Complex numbers and conjugates.
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3
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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
edited Nov 11 at 11:25
user10354138
6,7751623
asked Nov 11 at 11:04
Vittal Kamath
383
add a comment |
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
edited Nov 11 at 11:25
user10354138
6,7751623
asked Nov 11 at 11:04
Vittal Kamath
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
add a comment |
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
edited Nov 11 at 11:25
user10354138
6,7751623
asked Nov 11 at 11:04
Vittal Kamath
383
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
edited Nov 11 at 11:25
user10354138
6,7751623
asked Nov 11 at 11:04
Vittal Kamath
383
edited Nov 11 at 11:25
user10354138
6,7751623
asked Nov 11 at 11:04
Vittal Kamath
383
edited Nov 11 at 11:25
user10354138
6,7751623
edited Nov 11 at 11:25
user10354138
6,7751623
edited Nov 11 at 11:25
user10354138
6,7751623
6,7751623
asked Nov 11 at 11:04
Vittal Kamath
383
asked Nov 11 at 11:04
Vittal Kamath
383
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
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
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 $bar z z=|z|^2$.
answered Nov 11 at 11:16
gimusi
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
|
show 1 more comment
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
answered Nov 11 at 12:26
lab bhattacharjee
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
add a comment |
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 $bar z z=|z|^2$.
answered Nov 11 at 11:16
gimusi
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
|
show 1 more comment
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 $bar z z=|z|^2$.
answered Nov 11 at 11:16
gimusi
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
|
show 1 more comment
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 $bar z z=|z|^2$.
answered Nov 11 at 11:16
gimusi
90.6k74495
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 $bar z z=|z|^2$.
answered Nov 11 at 11:16
gimusi
90.6k74495
answered Nov 11 at 11:16
gimusi
90.6k74495
answered Nov 11 at 11:16
gimusi
90.6k74495
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
|
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 $2bar zz+frac3{iz}zfrac i i=sqrt{15}z$?
– 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
|
show 1 more comment
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
answered Nov 11 at 12:26
lab bhattacharjee
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
add a comment |
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
answered Nov 11 at 12:26
lab bhattacharjee
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
add a comment |
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
answered Nov 11 at 12:26
lab bhattacharjee
221k15155273
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
answered Nov 11 at 12:26
lab bhattacharjee
221k15155273
answered Nov 11 at 12:26
lab bhattacharjee
221k15155273
answered Nov 11 at 12:26
lab bhattacharjee
221k15155273
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
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