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 Nov 16 at 8:18
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 Nov 16 at 8:18
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 Nov 16 at 8:18
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 Nov 16 at 8:18
karun mathews
223
asked Nov 16 at 8:18
karun mathews
223
asked Nov 16 at 8:18
karun mathews
223
asked Nov 16 at 8:18
karun mathews
223
asked Nov 16 at 8:18
karun mathews
223
223
add a comment |
add a comment |
2 Answers
2
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1
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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 Nov 16 at 8:36
Atiq Rahman
313
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
add a comment |
up vote
2
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$|z^n|=|z|^n$.
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 Nov 16 at 8:29
Fred
42.8k1643
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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 Nov 16 at 8:36
Atiq Rahman
313
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
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 Nov 16 at 8:36
Atiq Rahman
313
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
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 Nov 16 at 8:36
Atiq Rahman
313
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 Nov 16 at 8:36
Atiq Rahman
313
answered Nov 16 at 8:36
Atiq Rahman
313
answered Nov 16 at 8:36
Atiq Rahman
313
answered Nov 16 at 8:36
Atiq Rahman
313
313
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
add a comment |
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
Oh, I had forgotten about Euler's form- now it makes sense thanks!
– karun mathews
Nov 16 at 8:54
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 Nov 16 at 8:29
Fred
42.8k1643
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 Nov 16 at 8:29
Fred
42.8k1643
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 Nov 16 at 8:29
Fred
42.8k1643
$|z^n|=|z|^n$.
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 Nov 16 at 8:29
Fred
42.8k1643
answered Nov 16 at 8:29
Fred
42.8k1643
answered Nov 16 at 8:29
Fred
42.8k1643
answered Nov 16 at 8:29
Fred
42.8k1643
42.8k1643
add a comment |
add a comment |
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