{2}}
Did you mean

=1−einθ1−eiθ=einθ2(e−inθ2−einθ2)eiθ2(e−iθ2−eiθ2)\displaystyle =\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i \theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})} (note brackets)?

Original post by Darth_Narwhale
I would guess that you first rationalise the fraction so the denominator is real by multiplying by the complex conjugate, then just equate the real and imaginary parts
At this point, it works out much easier if you go:

=ei(n−1)θ2(e−inθ2−einθ2)e−iθ2−eiθ2[br]\displaystyle = \frac{e^\frac{i(n-1) \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{-i\theta}{2}-e\frac{i\theta}{2}}[br]

Now, rather mulitiplying by the complex conjugate, notice that the bottom is now −2isin⁡(θ/2)-2i \sin(\theta /2) and so you can make it real simply by multipllying by i instead.
Original post by DFranklin
Did you mean

=1−einθ1−eiθ=einθ2(e−inθ2−einθ2)eiθ2(e−iθ2−eiθ2)\displaystyle =\frac{1-e^{in\theta}}{1-e^{i\theta}}=\frac{e^\frac{in \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{i \theta}{2}(e\frac{-i\theta}{2}-e\frac{i\theta}{2})} (note brackets)?

At this point, it works out much easier if you go:

=ei(n−1)θ2(e−inθ2−einθ2)e−iθ2−eiθ2[br]\displaystyle = \frac{e^\frac{i(n-1) \theta}{2}(e^\frac{-in\theta}{2}-e^\frac{in\theta}{2})}{e\frac{-i\theta}{2}-e\frac{i\theta}{2}}[br]

Now, rather mulitiplying by the complex conjugate, notice that the bottom is now −2isin⁡(θ/2)-2i \sin(\theta /2) and so you can make it real simply by multipllying by i instead.


Ah yeah small mistake with the brackets. Thanks DFranklin.

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