Question Number 194847 by cortano12 last updated on 17/Jul/23
$$\:\:\:\:\:\:\underbrace{\:} \\ $$
Commented by Tinku Tara last updated on 17/Jul/23
$${z}=?,\:\mathrm{without}\:\mathrm{knowing}\:{z}\:\mathrm{you}\:\mathrm{cannot} \\ $$$$\mathrm{find}\:{n} \\ $$
Commented by Frix last updated on 17/Jul/23
$${n}=\frac{\mathrm{19ln}\:\mathrm{2}}{\mathrm{2ln}\:{z}}+\frac{\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{2}{k}\pi}{\mathrm{ln}\:{z}}\mathrm{i}= \\ $$$$=\frac{\mathrm{19ln}\:\mathrm{2}}{\mathrm{2ln}\:{z}}+\frac{\left(\mathrm{8}{k}+\mathrm{3}\right)\pi}{\mathrm{4ln}\:{z}}\:\mathrm{with}\:{k}\in\mathbb{Z} \\ $$$$\Leftrightarrow \\ $$$${z}=\mathrm{2}^{\frac{\mathrm{19}}{\mathrm{2}{n}}} \mathrm{e}^{\mathrm{i}\frac{\left(\mathrm{8}{k}+\mathrm{3}\right)\pi}{\mathrm{4}{n}}} \:\mathrm{with}\:{k}\in\mathbb{Z}\:\wedge\:−\frac{\mathrm{4}{n}+\mathrm{3}}{\mathrm{8}}<{k}\leqslant\frac{\mathrm{4}{n}+\mathrm{3}}{\mathrm{8}} \\ $$