Complex numbers and conjugates.











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Given that $|z|=√3$, solve the equation $$2overline{z}+frac3{iz}=sqrt{15}.$$




How to solve this question without a calculator?






complex-numbers







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  • Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be.
    – Aleksa
    Nov 11 at 11:16












  • @Vittal Kamath, so what is the answer did you get?
    – Dhamnekar Winod
    Nov 11 at 12:12















up vote
3
down vote

favorite













Given that $|z|=√3$, solve the equation $$2overline{z}+frac3{iz}=sqrt{15}.$$




How to solve this question without a calculator?






complex-numbers







share|cite|improve this question
























  • Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be.
    – Aleksa
    Nov 11 at 11:16












  • @Vittal Kamath, so what is the answer did you get?
    – Dhamnekar Winod
    Nov 11 at 12:12













up vote
3
down vote

favorite









up vote
3
down vote

favorite












Given that $|z|=√3$, solve the equation $$2overline{z}+frac3{iz}=sqrt{15}.$$




How to solve this question without a calculator?






complex-numbers







share|cite|improve this question
















Given that $|z|=√3$, solve the equation $$2overline{z}+frac3{iz}=sqrt{15}.$$




How to solve this question without a calculator?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 11 at 11:25









user10354138

6,7751623




6,7751623










asked Nov 11 at 11:04









Vittal Kamath

383




383












  • Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be.
    – Aleksa
    Nov 11 at 11:16












  • @Vittal Kamath, so what is the answer did you get?
    – Dhamnekar Winod
    Nov 11 at 12:12


















  • Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be.
    – Aleksa
    Nov 11 at 11:16












  • @Vittal Kamath, so what is the answer did you get?
    – Dhamnekar Winod
    Nov 11 at 12:12
















Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be.
– Aleksa
Nov 11 at 11:16






Could you edit your question using MathJax? It's unclear what you're asking and where the division symbol should be.
– Aleksa
Nov 11 at 11:16














@Vittal Kamath, so what is the answer did you get?
– Dhamnekar Winod
Nov 11 at 12:12




@Vittal Kamath, so what is the answer did you get?
– Dhamnekar Winod
Nov 11 at 12:12










2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










HINT



Multiplying by $z$ we obtain



$$2bar z+frac3{iz}=sqrt{15} implies 2bar zz+frac3{iz}zfrac i i=sqrt{15}z$$



then recall that $bar z z=|z|^2$.






share|cite|improve this answer





















  • ,what are the next steps to arrive at final answer?
    – Dhamnekar Winod
    Nov 11 at 12:13












  • @DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
    – gimusi
    Nov 11 at 12:14










  • ,I got $6+ frac{3}{i}=sqrt{45}$
    – Dhamnekar Winod
    Nov 11 at 12:16












  • @DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
    – gimusi
    Nov 11 at 12:21










  • because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
    – Dhamnekar Winod
    Nov 11 at 12:31




















up vote
1
down vote













WLOG $z=sqrt3e^{it}impliesbar z=sqrt3e^{-it}$ where $t$ is real



$$sqrt{15}=2sqrt3e^{-it}+dfrac3{isqrt3e^{it}}=sqrt3(2-i)e^{-it}$$



$$iff e^{it}=dfrac{2-i}{sqrt5}$$



We are done.



We can go even further.



$$e^{it}=e^{-iarcsindfrac1{sqrt5}}$$



$$implies t=2npi -arcsindfrac1{sqrt5}$$ where $n$ is any integer






share|cite|improve this answer





















  • ,what is WLOG z means?
    – Dhamnekar Winod
    Nov 11 at 13:00










  • artofproblemsolving.com/wiki/…
    – lab bhattacharjee
    Nov 11 at 13:03










  • I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
    – Dhamnekar Winod
    Nov 11 at 13:11












  • math.stackexchange.com/questions/2660361/…
    – lab bhattacharjee
    Nov 11 at 13:12











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










HINT



Multiplying by $z$ we obtain



$$2bar z+frac3{iz}=sqrt{15} implies 2bar zz+frac3{iz}zfrac i i=sqrt{15}z$$



then recall that $bar z z=|z|^2$.






share|cite|improve this answer





















  • ,what are the next steps to arrive at final answer?
    – Dhamnekar Winod
    Nov 11 at 12:13












