Exponentiation with complex numbers
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If a complex number $z$ is multiplied with itself n times, will the result $z^n$ always be greater in magnitude than $z^{n-1}$ or z ?
Is there some formula to find the magnitude of $z^n$ , without finding $z^n$ first?
complex-numbers
asked 3 hours ago
karun mathews
223
add a comment |
up vote
0
down vote
favorite
If a complex number $z$ is multiplied with itself n times, will the result $z^n$ always be greater in magnitude than $z^{n-1}$ or z ?
Is there some formula to find the magnitude of $z^n$ , without finding $z^n$ first?
complex-numbers
asked 3 hours ago
karun mathews
223
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If a complex number $z$ is multiplied with itself n times, will the result $z^n$ always be greater in magnitude than $z^{n-1}$ or z ?
Is there some formula to find the magnitude of $z^n$ , without finding $z^n$ first?
complex-numbers
asked 3 hours ago
karun mathews
223
If a complex number $z$ is multiplied with itself n times, will the result $z^n$ always be greater in magnitude than $z^{n-1}$ or z ?
Is there some formula to find the magnitude of $z^n$ , without finding $z^n$ first?
complex-numbers
complex-numbers
asked 3 hours ago
karun mathews
223
asked 3 hours ago
karun mathews
223
asked 3 hours ago
karun mathews
223
asked 3 hours ago
karun mathews
223
asked 3 hours ago
karun mathews
223
223
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note that every complex number $z$ can be written as $re^{i theta}$ where $r$ is the magnitude of $z$.
Now it follows $z^n=r^n e^{in theta}$
$$=r^n left( cos n theta + i sin n theta right)$$
$$=r^n cos n theta + i r^n sin n theta$$
What's the magnitude of $z^n$?
Well, it's $sqrt{left(r^n cos n theta right)^2+ left( r^n sin n theta right)^2}=r^n$
Now ask yourself when $r^n$ is greater than $r^{n-1}$ or, simply $r$.
answered 3 hours ago
Atiq Rahman
261
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Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
add a comment |
up vote
2
down vote
$|z^n|=|z|^n$.
If $|z|>1$, then the sequence $(|z^n|)$ is strictly increasing,
if $|z|<1$, then the sequence $(|z^n|)$ is strictly decreasing
and if $|z|=1$, then the sequence $(|z^n|)$ is constant.
answered 3 hours ago
Fred
41.9k1642
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that every complex number $z$ can be written as $re^{i theta}$ where $r$ is the magnitude of $z$.
Now it follows $z^n=r^n e^{in theta}$
$$=r^n left( cos n theta + i sin n theta right)$$
$$=r^n cos n theta + i r^n sin n theta$$
What's the magnitude of $z^n$?
Well, it's $sqrt{left(r^n cos n theta right)^2+ left( r^n sin n theta right)^2}=r^n$
Now ask yourself when $r^n$ is greater than $r^{n-1}$ or, simply $r$.
answered 3 hours ago
Atiq Rahman
261
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
add a comment |
up vote
1
down vote
accepted
Note that every complex number $z$ can be written as $re^{i theta}$ where $r$ is the magnitude of $z$.
Now it follows $z^n=r^n e^{in theta}$
$$=r^n left( cos n theta + i sin n theta right)$$
$$=r^n cos n theta + i r^n sin n theta$$
What's the magnitude of $z^n$?
Well, it's $sqrt{left(r^n cos n theta right)^2+ left( r^n sin n theta right)^2}=r^n$
Now ask yourself when $r^n$ is greater than $r^{n-1}$ or, simply $r$.
answered 3 hours ago
Atiq Rahman
261
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that every complex number $z$ can be written as $re^{i theta}$ where $r$ is the magnitude of $z$.
Now it follows $z^n=r^n e^{in theta}$
$$=r^n left( cos n theta + i sin n theta right)$$
$$=r^n cos n theta + i r^n sin n theta$$
What's the magnitude of $z^n$?
Well, it's $sqrt{left(r^n cos n theta right)^2+ left( r^n sin n theta right)^2}=r^n$
Now ask yourself when $r^n$ is greater than $r^{n-1}$ or, simply $r$.
answered 3 hours ago
Atiq Rahman
261
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that every complex number $z$ can be written as $re^{i theta}$ where $r$ is the magnitude of $z$.
Now it follows $z^n=r^n e^{in theta}$
$$=r^n left( cos n theta + i sin n theta right)$$
$$=r^n cos n theta + i r^n sin n theta$$
What's the magnitude of $z^n$?
Well, it's $sqrt{left(r^n cos n theta right)^2+ left( r^n sin n theta right)^2}=r^n$
Now ask yourself when $r^n$ is greater than $r^{n-1}$ or, simply $r$.
answered 3 hours ago
Atiq Rahman
261
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Atiq Rahman
261
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Atiq Rahman
261
answered 3 hours ago
Atiq Rahman
261
261
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Atiq Rahman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
add a comment |
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
3 hours ago
add a comment |
up vote
2
down vote
$|z^n|=|z|^n$.
If $|z|>1$, then the sequence $(|z^n|)$ is strictly increasing,
if $|z|<1$, then the sequence $(|z^n|)$ is strictly decreasing
and if $|z|=1$, then the sequence $(|z^n|)$ is constant.
answered 3 hours ago
Fred
41.9k1642
add a comment |
up vote
2
down vote
$|z^n|=|z|^n$.
If $|z|>1$, then the sequence $(|z^n|)$ is strictly increasing,
if $|z|<1$, then the sequence $(|z^n|)$ is strictly decreasing
and if $|z|=1$, then the sequence $(|z^n|)$ is constant.
answered 3 hours ago
Fred
41.9k1642
add a comment |
up vote
2
down vote
up vote
2
down vote
$|z^n|=|z|^n$.
If $|z|>1$, then the sequence $(|z^n|)$ is strictly increasing,
if $|z|<1$, then the sequence $(|z^n|)$ is strictly decreasing
and if $|z|=1$, then the sequence $(|z^n|)$ is constant.
answered 3 hours ago
Fred
41.9k1642
$|z^n|=|z|^n$.
If $|z|>1$, then the sequence $(|z^n|)$ is strictly increasing,
if $|z|<1$, then the sequence $(|z^n|)$ is strictly decreasing
and if $|z|=1$, then the sequence $(|z^n|)$ is constant.
answered 3 hours ago
Fred
41.9k1642
answered 3 hours ago
Fred
41.9k1642
answered 3 hours ago
Fred
41.9k1642
answered 3 hours ago
Fred
41.9k1642
41.9k1642
add a comment |
add a comment |
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