Proof for $z\neq 0$ , z is any complex number?

Question

$\sum _{k=1}^{n}\mathrm{sin}kz=\frac{\mathrm{sin}\frac{n+1}{{2}}\cdot \mathrm{sin}\frac{nz}{2}}{\mathrm{sin}\frac{z}{2}}$ Proof for $z\neq 0$

Answer 1

HINT.

$sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$

left sum is $\frac{e^{iz}\frac{e^{inz}-1}{e^{iz} -1} - e^{-iz}\frac{e^{-inz}-1}{e^{-iz} -1} }{2i}$

Try to prove this equation now

Answer 2

For notational simplification, let us use arguments $2z$ instead of $z$ and let $w=\cos z+i\sin z$.

Consider the geometric sum $$\sum_{k=0}^n w^{2k}=\frac{w^{2n+2}-1}{w^2-1}=\frac{(w^{2n+2}-1)(w^{-2}-1)}{(w^2-1)(w^{-2}-1)}=w^n\frac{(w^{n+1}-w^{-n-1})(w^{-1}-w^1)}{(2-w^2-w^{-2})}.$$

(The factor $w^{-2}-1$ is the conjugate of $w^2-1$.)

In terms of $z$, this is

$$\sum_{k=0}^n(\cos(2kz)+i\sin(2kz))=\left(\cos(kn)+i\sin\left(kn\right)\right)\frac{2\sin\left((n+1)z\right)\,2\sin(z)}{2(1-\cos(2z))}$$

Simplifying and splitting the real and imaginary parts, we get two identities for the same price

$$\sum_{k=0}^n\cos(2kz)=\frac{\cos(nz)\sin((n+1)z)}{\sin(z)},\\ \sum_{k=0}^n\sin(2kz)=\frac{\sin(nz)\sin((n+1)z)}{\sin(z)}.$$

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