  • @DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
    – gimusi
    Nov 11 at 12:14










  • ,I got $6+ frac{3}{i}=sqrt{45}$
    – Dhamnekar Winod
    Nov 11 at 12:16












  • @DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
    – gimusi
    Nov 11 at 12:21










  • because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
    – Dhamnekar Winod
    Nov 11 at 12:31

















up vote
4
down vote



accepted










HINT



Multiplying by $z$ we obtain



$$2bar z+frac3{iz}=sqrt{15} implies 2bar zz+frac3{iz}zfrac i i=sqrt{15}z$$



then recall that $bar z z=|z|^2$.






share|cite|improve this answer





















  • ,what are the next steps to arrive at final answer?
    – Dhamnekar Winod
    Nov 11 at 12:13












  • @DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
    – gimusi
    Nov 11 at 12:14










  • ,I got $6+ frac{3}{i}=sqrt{45}$
    – Dhamnekar Winod
    Nov 11 at 12:16












  • @DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
    – gimusi
    Nov 11 at 12:21










  • because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
    – Dhamnekar Winod
    Nov 11 at 12:31















up vote
4
down vote



accepted







up vote
4
down vote



accepted






HINT



Multiplying by $z$ we obtain



$$2bar z+frac3{iz}=sqrt{15} implies 2bar zz+frac3{iz}zfrac i i=sqrt{15}z$$



then recall that $bar z z=|z|^2$.






share|cite|improve this answer












HINT



Multiplying by $z$ we obtain



$$2bar z+frac3{iz}=sqrt{15} implies 2bar zz+frac3{iz}zfrac i i=sqrt{15}z$$



then recall that $bar z z=|z|^2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 11 at 11:16









gimusi

90.6k74495




90.6k74495












  • ,what are the next steps to arrive at final answer?
    – Dhamnekar Winod
    Nov 11 at 12:13












  • @DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
    – gimusi
    Nov 11 at 12:14










  • ,I got $6+ frac{3}{i}=sqrt{45}$
    – Dhamnekar Winod
    Nov 11 at 12:16












  • @DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
    – gimusi
    Nov 11 at 12:21










  • because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
    – Dhamnekar Winod
    Nov 11 at 12:31




















  • ,what are the next steps to arrive at final answer?
    – Dhamnekar Winod
    Nov 11 at 12:13












  • @DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
    – gimusi
    Nov 11 at 12:14










  • ,I got $6+ frac{3}{i}=sqrt{45}$
    – Dhamnekar Winod
    Nov 11 at 12:16












  • @DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
    – gimusi
    Nov 11 at 12:21










  • because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
    – Dhamnekar Winod
    Nov 11 at 12:31


















,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13






,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13














@DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
– gimusi
Nov 11 at 12:14




@DhamnekarWinod What did you obtain from here $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
– gimusi
Nov 11 at 12:14












,I got $6+ frac{3}{i}=sqrt{45}$
– Dhamnekar Winod
Nov 11 at 12:16






,I got $6+ frac{3}{i}=sqrt{45}$
– Dhamnekar Winod
Nov 11 at 12:16














@DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
– gimusi
Nov 11 at 12:21




@DhamnekarWinod Why that? We have $bar z z=|z|^2$ and here we are ok, then $frac 3 i =-3i$ but at the RHS we should have $sqrt{15}z$.
– gimusi
Nov 11 at 12:21












because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
– Dhamnekar Winod
Nov 11 at 12:31






because|z|=$sqrt{3}$. If this is wrong,then $z=frac{6-3i}{sqrt{15}}$
– Dhamnekar Winod
Nov 11 at 12:31












up vote
1
down vote













WLOG $z=sqrt3e^{it}impliesbar z=sqrt3e^{-it}$ where $t$ is real



$$sqrt{15}=2sqrt3e^{-it}+dfrac3{isqrt3e^{it}}=sqrt3(2-i)e^{-it}$$



$$iff e^{it}=dfrac{2-i}{sqrt5}$$



We are done.



