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Question Number 110197 by redmiiuser last updated on 27/Aug/20
(−1)^π =? (−1)^((22)/7) =(−1)^(3+(1/7)) =(−1)^3 .(−1)^(1/7) =−(−1)^(1/7) let (−1)^π =t ⇒−(−1)^(1/7) =t ⇒(−1)^(1/7) =−t ⇒(−1)=(−t)^7 ⇒−1=−t^7 ⇒1=t^7 hence t=1 ∴(−1)^π =1 i request all math professionals to check this and if any error then pls comment.
$$\left(−\mathrm{1}\right)^{\pi} =? \\ $$$$\left(−\mathrm{1}\right)^{\frac{\mathrm{22}}{\mathrm{7}}} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{3}} .\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$=−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$${let}\:\left(−\mathrm{1}\right)^{\pi} ={t} \\ $$$$\Rightarrow−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} ={t} \\ $$$$\Rightarrow\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} =−{t} \\ $$$$\Rightarrow\left(−\mathrm{1}\right)=\left(−{t}\right)^{\mathrm{7}} \\ $$$$\Rightarrow−\mathrm{1}=−{t}^{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{1}={t}^{\mathrm{7}} \\ $$$${hence}\:{t}=\mathrm{1} \\ $$$$\therefore\left(−\mathrm{1}\right)^{\pi} =\mathrm{1} \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{request}}\:\boldsymbol{{all}}\:\boldsymbol{{math}} \\ $$$$\boldsymbol{{professionals}}\:\boldsymbol{{to}}\: \\ $$$$\boldsymbol{{check}}\:\boldsymbol{{this}}\:\boldsymbol{{and}}\:\boldsymbol{{if}} \\ $$$$\boldsymbol{{any}}\:\boldsymbol{{error}}\:\boldsymbol{{then}}\:\boldsymbol{{pls}} \\ $$$$\boldsymbol{{comment}}. \\ $$
Commented by Dwaipayan Shikari last updated on 27/Aug/20
(−1)^π =e^(π^2 i) =cos(π^2 )+isin(π^2 ) I think it is an imaginary solution
$$\left(−\mathrm{1}\right)^{\pi} ={e}^{\pi^{\mathrm{2}} {i}} ={cos}\left(\pi^{\mathrm{2}} \right)+{isin}\left(\pi^{\mathrm{2}} \right) \\ $$$${I}\:{think}\:{it}\:{is}\:{an}\:{imaginary}\:{solution}\: \\ $$$$ \\ $$
Commented by Aziztisffola last updated on 27/Aug/20
why you put (−1)^π =t and −(−1)^(1/7) =t you suppose that (−1)^π =−(−1)^(1/7) but (−1)^π ≠−(−1)^(1/7) .
$$\mathrm{why}\:\mathrm{you}\:\mathrm{put}\:\left(−\mathrm{1}\right)^{\pi} ={t}\:\mathrm{and}\:−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} ={t} \\ $$$$\:\mathrm{you}\:\mathrm{suppose}\:\mathrm{that}\:\left(−\mathrm{1}\right)^{\pi} =−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$\:\mathrm{but}\:\left(−\mathrm{1}\right)^{\pi} \neq−\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} . \\ $$
Commented by Rasheed.Sindhi last updated on 27/Aug/20
π is an irrational number and is not equal to ((22)/7)(rational) You can′t use 22/7 in the place of π. (22/7 is only approximated value of π)
$$\pi\:{is}\:{an}\:{irrational}\:{number}\:{and} \\ $$$${is}\:{not}\:{equal}\:{to}\:\frac{\mathrm{22}}{\mathrm{7}}\left({rational}\right) \\ $$$${You}\:{can}'{t}\:{use}\:\mathrm{22}/\mathrm{7}\:{in}\:{the}\:{place} \\ $$$${of}\:\pi. \\ $$$$\left(\mathrm{22}/\mathrm{7}\:{is}\:{only}\:{approximated}\right. \\ $$$$\left.{value}\:{of}\:\pi\right) \\ $$
Commented by JDamian last updated on 27/Aug/20
(−1)^((22)/7) =[(−1)^(22) ]^(1/7) =(1)^(1/7) =1^(1/7) Why do you wrongly assume π=((22)/7)?
$$\left(−\mathrm{1}\right)^{\frac{\mathrm{22}}{\mathrm{7}}} =\left[\left(−\mathrm{1}\right)^{\mathrm{22}} \right]^{\frac{\mathrm{1}}{\mathrm{7}}} =\left(\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{7}}} =\mathrm{1}^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$ \\ $$$${Why}\:{do}\:{you}\:{wrongly}\:{assume}\:\pi=\frac{\mathrm{22}}{\mathrm{7}}? \\ $$
Answered by mr W last updated on 27/Aug/20
−1=e^(πi) (−1)^π =e^(π^2 i) =cos π^2 +i sin π^2
$$−\mathrm{1}={e}^{\pi{i}} \\ $$$$\left(−\mathrm{1}\right)^{\pi} ={e}^{\pi^{\mathrm{2}} {i}} =\mathrm{cos}\:\pi^{\mathrm{2}} +{i}\:\mathrm{sin}\:\pi^{\mathrm{2}} \\ $$
Commented by redmiiuser last updated on 27/Aug/20
but sir whats the problem in the above process
$${but}\:{sir}\:{whats}\:{the}\:{problem} \\ $$$${in}\:{the}\:{above}\:{process} \\ $$
Commented by redmiiuser last updated on 27/Aug/20
why?
$${why}? \\ $$
Answered by 1549442205PVT last updated on 27/Aug/20
Set a=(−1)^π ⇒lna=πlni^2 =2πln(i) =2π[ln1+i((π/2)+2kπ)]=iπ^2 (4k+1) ⇒a=(−1)^π =e^(iπ^2 (4k+1)) (k∈Z) (since ln(i)=ln(1)+i((π/2)+2kπ),k∈Z)
$$\mathrm{Set}\:\mathrm{a}=\left(−\mathrm{1}\right)^{\pi} \Rightarrow\mathrm{lna}=\pi\mathrm{lni}^{\mathrm{2}} =\mathrm{2}\pi\mathrm{ln}\left(\mathrm{i}\right) \\ $$$$=\mathrm{2}\pi\left[\mathrm{ln1}+\mathrm{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi\right)\right]=\mathrm{i}\pi^{\mathrm{2}} \left(\mathrm{4k}+\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{a}=\left(−\mathrm{1}\right)^{\pi} =\mathrm{e}^{\mathrm{i}\pi^{\mathrm{2}} \left(\mathrm{4k}+\mathrm{1}\right)} \left(\mathrm{k}\in\mathbb{Z}\right) \\ $$$$\left(\mathrm{since}\:\mathrm{ln}\left(\mathrm{i}\right)=\mathrm{ln}\left(\mathrm{1}\right)+\mathrm{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2k}\pi\right),\mathrm{k}\in\mathbb{Z}\right) \\ $$

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