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

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

$$\zeta \left(\mathrm{z}\right)=\sum _{\mathrm{n}⩾1}\frac{1}{{\mathrm{n}}^{\mathrm{z}}}$$$$\mathrm{Re}\left(\zeta \left(\mathrm{z}\right)\right)=\mathrm{Re}\left\{\sum _{\mathrm{n}⩾1}\frac{1}{{\mathrm{n}}^{\mathrm{Re}\left(\mathrm{z}\right)+\mathrm{iIm}\left(\mathrm{z}\right)}}\right\}$$$$=\sum _{\mathrm{n}⩾1}\mathrm{Re}\left\{\frac{{\mathrm{n}}^{-\mathrm{iIm}\left(\mathrm{z}\right)}}{{\mathrm{n}}^{\mathrm{Re}\left(\mathrm{z}\right)}}\right\}=\sum _{\mathrm{n}⩾1}\mathrm{Re}\left\{\frac{{\mathrm{e}}^{-\mathrm{iIm}\left(\mathrm{z}\right)\ln \left(\mathrm{n}\right)}}{{\mathrm{n}}^{\mathrm{Re}\left(\mathrm{z}\right)}}\right\}$$$$=\sum _{\mathrm{n}⩾1}\mathrm{Re}\left\{\frac{\cos \left(\ln \left(\mathrm{n}\right)\mathrm{Im}\left(\mathrm{z}\right)\right)-\mathrm{isin}\left(\ln \left(\mathrm{n}\right)\mathrm{Im}\left(\mathrm{z}\right)\right)}{{\mathrm{n}}^{\mathrm{Re}\left(\mathrm{z}\right)}}\right\}$$$$=\sum _{\mathrm{n}⩾1}\frac{\cos (\ln \left(\mathrm{n}\right).\mathrm{Im}\left(\mathrm{z}\right))}{{\mathrm{n}}^{\mathrm{Re}\left(\mathrm{z}\right)}}$$$$$$

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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