$|z|=z^5$ How many solutions does this equation have?
Multiple choice question : How many solutions does the equation $|z|=z^5$ have?
A - $1$ solution
B - $2$ solutions
C - $5$ solutions
D - $6$ solutions
These were the $4$ possible answers.
I started by allowing $z^5in Bbb R$.
$|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.
I found $2$ obvious solutions, $z=1$ and $z=0$
However I couldn't find any other complex solutions; and the B answer was not correct.
I have also tried to expand with $z=a+ib$ but couldn't find anything, still.
The correct answer was D but I couldn't understand why.
Excuse my English, I am not used to doing maths in English.
complex-analysis complex-numbers
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
asked Jan 6 at 17:21
CaioCaio
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Multiple choice question : How many solutions does the equation $|z|=z^5$ have?
A - $1$ solution
B - $2$ solutions
C - $5$ solutions
D - $6$ solutions
These were the $4$ possible answers.
I started by allowing $z^5in Bbb R$.
$|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.
I found $2$ obvious solutions, $z=1$ and $z=0$
However I couldn't find any other complex solutions; and the B answer was not correct.
I have also tried to expand with $z=a+ib$ but couldn't find anything, still.
The correct answer was D but I couldn't understand why.
Excuse my English, I am not used to doing maths in English.
complex-analysis complex-numbers
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
asked Jan 6 at 17:21
CaioCaio
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Multiple choice question : How many solutions does the equation $|z|=z^5$ have?
A - $1$ solution
B - $2$ solutions
C - $5$ solutions
D - $6$ solutions
These were the $4$ possible answers.
I started by allowing $z^5in Bbb R$.
$|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.
I found $2$ obvious solutions, $z=1$ and $z=0$
However I couldn't find any other complex solutions; and the B answer was not correct.
I have also tried to expand with $z=a+ib$ but couldn't find anything, still.
The correct answer was D but I couldn't understand why.
Excuse my English, I am not used to doing maths in English.
complex-analysis complex-numbers
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
asked Jan 6 at 17:21
CaioCaio
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Multiple choice question : How many solutions does the equation $|z|=z^5$ have?
A - $1$ solution
B - $2$ solutions
C - $5$ solutions
D - $6$ solutions
These were the $4$ possible answers.
I started by allowing $z^5in Bbb R$.
$|z|=|z|^5$ which led me to $|z|=1$ or $|z|=0$.
I found $2$ obvious solutions, $z=1$ and $z=0$
However I couldn't find any other complex solutions; and the B answer was not correct.
I have also tried to expand with $z=a+ib$ but couldn't find anything, still.
The correct answer was D but I couldn't understand why.
Excuse my English, I am not used to doing maths in English.
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
asked Jan 6 at 17:21
CaioCaio
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
asked Jan 6 at 17:21
CaioCaio
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
edited Jan 6 at 17:39
José Carlos Santos
152k22123225
152k22123225
asked Jan 6 at 17:21
CaioCaio
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 6 at 17:21
CaioCaio
61
asked Jan 6 at 17:21
CaioCaio
61
61
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Caio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
add a comment |
It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
add a comment |
$$
eqalign{
& left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
& Rightarrow quad left{ {matrix{
{A = A^{,5} } cr
{e^{,i,5alpha } = 1} cr
} } right.quad Rightarrow quad left{ {matrix{
{A = 0,1} cr
{alpha = 2kpi /5} cr
} } right.quad Rightarrow cr
& Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
$$
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
thank you all this was very helpful
– Caio
Jan 6 at 18:01
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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active
oldest
votes
Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
add a comment |
Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
add a comment |
Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
Any $5$-th root of $1$ is also a solution. With zero, this makes $6$ solutions in total.
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Jan 6 at 17:25
mathcounterexamples.netmathcounterexamples.net
25.3k21953
25.3k21953
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
add a comment |
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
You mean 4th root, no ? btw, $z^5$ must be real, so not that much solution...
– NewMath
Jan 6 at 17:29
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
@NewMath For any $5$-th root $z$ of $1$ you have $z^5=1$ and $vert z vert =1$. Hence this is a root of the equation provided.
