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Question Number 60039 by MJS last updated on 17/May/19
find all solutions for z∈C z^i =(1/2)−(1/2)i z^(1−i) =1+i
$$\mathrm{find}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$${z}^{\mathrm{i}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$$${z}^{\mathrm{1}−\mathrm{i}} =\mathrm{1}+\mathrm{i} \\ $$
Commented by maxmathsup by imad last updated on 18/May/19
we have z^i =e^(iln(z)) so if z=r e^(iθ) we get ln(z) =ln(r)+iθ ⇒z^i =e^(i{ln(r)+iθ}) =e^(−θ) e^(iln(r)) =e^(−θ) {cos(ln(r) +isin(ln(r)} z^i =(1/2) −(1/2)i ⇒ e^(−θ) cos(ln(r))=(1/2) and e^(−θ) sin(ln(r))=−(1/2) ⇒ e^(−2θ) {cos^2 (lnr) +sin^2 (ln(r))} =(1/4) +(1/4) =(1/2) ⇒e^(−2θ) =(1/2) ⇒−2θ =−ln(2) ⇒θ =((ln(2))/2) also we have ((sin(lnr))/(cos(lnr))) =−1 ⇒tan(lnr) =−1 ⇒ ln(r) =arctan(−1) =−(π/4) ⇒r =e^(−(π/4)) ⇒z =e^(−(π/4)) e^(i((ln(2))/2)) .
$$\:{we}\:{have}\:{z}^{{i}} \:={e}^{{iln}\left({z}\right)} \:\:{so}\:{if}\:{z}={r}\:{e}^{{i}\theta} \:\:{we}\:{get}\: \\ $$$${ln}\left({z}\right)\:={ln}\left({r}\right)+{i}\theta\:\Rightarrow{z}^{{i}} \:={e}^{{i}\left\{{ln}\left({r}\right)+{i}\theta\right\}} \:={e}^{−\theta} \:{e}^{{iln}\left({r}\right)} ={e}^{−\theta} \left\{{cos}\left({ln}\left({r}\right)\:+{isin}\left({ln}\left({r}\right)\right\}\right.\right. \\ $$$${z}^{{i}} =\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}{i}\:\Rightarrow\:{e}^{−\theta} {cos}\left({ln}\left({r}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{e}^{−\theta} \:{sin}\left({ln}\left({r}\right)\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${e}^{−\mathrm{2}\theta} \:\left\{{cos}^{\mathrm{2}} \left({lnr}\right)\:+{sin}^{\mathrm{2}} \left({ln}\left({r}\right)\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{e}^{−\mathrm{2}\theta} =\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow−\mathrm{2}\theta\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\theta\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\:{also}\:{we}\:{have}\:\:\:\:\frac{{sin}\left({lnr}\right)}{{cos}\left({lnr}\right)}\:=−\mathrm{1}\:\Rightarrow{tan}\left({lnr}\right)\:=−\mathrm{1}\:\Rightarrow \\ $$$${ln}\left({r}\right)\:={arctan}\left(−\mathrm{1}\right)\:=−\frac{\pi}{\mathrm{4}}\:\Rightarrow{r}\:={e}^{−\frac{\pi}{\mathrm{4}}} \:\Rightarrow{z}\:={e}^{−\frac{\pi}{\mathrm{4}}} \:{e}^{{i}\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}} \:\:. \\ $$
Commented by maxmathsup by imad last updated on 18/May/19
in my solution i have taken the principal determination of log.
$${in}\:{my}\:{solution}\:{i}\:{have}\:{taken}\:{the}\:{principal}\:{determination}\:{of}\:{log}. \\ $$
Commented by maxmathsup by imad last updated on 19/May/19
all solution for this equation are Z_n =e^(−(π/4)) e^(i(((ln(2))/2) +2nπ))
$${all}\:{solution}\:{for}\:{this}\:{equation}\:{are}\:{Z}_{{n}} ={e}^{−\frac{\pi}{\mathrm{4}}} \:{e}^{{i}\left(\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:+\mathrm{2}{n}\pi\right)} \\ $$
Answered by tanmay last updated on 17/May/19
(A+iB)^(P+iQ) =e^((P+iQ)Log(A+iB)) Log(A+iB) A=rcosθ B=rsinθ Log(re^(iθ) ) =Log_e (re^(i(2kπ+θ)) =log_e r+i(2kπ+θ) =(1/2)log_e (A^2 +B^2 )+i[2kπ+tan^(−1) ((B/A))] e^((P+iQ)[(1/2)log(A^2 +B^2 )+i{2kπ+tan^(−1) ((B/A))}]) e^p ×e^([{−Q(2kπ+tan^(−1) ((B/A))}+i(Q/2)log(A^2 +B^2 )]) e^P ×e^(−Q(2kπ+tan^(−1) ((B/A))) ×e^(i×(Q/2)log(A^2 +B^2 )) e^(P−Q(2kπ+tan^(−1) ((B/A))) ×[cos(((Qlog(A^2 +B^2 ))/2))+isin(((Qlog(A^2 +B^2 ))/2))] wait sir... now put P=0 and Q=1 e^(−(2kπ+tan^(−1) ((B/A)))) ×[cos(((log(A^2 +B^2 ))/2))+isin(((log(A^2 +B^2 )/2))] still[to go...
