Showing that $|z^{n+1}-z^n|=|z-1|$
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I need to show that $|z^{n+1}-z^n|=|z-1|$ where $z$ is a complex number, and the series ${ a_{n} } ^infty _{n=1}$ is defined as $a_n=z^n$. It is assumed that $|z|=1$.
I'm having trouble showing that :/
sequences-and-series complex-numbers
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
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add a comment |
$begingroup$
I need to show that $|z^{n+1}-z^n|=|z-1|$ where $z$ is a complex number, and the series ${ a_{n} } ^infty _{n=1}$ is defined as $a_n=z^n$. It is assumed that $|z|=1$.
I'm having trouble showing that :/
sequences-and-series complex-numbers
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
$endgroup$
2
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$$|ab|=|a||b|$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:06
add a comment |
$begingroup$
I need to show that $|z^{n+1}-z^n|=|z-1|$ where $z$ is a complex number, and the series ${ a_{n} } ^infty _{n=1}$ is defined as $a_n=z^n$. It is assumed that $|z|=1$.
I'm having trouble showing that :/
sequences-and-series complex-numbers
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
$endgroup$
I need to show that $|z^{n+1}-z^n|=|z-1|$ where $z$ is a complex number, and the series ${ a_{n} } ^infty _{n=1}$ is defined as $a_n=z^n$. It is assumed that $|z|=1$.
I'm having trouble showing that :/
sequences-and-series complex-numbers
sequences-and-series complex-numbers
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
asked Jan 5 at 10:04
Kasper LarsenKasper Larsen
315
315
2
$begingroup$
$$|ab|=|a||b|$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:06
add a comment |
2
$begingroup$
$$|ab|=|a||b|$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:06
2
2
$begingroup$
$$|ab|=|a||b|$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:06
$begingroup$
$$|ab|=|a||b|$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:06
add a comment |
1 Answer
1
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$begingroup$
Note that
$$|z^{n+1} - z^n| = |z^n(z - 1)| = |z^n| cdot |z - 1| = |z|^n cdot |z - 1| = |z - 1|.$$
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
$endgroup$
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$|z^{n+1} - z^n| = |z^n(z - 1)| = |z^n| cdot |z - 1| = |z|^n cdot |z - 1| = |z - 1|.$$
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
$endgroup$
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
add a comment |
$begingroup$
Note that
$$|z^{n+1} - z^n| = |z^n(z - 1)| = |z^n| cdot |z - 1| = |z|^n cdot |z - 1| = |z - 1|.$$
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
$endgroup$
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
add a comment |
$begingroup$
Note that
$$|z^{n+1} - z^n| = |z^n(z - 1)| = |z^n| cdot |z - 1| = |z|^n cdot |z - 1| = |z - 1|.$$
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
$endgroup$
Note that
$$|z^{n+1} - z^n| = |z^n(z - 1)| = |z^n| cdot |z - 1| = |z|^n cdot |z - 1| = |z - 1|.$$
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
answered Jan 5 at 10:07
Theo BenditTheo Bendit
20.5k12354
20.5k12354
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
add a comment |
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
$begingroup$
I can't believe I didn't see that. Thank you.
$endgroup$
– Kasper Larsen
Jan 5 at 10:10
add a comment |
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$begingroup$
$$|ab|=|a||b|$$
$endgroup$
– lab bhattacharjee
Jan 5 at 10:06