Question-78465

Question Number 78465 by TawaTawa last updated on 17/Jan/20

Answered by mind is power last updated on 17/Jan/20

ζ(z)=Σ_(n≥1) (1/n^z ) Re(ζ(z))=Re{Σ_(n≥1) (1/n^(Re(z)+iIm(z)) )} =Σ_(n≥1) Re{(n^(−iIm(z)) /n^(Re(z)) )}=Σ_(n≥1) Re{(e^(−iIm(z)ln(n)) /n^(Re(z)) ) } =Σ_(n≥1) Re{((cos(ln(n)Im(z))−isin(ln(n)Im(z)))/n^(Re(z)) )} =Σ_(n≥1) ((cos(ln(n).Im(z)))/n^(Re(z)) )

$$\zeta\left(\mathrm{z}\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{z}} } \\ $$$$\mathrm{Re}\left(\zeta\left(\mathrm{z}\right)\right)=\mathrm{Re}\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)+\mathrm{iIm}\left(\mathrm{z}\right)} }\right\} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{Re}\left\{\frac{\mathrm{n}^{−\mathrm{iIm}\left(\mathrm{z}\right)} }{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} }\right\}=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{Re}\left\{\frac{\mathrm{e}^{−\mathrm{iIm}\left(\mathrm{z}\right)\mathrm{ln}\left(\mathrm{n}\right)} }{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} }\:\right\} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{Re}\left\{\frac{\mathrm{cos}\left(\mathrm{ln}\left(\mathrm{n}\right)\mathrm{Im}\left(\mathrm{z}\right)\right)−\mathrm{isin}\left(\mathrm{ln}\left(\mathrm{n}\right)\mathrm{Im}\left(\mathrm{z}\right)\right)}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} }\right\} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{cos}\left(\mathrm{ln}\left(\mathrm{n}\right).\mathrm{Im}\left(\mathrm{z}\right)\right)}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)} } \\ $$$$ \\ $$

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

$$\mathrm{thanx}\:\mathrm{sir}\:\mathrm{you}\:\mathrm{Too} \\ $$

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