We can go even further.



$$e^{it}=e^{-iarcsindfrac1{sqrt5}}$$



$$implies t=2npi -arcsindfrac1{sqrt5}$$ where $n$ is any integer






share|cite|improve this answer





















  • ,what is WLOG z means?
    – Dhamnekar Winod
    Nov 11 at 13:00










  • artofproblemsolving.com/wiki/…
    – lab bhattacharjee
    Nov 11 at 13:03










  • I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
    – Dhamnekar Winod
    Nov 11 at 13:11












  • math.stackexchange.com/questions/2660361/…
    – lab bhattacharjee
    Nov 11 at 13:12















up vote
1
down vote













WLOG $z=sqrt3e^{it}impliesbar z=sqrt3e^{-it}$ where $t$ is real



$$sqrt{15}=2sqrt3e^{-it}+dfrac3{isqrt3e^{it}}=sqrt3(2-i)e^{-it}$$



$$iff e^{it}=dfrac{2-i}{sqrt5}$$



We are done.



We can go even further.



$$e^{it}=e^{-iarcsindfrac1{sqrt5}}$$



$$implies t=2npi -arcsindfrac1{sqrt5}$$ where $n$ is any integer






share|cite|improve this answer





















  • ,what is WLOG z means?
    – Dhamnekar Winod
    Nov 11 at 13:00










  • artofproblemsolving.com/wiki/…
    – lab bhattacharjee
    Nov 11 at 13:03










  • I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
    – Dhamnekar Winod
    Nov 11 at 13:11












  • math.stackexchange.com/questions/2660361/…
    – lab bhattacharjee
    Nov 11 at 13:12













up vote
1
down vote










up vote
1
down vote









WLOG $z=sqrt3e^{it}impliesbar z=sqrt3e^{-it}$ where $t$ is real



$$sqrt{15}=2sqrt3e^{-it}+dfrac3{isqrt3e^{it}}=sqrt3(2-i)e^{-it}$$



$$iff e^{it}=dfrac{2-i}{sqrt5}$$



We are done.



We can go even further.



$$e^{it}=e^{-iarcsindfrac1{sqrt5}}$$



$$implies t=2npi -arcsindfrac1{sqrt5}$$ where $n$ is any integer






share|cite|improve this answer












WLOG $z=sqrt3e^{it}impliesbar z=sqrt3e^{-it}$ where $t$ is real



$$sqrt{15}=2sqrt3e^{-it}+dfrac3{isqrt3e^{it}}=sqrt3(2-i)e^{-it}$$



$$iff e^{it}=dfrac{2-i}{sqrt5}$$



We are done.



We can go even further.



$$e^{it}=e^{-iarcsindfrac1{sqrt5}}$$



$$implies t=2npi -arcsindfrac1{sqrt5}$$ where $n$ is any integer







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 11 at 12:26









lab bhattacharjee

221k15155273




221k15155273












  • ,what is WLOG z means?
    – Dhamnekar Winod
    Nov 11 at 13:00










  • artofproblemsolving.com/wiki/…
    – lab bhattacharjee
    Nov 11 at 13:03










  • I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
    – Dhamnekar Winod
    Nov 11 at 13:11












  • math.stackexchange.com/questions/2660361/…
    – lab bhattacharjee
    Nov 11 at 13:12


















  • ,what is WLOG z means?
    – Dhamnekar Winod
    Nov 11 at 13:00










  • artofproblemsolving.com/wiki/…
    – lab bhattacharjee
    Nov 11 at 13:03










  • I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
    – Dhamnekar Winod
    Nov 11 at 13:11












  • math.stackexchange.com/questions/2660361/…
    – lab bhattacharjee
    Nov 11 at 13:12
















,what is WLOG z means?
– Dhamnekar Winod
Nov 11 at 13:00




,what is WLOG z means?
– Dhamnekar Winod
Nov 11 at 13:00












artofproblemsolving.com/wiki/…
– lab bhattacharjee
Nov 11 at 13:03




artofproblemsolving.com/wiki/…
– lab bhattacharjee
Nov 11 at 13:03












I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
– Dhamnekar Winod
Nov 11 at 13:11






I know $e^{ipi}=-1$. So what is $e^{it}?$ and why did you multiply it by z?
– Dhamnekar Winod
Nov 11 at 13:11














math.stackexchange.com/questions/2660361/…
– lab bhattacharjee
Nov 11 at 13:12




math.stackexchange.com/questions/2660361/…
– lab bhattacharjee
Nov 11 at 13:12


















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