Question Number 78465 by TawaTawa last updated on 17/Jan/20
Answered by mind is power last updated on 17/Jan/20
$$\zeta\left(\mathrm{z}\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{z}} } \\ $$$$\mathrm{Re}\left(\zeta\left(\mathrm{z}\right)\right)=\mathrm{Re}\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)+\mathrm{iIm}\left(\mathrm{z}\right)} }\right\} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{Re}\left\{\frac{\mathrm{n}^{−\mathrm{iIm}\left(\mathrm{z}\right)} }{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} }\right\}=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{Re}\left\{\frac{\mathrm{e}^{−\mathrm{iIm}\left(\mathrm{z}\right)\mathrm{ln}\left(\mathrm{n}\right)} }{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} }\:\right\} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{Re}\left\{\frac{\mathrm{cos}\left(\mathrm{ln}\left(\mathrm{n}\right)\mathrm{Im}\left(\mathrm{z}\right)\right)−\mathrm{isin}\left(\mathrm{ln}\left(\mathrm{n}\right)\mathrm{Im}\left(\mathrm{z}\right)\right)}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} }\right\} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{ln}\left(\mathrm{n}\right).\mathrm{Im}\left(\mathrm{z}\right)\right)}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} } \\ $$$$ \\ $$
Commented by TawaTawa last updated on 18/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mind is power last updated on 18/Jan/20
$$\mathrm{thanx}\:\mathrm{sir}\:\mathrm{you}\:\mathrm{Too} \\ $$