Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 196560 by peter frank last updated on 27/Aug/23

Commented by Spillover last updated on 27/Aug/23

Check your solution https://m.facebook.com/story.php?story_fbid=pfbid0QHNhoZjesQzGqy7jHAgbvRSRUx25yXHezsGoh2qHLCapd9KB4z5PFi88czsFYxZ6l&id=100084816875916&mibextid=Nif5oz

Answered by MM42 last updated on 27/Aug/23

$$e^{\frac{(4k+1)\pi }{2}ix}=e^{\pi }\Rightarrow x=\frac{-2i}{4k+1}✓$$

Commented by Frix last updated on 27/Aug/23

$${\mathrm{i}}^{-\frac{2\mathrm{i}}{4k+1}}={\left({\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\right)}^{-\frac{2\mathrm{i}}{4k+1}}={\mathrm{e}}^{\frac{\pi }{4k+1}}\ne {\mathrm{e}}^{\pi }\mathrm{\forall }k\ne 0$$$$$$

Commented by MM42 last updated on 28/Aug/23

$$alright$$$$e^{\frac{i\pi }{2}x}\times e^{2k\pi i}=e^{\pi }\Rightarrow \frac{i}{2}x=1-2ki\Rightarrow x=4k-2i$$

Answered by Frix last updated on 27/Aug/23

$${\mathrm{i}}^x={\mathrm{e}}^{\mathrm{i}\frac{\pi x}{2}}={\mathrm{e}}^{\pi }$$$$x=4n-2\mathrm{i}$$$$\mathrm{Test}:{\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}(4n-2\mathrm{i})}={\mathrm{e}}^{\pi +2n\pi \mathrm{i}}={\mathrm{e}}^{2n\pi \mathrm{i}}{\mathrm{e}}^{\pi }={1\mathrm{e}}^{\pi }$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com