Complex numbers and conjugates.
Given that $|z|=√3$, solve the equation $$2overlinez+frac3iz=sqrt15.$$
How to solve this question without a calculator?
complex-numbers
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Given that $|z|=√3$, solve the equation $$2overlinez+frac3iz=sqrt15.$$
How to solve this question without a calculator?
complex-numbers
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
add a comment |
Given that $|z|=√3$, solve the equation $$2overlinez+frac3iz=sqrt15.$$
How to solve this question without a calculator?
complex-numbers
Given that $|z|=√3$, solve the equation $$2overlinez+frac3iz=sqrt15.$$
How to solve this question without a calculator?
complex-numbers
complex-numbers
edited Nov 11 at 11:25
user10354138
7,3592824
7,3592824
asked Nov 11 at 11:04
Vittal Kamath
385
385
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
HINT
Multiplying by $z$ we obtain
$$2bar z+frac3iz=sqrt15 implies 2bar zz+frac3izzfrac i i=sqrt15z$$
then recall that $bar z z=|z|^2$.
,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13
@DhamnekarWinod What did you obtain from here $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
,I got $6+ frac3i=sqrt45$
– 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 $sqrt15z$.
– gimusi
Nov 11 at 12:21
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– Dhamnekar Winod
Nov 11 at 12:31
|
show 1 more comment
WLOG $z=sqrt3e^itimpliesbar z=sqrt3e^-it$ where $t$ is real
$$sqrt15=2sqrt3e^-it+dfrac3isqrt3e^it=sqrt3(2-i)e^-it$$
$$iff e^it=dfrac2-isqrt5$$
We are done.
We can go even further.
$$e^it=e^-iarcsindfrac1sqrt5$$
$$implies t=2npi -arcsindfrac1sqrt5$$ where $n$ is any integer
,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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
Multiplying by $z$ we obtain
$$2bar z+frac3iz=sqrt15 implies 2bar zz+frac3izzfrac i i=sqrt15z$$
then recall that $bar z z=|z|^2$.
,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13
@DhamnekarWinod What did you obtain from here $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
,I got $6+ frac3i=sqrt45$
– 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 $sqrt15z$.
– gimusi
Nov 11 at 12:21
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– Dhamnekar Winod
Nov 11 at 12:31
|
show 1 more comment
HINT
Multiplying by $z$ we obtain
$$2bar z+frac3iz=sqrt15 implies 2bar zz+frac3izzfrac i i=sqrt15z$$
then recall that $bar z z=|z|^2$.
,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13
@DhamnekarWinod What did you obtain from here $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
,I got $6+ frac3i=sqrt45$
– 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 $sqrt15z$.
– gimusi
Nov 11 at 12:21
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– Dhamnekar Winod
Nov 11 at 12:31
|
show 1 more comment
HINT
Multiplying by $z$ we obtain
$$2bar z+frac3iz=sqrt15 implies 2bar zz+frac3izzfrac i i=sqrt15z$$
then recall that $bar z z=|z|^2$.
HINT
Multiplying by $z$ we obtain
$$2bar z+frac3iz=sqrt15 implies 2bar zz+frac3izzfrac i i=sqrt15z$$
then recall that $bar z z=|z|^2$.
answered Nov 11 at 11:16
gimusi
1
1
,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13
@DhamnekarWinod What did you obtain from here $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
,I got $6+ frac3i=sqrt45$
– 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 $sqrt15z$.
– gimusi
Nov 11 at 12:21
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– Dhamnekar Winod
Nov 11 at 12:31
|
show 1 more comment
,what are the next steps to arrive at final answer?
– Dhamnekar Winod
Nov 11 at 12:13
@DhamnekarWinod What did you obtain from here $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
,I got $6+ frac3i=sqrt45$
– 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 $sqrt15z$.
– gimusi
Nov 11 at 12:21
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– 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 $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
@DhamnekarWinod What did you obtain from here $2bar zz+frac3izzfrac i i=sqrt15z$?
– gimusi
Nov 11 at 12:14
,I got $6+ frac3i=sqrt45$
– Dhamnekar Winod
Nov 11 at 12:16
,I got $6+ frac3i=sqrt45$
– 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 $sqrt15z$.
– 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 $sqrt15z$.
– gimusi
Nov 11 at 12:21
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– Dhamnekar Winod
Nov 11 at 12:31
because|z|=$sqrt3$. If this is wrong,then $z=frac6-3isqrt15$
– Dhamnekar Winod
Nov 11 at 12:31
|
show 1 more comment
WLOG $z=sqrt3e^itimpliesbar z=sqrt3e^-it$ where $t$ is real
$$sqrt15=2sqrt3e^-it+dfrac3isqrt3e^it=sqrt3(2-i)e^-it$$
$$iff e^it=dfrac2-isqrt5$$
We are done.
We can go even further.
$$e^it=e^-iarcsindfrac1sqrt5$$
$$implies t=2npi -arcsindfrac1sqrt5$$ where $n$ is any integer
,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
add a comment |
WLOG $z=sqrt3e^itimpliesbar z=sqrt3e^-it$ where $t$ is real
$$sqrt15=2sqrt3e^-it+dfrac3isqrt3e^it=sqrt3(2-i)e^-it$$
$$iff e^it=dfrac2-isqrt5$$
We are done.
We can go even further.
$$e^it=e^-iarcsindfrac1sqrt5$$
$$implies t=2npi -arcsindfrac1sqrt5$$ where $n$ is any integer
,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
add a comment |
WLOG $z=sqrt3e^itimpliesbar z=sqrt3e^-it$ where $t$ is real
$$sqrt15=2sqrt3e^-it+dfrac3isqrt3e^it=sqrt3(2-i)e^-it$$
$$iff e^it=dfrac2-isqrt5$$
We are done.
We can go even further.
$$e^it=e^-iarcsindfrac1sqrt5$$
$$implies t=2npi -arcsindfrac1sqrt5$$ where $n$ is any integer
WLOG $z=sqrt3e^itimpliesbar z=sqrt3e^-it$ where $t$ is real
$$sqrt15=2sqrt3e^-it+dfrac3isqrt3e^it=sqrt3(2-i)e^-it$$
$$iff e^it=dfrac2-isqrt5$$
We are done.
We can go even further.
$$e^it=e^-iarcsindfrac1sqrt5$$
$$implies t=2npi -arcsindfrac1sqrt5$$ where $n$ is any integer
answered Nov 11 at 12:26
lab bhattacharjee
222k15156274
222k15156274
,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
add a comment |
,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
add a comment |
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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