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Question Number 184481 by yaslm last updated on 07/Jan/23

Answered by mahdipoor last updated on 07/Jan/23

$$\mathrm{i}=e^0(cos\frac{\pi }{2}+\mathrm{i}sin\frac{\pi }{2})=e^{\frac{\pi }{2}\mathrm{i}}\Rightarrow$$$$Z_1={\left({\mathrm{i}}^{\frac{100}{\mathrm{i}}}\right)}^{0.5}={\left(e^{\left(\frac{\pi }{2}\mathrm{i}\right)\left(\frac{100}{\mathrm{i}}\right)}\right)}^{0.5}={\left(e^{\frac{100\pi }{2}}\right)}^{0.5}={\left(e^{50\pi }\right)}^{0.5}$$$$Z_1=\pm e^{25\pi }$$$$Z_2=e^{\frac{\pi \sqrt{3}}{2}\mathrm{i}}=\mathrm{i}sin\left(\frac{\pi \sqrt{3}}{2}\right)$$$$Z_3=e^{\frac{9\pi }{4}\mathrm{i}}=\mathrm{i}sin\left(\frac{9\pi }{4}\right)$$

Commented by yaslm last updated on 07/Jan/23

thanks complete sir

Commented by Frix last updated on 07/Jan/23

$$\mathrm{If}{\left({\mathrm{e}}^{50\pi }\right)}^{0.5}=\pm {\mathrm{e}}^{25\pi }\mathrm{then}\mathrm{why}{\mathrm{e}}^{\frac{\pi }{2}\mathrm{i}}\mathrm{which}\mathrm{can}$$$$\mathrm{be}\mathrm{written}\mathrm{as}{\left({\mathrm{e}}^{\pi \mathrm{i}}\right)}^{0.5}\ne \pm {\mathrm{e}}^{\frac{\pi }{2}\mathrm{i}}?\mathrm{Same}\mathrm{for}{\mathrm{e}}^{\frac{\pi \sqrt{3}}{2}\mathrm{i}}.$$$$\mathrm{Even}\mathrm{worse}:{\mathrm{e}}^{\frac{9\pi }{4}\mathrm{i}}={\left({\mathrm{e}}^{9\pi \mathrm{i}}\right)}^{0.25}=\{\begin{array}{l}\pm {\mathrm{e}}^{\frac{9\pi }{4}\mathrm{i}}\\ \pm {\mathrm{ie}}^{\frac{9\pi }{4}\mathrm{i}}\end{array}$$$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{not}\mathrm{using}\mathrm{the}\mathrm{same}\mathrm{rule}\mathrm{in}\mathrm{each}\mathrm{case}...$$

Answered by Frix last updated on 07/Jan/23

$$z=r{\mathrm{e}}^{\mathrm{i}\theta };r=\mid z\mid \Rightarrow r\in {\mathbb{R}}_0^{+};-\pi <\theta ⩽\pi$$$$z^c=r^c{\mathrm{e}}^{\mathrm{i}c\theta }\mathrm{\forall }c\in \mathbb{Z}\mathrm{which}\mathrm{is}\mathrm{always}\mathrm{unique}$$$$$$$$Z_1=\sqrt{{\mathrm{i}}^{-100\mathrm{i}}}=\sqrt{{\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{-100\mathrm{i}}}=\sqrt{{\mathrm{e}}^{-50\pi {\mathrm{i}}^2}}=\sqrt{{\mathrm{e}}^{50\pi }}={\mathrm{e}}^{25\pi }$$$$Z_2={\mathrm{i}}^{\sqrt{3}}={\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{\sqrt{3}}={\mathrm{e}}^{\mathrm{i}\frac{\pi \sqrt{3}}{2}}=\cos \frac{\pi \sqrt{3}}{2}+\mathrm{i}\sin \frac{\pi \sqrt{3}}{2}$$$$Z_3={\mathrm{i}}^{\frac{9}{2}}={\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{\frac{9}{2}}={\mathrm{e}}^{\mathrm{i}\frac{9\pi }{4}}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}$$

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