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Question Number 225503 by Jyrgen last updated on 31/Oct/25
find x∈C for r∈R\{0} (−r)^x =r
$${find}\:{x}\in\mathbb{C}\:{for}\:{r}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(−{r}\right)^{{x}} ={r} \\ $$
Answered by MrAjder last updated on 02/Nov/25
(−r)^x =r e^(x log(−r)) =r x log(−r)=log(r) x=((log(r))/(log(−r))) (r>0)∀(r<0) (r>0)_(Case 1) :log(r)=ln(r)+2kπi,k∈Z log(−r)=ln(r)+i(2m−1)π,m∈Z x=((ln(r)2kπi)/(ln(r)+i(2m+1)π)) (r<0)_(Case 2) :r=−s,s>0 s^x =−s x log(s)=log(−s) x(ln(s)+2kπi)=ln(s)+i(2m+1)π x=((ln(s)+i(2m+1)π)/(ln(s)+2kπi)) x=((ln(−r)+i(2m+1)π)/(ln(−r)+2kπi)) determinant (((∴x= { ((((ln(r)+2kπi)/(ln(r)+i(2m+1)π)) if r>0)),((((ln(−r)−i(2m+1)π)/(ln(−r)+2kπi)) if r<0)) :}k,m∈Z)))
$$\left(−{r}\right)^{{x}} ={r} \\ $$$${e}^{{x}\:\mathrm{log}\left(−{r}\right)} ={r} \\ $$$${x}\:\mathrm{log}\left(−{r}\right)=\mathrm{log}\left({r}\right) \\ $$$${x}=\frac{\mathrm{log}\left({r}\right)}{\mathrm{log}\left(−{r}\right)} \\ $$$$\left({r}>\mathrm{0}\right)\forall\left({r}<\mathrm{0}\right) \\ $$$$\underset{\mathrm{Case}\:\mathrm{1}} {\underbrace{\left({r}>\mathrm{0}\right)}}\::\mathrm{log}\left({r}\right)=\mathrm{ln}\left({r}\right)+\mathrm{2}{k}\pi{i},{k}\in\mathbb{Z} \\ $$$$\mathrm{log}\left(−{r}\right)=\mathrm{ln}\left({r}\right)+{i}\left(\mathrm{2}{m}−\mathrm{1}\right)\pi,{m}\in\mathbb{Z} \\ $$$${x}=\frac{\mathrm{ln}\left({r}\right)\mathrm{2}{k}\pi{i}}{\mathrm{ln}\left({r}\right)+{i}\left(\mathrm{2}{m}+\mathrm{1}\right)\pi} \\ $$$$\underset{\mathrm{Case}\:\mathrm{2}} {\underbrace{\left({r}<\mathrm{0}\right)}}\::{r}=−{s},{s}>\mathrm{0} \\ $$$${s}^{{x}} =−{s} \\ $$$${x}\:\mathrm{log}\left({s}\right)=\mathrm{log}\left(−{s}\right) \\ $$$${x}\left(\mathrm{ln}\left({s}\right)+\mathrm{2}{k}\pi{i}\right)=\mathrm{ln}\left({s}\right)+{i}\left(\mathrm{2}{m}+\mathrm{1}\right)\pi \\ $$$${x}=\frac{\mathrm{ln}\left({s}\right)+{i}\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}{\mathrm{ln}\left({s}\right)+\mathrm{2}{k}\pi{i}} \\ $$$${x}=\frac{\mathrm{ln}\left(−{r}\right)+{i}\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}{\mathrm{ln}\left(−{r}\right)+\mathrm{2}{k}\pi{i}} \\ $$$$\begin{array}{|c|}{\therefore{x}=\begin{cases}{\frac{\mathrm{ln}\left({r}\right)+\mathrm{2}{k}\pi{i}}{\mathrm{ln}\left({r}\right)+{i}\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}\:\:\:\:\:\:\mathrm{if}\:{r}>\mathrm{0}}\\{\frac{\mathrm{ln}\left(−{r}\right)−{i}\left(\mathrm{2}{m}+\mathrm{1}\right)\pi}{\mathrm{ln}\left(−{r}\right)+\mathrm{2}{k}\pi{i}}\:\mathrm{if}\:{r}<\mathrm{0}}\end{cases}{k},{m}\in\mathbb{Z}}\\\hline\end{array} \\ $$
Commented by mr W last updated on 02/Nov/25
it seems to be alright. but can you check the case r=1? (−1)^x =1 obvioursly x=even numbers are valid solutions. these are included in your answer x=((ln (r)+2kπi)/(ln (r)+(2m+1)πi)). but according to your answer, x=(2/3), (2/5), (4/3) ... were also valid roots. that means: (−1)^(3/2) =1, (−1)^(5/2) =1, (−1)^(4/3) =1, ... isn′t it strange? we can also check the case r=e. (−e)^x =e acc. to your answer x=((ln (r)+2kπi)/(ln (r)+(2m+1)πi)), taking k=0, m=1, we get: x=(1/(1+3πi)) were a valid root, but (−e)^(1/(1+3πi)) ≠e.
