Question-196560

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^((((4k+1)π)/2)ix) =e^π ⇒x=((−2i)/(4k+1)) ✓

$${e}^{\frac{\left(\mathrm{4}{k}+\mathrm{1}\right)\pi}{\mathrm{2}}{ix}} ={e}^{\pi} \Rightarrow{x}=\frac{−\mathrm{2}{i}}{\mathrm{4}{k}+\mathrm{1}}\:\checkmark\: \\ $$

Commented by Frix last updated on 27/Aug/23

i^(−((2i)/(4k+1))) =(e^(i(π/2)) )^(−((2i)/(4k+1))) =e^(π/(4k+1)) ≠e^π ∀k≠0

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

Commented by MM42 last updated on 28/Aug/23

alright e^(((iπ)/2)x) ×e^(2kπi) =e^π ⇒(i/2)x=1−2ki⇒x=4k−2i

$${alright} \\ $$$${e}^{\frac{{i}\pi}{\mathrm{2}}{x}} ×{e}^{\mathrm{2}{k}\pi{i}} ={e}^{\pi} \Rightarrow\frac{{i}}{\mathrm{2}}{x}=\mathrm{1}−\mathrm{2}{ki}\Rightarrow{x}=\mathrm{4}{k}−\mathrm{2}{i} \\ $$

Answered by Frix last updated on 27/Aug/23

i^x =e^(i((πx)/2)) =e^π x=4n−2i Test: e^(i(π/2)(4n−2i)) =e^(π+2nπi) =e^(2nπi) e^π =1e^π

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

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