– mathcounterexamples.net
Jan 6 at 17:31
add a comment |
It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
add a comment |
It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
add a comment |
It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
It is indeed true that$$lvert zrvert=z^5implieslvert zrvert=0veelvert zrvert=1.$$However, I don't understand your proof. You can prove it as follows:begin{align}lvert zrvert=z^5implies&bigllvertlvert zrvertbigllvert=lvert z^5rvert\iff&lvert zrvert=lvert zrvert^5\iff&lvert zrvert=0veelvert zrvert^4=1\iff&lvert zrvert=0veelvert zrvert=1.end{align}If $lvert zrvert=0$, then $z=0$ and $0$ is indeed a solution. And if $lvert zrvert=1$, the equation becomes $z^5=1$. So, take the $5$ fifth roots of $1$.
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
edited Jan 6 at 17:34
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
answered Jan 6 at 17:28
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
add a comment |
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
Shouldn't it be $|z|=|z|^5iff |z|=0vee |z|^4=1$ instead of $|z|=1$ ?
– NewMath
Jan 6 at 17:31
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Jan 6 at 17:35
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
@NewMath While indeed $|z|=|z|^5iff |z|=0vee |z|^4=1$, you want to solve the stronger $|z|=z^5$.
– Hagen von Eitzen
Jan 6 at 17:48
add a comment |
$$
eqalign{
& left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
& Rightarrow quad left{ {matrix{
{A = A^{,5} } cr
{e^{,i,5alpha } = 1} cr
} } right.quad Rightarrow quad left{ {matrix{
{A = 0,1} cr
{alpha = 2kpi /5} cr
} } right.quad Rightarrow cr
& Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
$$
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
thank you all this was very helpful
– Caio
Jan 6 at 18:01
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
add a comment |
$$
eqalign{
& left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
& Rightarrow quad left{ {matrix{
{A = A^{,5} } cr
{e^{,i,5alpha } = 1} cr
} } right.quad Rightarrow quad left{ {matrix{
{A = 0,1} cr
{alpha = 2kpi /5} cr
} } right.quad Rightarrow cr
& Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
$$
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
thank you all this was very helpful
– Caio
Jan 6 at 18:01
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
add a comment |
$$
eqalign{
& left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
& Rightarrow quad left{ {matrix{
{A = A^{,5} } cr
{e^{,i,5alpha } = 1} cr
} } right.quad Rightarrow quad left{ {matrix{
{A = 0,1} cr
{alpha = 2kpi /5} cr
} } right.quad Rightarrow cr
& Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
$$
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
$$
eqalign{
& left| z right| = z^{,5} ,quad mathop Rightarrow limits^{z = Ae^{,i,alpha } } quad A = A^{,5} e^{,i,5alpha } quad Rightarrow cr
& Rightarrow quad left{ {matrix{
{A = A^{,5} } cr
{e^{,i,5alpha } = 1} cr
} } right.quad Rightarrow quad left{ {matrix{
{A = 0,1} cr
{alpha = 2kpi /5} cr
} } right.quad Rightarrow cr
& Rightarrow quad z = 0,e^{,2kpi /5} ;left| {;k = 0, cdots ,4} right.quad Rightarrow quad 6quad sol. cr}
$$
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
answered Jan 6 at 17:36
G CabG Cab
18.1k31237
18.1k31237
thank you all this was very helpful
– Caio
Jan 6 at 18:01
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
add a comment |
thank you all this was very helpful
– Caio
Jan 6 at 18:01
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
thank you all this was very helpful
– Caio
Jan 6 at 18:01
thank you all this was very helpful
– Caio
Jan 6 at 18:01
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
@Caio: wish it is also clear which is the step-wise approach you should take not to entangle in the common "traps" of complex calculus
– G Cab
Jan 6 at 18:22
add a comment |
Caio is a new contributor. Be nice, and check out our Code of Conduct.
Caio is a new contributor. Be nice, and check out our Code of Conduct.
Caio is a new contributor. Be nice, and check out our Code of Conduct.
Caio is a new contributor. Be nice, and check out our Code of Conduct.
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