$${it}\:{seems}\:{to}\:{be}\:{alright}. \\ $$$${but}\:{can}\:{you}\:{check}\:{the}\:{case}\:{r}=\mathrm{1}? \\ $$$$\left(−\mathrm{1}\right)^{{x}} =\mathrm{1} \\ $$$${obvioursly}\:{x}={even}\:{numbers}\:{are} \\ $$$${valid}\:{solutions}.\:{these}\:{are}\:{included} \\ $$$${in}\:{your}\:{answer}\:{x}=\frac{\mathrm{ln}\:\left({r}\right)+\mathrm{2}{k}\pi{i}}{\mathrm{ln}\:\left({r}\right)+\left(\mathrm{2}{m}+\mathrm{1}\right)\pi{i}}. \\ $$$${but}\:{according}\:{to}\:{your}\:{answer},\: \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{5}},\:\frac{\mathrm{4}}{\mathrm{3}}\:…\:{were}\:{also}\:{valid}\:{roots}. \\ $$$${that}\:{means}: \\ $$$$\left(−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{1},\:\left(−\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} =\mathrm{1},\:\:\left(−\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} =\mathrm{1},\:… \\ $$$${isn}'{t}\:{it}\:{strange}? \\ $$$${we}\:{can}\:{also}\:{check}\:{the}\:{case}\:{r}={e}. \\ $$$$\left(−{e}\right)^{{x}} ={e} \\ $$$${acc}.\:{to}\:{your}\:{answer}\: \\ $$$${x}=\frac{\mathrm{ln}\:\left({r}\right)+\mathrm{2}{k}\pi{i}}{\mathrm{ln}\:\left({r}\right)+\left(\mathrm{2}{m}+\mathrm{1}\right)\pi{i}},\: \\ $$$${taking}\:{k}=\mathrm{0},\:{m}=\mathrm{1},\:{we}\:{get}: \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}\pi{i}}\:{were}\:{a}\:{valid}\:{root}, \\ $$$${but}\:\left(−{e}\right)^{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}\pi{i}}} \neq{e}. \\ $$
Answered by Frix last updated on 02/Nov/25
Obviously r≠−1 because 1^x ≠1∀x x=p+qi; ∣r∣=ρ Case 1: r>0 (ρe^(iπ) )^(p+qi) =ρe^(2nπi) ⇒ p=((2nπ^2 +ln^2 ρ)/(π^2 +ln^2 ρ))∧q=(((2n−1)πln ρ)/(π^2 +ln^2 ρ)) Case 2: r<0 ρ^(p+qi) =ρe^((2n+1)πi) ⇒ p=1∧q=(((2n+1)π)/(ln ρ)) Examples r=1 (−1)^x =1 x=2n r=e (−e)^x =e x=((2nπ^2 +1)/(π^2 +1))+(((2n−1)π)/(π^2 +1))i r=−e e^x =−e x=1+(2n+1)πi
$$\mathrm{Obviously}\:{r}\neq−\mathrm{1}\:\mathrm{because}\:\mathrm{1}^{{x}} \neq\mathrm{1}\forall{x} \\ $$$${x}={p}+{q}\mathrm{i};\:\mid{r}\mid=\rho \\ $$$$\mathrm{Case}\:\mathrm{1}:\:{r}>\mathrm{0} \\ $$$$\left(\rho\mathrm{e}^{\mathrm{i}\pi} \right)^{{p}+{q}\mathrm{i}} =\rho\mathrm{e}^{\mathrm{2}{n}\pi\mathrm{i}} \\ $$$$\Rightarrow\:{p}=\frac{\mathrm{2}{n}\pi^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \:\rho}{\pi^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \:\rho}\wedge{q}=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi\mathrm{ln}\:\rho}{\pi^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \:\rho} \\ $$$$\mathrm{Case}\:\mathrm{2}:\:{r}<\mathrm{0} \\ $$$$\rho^{{p}+{q}\mathrm{i}} =\rho\mathrm{e}^{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\mathrm{i}} \\ $$$$\Rightarrow\:{p}=\mathrm{1}\wedge{q}=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{ln}\:\rho} \\ $$$$ \\ $$$$\mathrm{Examples} \\ $$$${r}=\mathrm{1}\:\:\:\:\:\left(−\mathrm{1}\right)^{{x}} =\mathrm{1}\:\:\:\:\:{x}=\mathrm{2}{n} \\ $$$${r}=\mathrm{e}\:\:\:\:\:\left(−\mathrm{e}\right)^{{x}} =\mathrm{e}\:\:\:\:\:{x}=\frac{\mathrm{2}{n}\pi^{\mathrm{2}} +\mathrm{1}}{\pi^{\mathrm{2}} +\mathrm{1}}+\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{\pi^{\mathrm{2}} +\mathrm{1}}\mathrm{i} \\ $$$${r}=−\mathrm{e}\:\:\:\:\:\mathrm{e}^{{x}} =−\mathrm{e}\:\:\:\:\:{x}=\mathrm{1}+\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\mathrm{i} \\ $$

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