$$\left({A}+{iB}\right)^{{P}+{iQ}} \\ $$$$={e}^{\left({P}+{iQ}\right){Log}\left({A}+{iB}\right)} \\ $$$${Log}\left({A}+{iB}\right) \\ $$$${A}={rcos}\theta\:\:\:\:{B}={rsin}\theta \\ $$$${Log}\left({re}^{{i}\theta} \right) \\ $$$$={Log}_{{e}} \left({re}^{{i}\left(\mathrm{2}{k}\pi+\theta\right)} \right. \\ $$$$={log}_{{e}} {r}+{i}\left(\mathrm{2}{k}\pi+\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}_{{e}} \left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)+{i}\left[\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)\right] \\ $$$${e}^{\left({P}+{iQ}\right)\left[\frac{\mathrm{1}}{\mathrm{2}}{log}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)+{i}\left\{\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)\right\}\right]} \\ $$$${e}^{{p}} ×{e}^{\left[\left\{−{Q}\left(\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)\right\}+{i}\frac{{Q}}{\mathrm{2}}{log}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)\right]\right.} \\ $$$${e}^{{P}} ×{e}^{−{Q}\left(\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)\right.} ×{e}^{{i}×\frac{{Q}}{\mathrm{2}}{log}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)} \\ $$$${e}^{{P}−{Q}\left(\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)\right.} ×\left[{cos}\left(\frac{{Qlog}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)}{\mathrm{2}}\right)+{isin}\left(\frac{{Qlog}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)}{\mathrm{2}}\right)\right] \\ $$$${wait}\:{sir}… \\ $$$${now}\:{put}\:{P}=\mathrm{0}\:\:{and}\:{Q}=\mathrm{1} \\ $$$${e}^{−\left(\mathrm{2}{k}\pi+{tan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)\right)} ×\left[{cos}\left(\frac{{log}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)}{\mathrm{2}}\right)+{isin}\left(\frac{{log}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right.}{\mathrm{2}}\right)\right] \\ $$$${still}\left[{to}\:{go}…\right. \\ $$
Commented by MJS last updated on 18/May/19
nice until here. thanks a lot! I′m really not sure how to solve these, will look into it soon...
$$\mathrm{nice}\:\mathrm{until}\:\mathrm{here}.\:\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}! \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{really}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these},\:\mathrm{will} \\ $$$$\mathrm{look}\:\mathrm{into}\:\mathrm{it}\:\mathrm{soon}… \\ $$
Answered by Smail last updated on 18/May/19
(1)z^i =(1/2)−(1/2)i=(1/( (√2)))e^(i(π/4)) =2^(−(1/2)) e^(i((π/4)+2kπ)) z=2^(−(1/(2i))) e^((π/4)+2kπ) =2^(i/2) e^((π/4)+2kπ) =e^((i/2)ln2) ×e^((π/4)+2kπ) z=e^((π/4)+2kπ) e^(i((ln2)/2)) =e^((π/4)+2kπ) 2^(i/2) (2)z^(1−i) =1+i=(√2)e^(i((π/4)+2kπ)) =2^(1/2) e^(i((π/4)+2kπ)) z=2^(1/(2(1−i))) e^((i/(1−i))(π/4+2kπ)) =2^((1+i)/4) ×e^(((i−1)/2)((π/4)+2kπ)) =2^(1/4) ×e^((i/4)ln2) ×e^((i/2)((π/4)+2kπ)) ×e^(−(1/2)((π/4)+2kπ)) z=(2)^(1/4) e^(−((π/4+2kπ)/2)) ×e^(i((π/4)+2kπ+((ln2)/4)))
$$\left(\mathrm{1}\right){z}^{{i}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{i}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} =\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi\right)} \\ $$$${z}=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}{i}}} {e}^{\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi} =\mathrm{2}^{\frac{{i}}{\mathrm{2}}} {e}^{\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi} \\ $$$$={e}^{\frac{{i}}{\mathrm{2}}{ln}\mathrm{2}} ×{e}^{\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi} \\ $$$${z}={e}^{\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi} {e}^{{i}\frac{{ln}\mathrm{2}}{\mathrm{2}}} ={e}^{\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi} \mathrm{2}^{{i}/\mathrm{2}} \\ $$$$\left(\mathrm{2}\right){z}^{\mathrm{1}−{i}} =\mathrm{1}+{i}=\sqrt{\mathrm{2}}{e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi\right)} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi\right)} \\ $$$${z}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{i}\right)}} {e}^{\frac{{i}}{\mathrm{1}−{i}}\left(\pi/\mathrm{4}+\mathrm{2}{k}\pi\right)} =\mathrm{2}^{\frac{\mathrm{1}+{i}}{\mathrm{4}}} ×{e}^{\frac{{i}−\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi\right)} \\ $$$$=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} ×{e}^{\frac{{i}}{\mathrm{4}}{ln}\mathrm{2}} ×{e}^{\frac{{i}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi\right)} ×{e}^{−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi\right)} \\ $$$${z}=\sqrt[{\mathrm{4}}]{\mathrm{2}}{e}^{−\frac{\pi/\mathrm{4}+\mathrm{2}{k}\pi}{\mathrm{2}}} ×{e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\mathrm{2}{k}\pi+\frac{{ln}\mathrm{2}}{\mathrm{4}}\right)} \\ $$
Commented by MJS last updated on 18/May/19
thank you so far. but which values of k are appropriate?
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{far}.\:\mathrm{but}\:\mathrm{which}\:\mathrm{values}\:\mathrm{of}\:{k}\:\mathrm{are} \\ $$$$\mathrm{appropriate}? \\ $$

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