How can we prove that $z^n+1 to 0_mathbbC$ as $nto infty$?
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In the book of Function of One Complex Variable by Conway, at page 31, it is given that
However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $nÃÂ to infty$ if $|z| < 1$, but in the case of complex numbers, if $|z = a + bi| = a^2 + b^2 < 1$, then, for example,
$$z^2 = a^2 - b^2 + (2ab)i,$$
and we can see that it might be the case that $2ab geq b$, so if we see z = (a,b), then $z^2 = (a^2 - b^2, 2ab)$, and the statement $z^n to 0_mathbbC$ means both of the components of $z^n$ goes to $0_mathbbR.
However, even when we square the number $z$, the components does not need to decrease, so how can we be sure that if $|z| < 1$, then $z^n to 0_mathbbC$ as $nto infty$ ?
tl:dr;
How can we prove that $z^n+1 to 0_mathbbC$ as $nto infty$ ?
sequences-and-series complex-analysis power-series
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show 1 more comment
up vote
-1
down vote
favorite
In the book of Function of One Complex Variable by Conway, at page 31, it is given that
However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $nÃÂ to infty$ if $|z| < 1$, but in the case of complex numbers, if $|z = a + bi| = a^2 + b^2 < 1$, then, for example,
$$z^2 = a^2 - b^2 + (2ab)i,$$
and we can see that it might be the case that $2ab geq b$, so if we see z = (a,b), then $z^2 = (a^2 - b^2, 2ab)$, and the statement $z^n to 0_mathbbC$ means both of the components of $z^n$ goes to $0_mathbbR.
However, even when we square the number $z$, the components does not need to decrease, so how can we be sure that if $|z| < 1$, then $z^n to 0_mathbbC$ as $nto infty$ ?
tl:dr;
How can we prove that $z^n+1 to 0_mathbbC$ as $nto infty$ ?
sequences-and-series complex-analysis power-series
2
Hint: $$|z^n|=|z|^n$$
â Did
Aug 18 at 7:23
Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
â drhab
Aug 18 at 7:28
@drhab Yes, that is correct; it is just a typo.
â onurcanbektas
Aug 18 at 7:29
@Did Thanks for the hint. I knew that, but when you write it, I thought I could use $epsilon - delta$ definition, which directly leads to the result that I'm looking for.
â onurcanbektas
Aug 18 at 7:30
1
What is the reason for the down vote ?
â onurcanbektas
Aug 18 at 7:35
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In the book of Function of One Complex Variable by Conway, at page 31, it is given that
However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $nÃÂ to infty$ if $|z| < 1$, but in the case of complex numbers, if $|z = a + bi| = a^2 + b^2 < 1$, then, for example,
$$z^2 = a^2 - b^2 + (2ab)i,$$
and we can see that it might be the case that $2ab geq b$, so if we see z = (a,b), then $z^2 = (a^2 - b^2, 2ab)$, and the statement $z^n to 0_mathbbC$ means both of the components of $z^n$ goes to $0_mathbbR.
However, even when we square the number $z$, the components does not need to decrease, so how can we be sure that if $|z| < 1$, then $z^n to 0_mathbbC$ as $nto infty$ ?
tl:dr;
How can we prove that $z^n+1 to 0_mathbbC$ as $nto infty$ ?
sequences-and-series complex-analysis power-series
In the book of Function of One Complex Variable by Conway, at page 31, it is given that
However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $nÃÂ to infty$ if $|z| < 1$, but in the case of complex numbers, if $|z = a + bi| = a^2 + b^2 < 1$, then, for example,
$$z^2 = a^2 - b^2 + (2ab)i,$$
and we can see that it might be the case that $2ab geq b$, so if we see z = (a,b), then $z^2 = (a^2 - b^2, 2ab)$, and the statement $z^n to 0_mathbbC$ means both of the components of $z^n$ goes to $0_mathbbR.
However, even when we square the number $z$, the components does not need to decrease, so how can we be sure that if $|z| < 1$, then $z^n to 0_mathbbC$ as $nto infty$ ?
tl:dr;
How can we prove that $z^n+1 to 0_mathbbC$ as $nto infty$ ?
sequences-and-series complex-analysis power-series
edited Aug 18 at 7:39
asked Aug 18 at 7:20
onurcanbektas
3,1221834
3,1221834
2
Hint: $$|z^n|=|z|^n$$
â Did
Aug 18 at 7:23
Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
â drhab
Aug 18 at 7:28
@drhab Yes, that is correct; it is just a typo.
â onurcanbektas
Aug 18 at 7:29
@Did Thanks for the hint. I knew that, but when you write it, I thought I could use $epsilon - delta$ definition, which directly leads to the result that I'm looking for.
â onurcanbektas
Aug 18 at 7:30
1
What is the reason for the down vote ?
â onurcanbektas
Aug 18 at 7:35
 |Â
show 1 more comment
2
Hint: $$|z^n|=|z|^n$$
â Did
Aug 18 at 7:23
Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
â drhab
Aug 18 at 7:28
@drhab Yes, that is correct; it is just a typo.
â onurcanbektas
Aug 18 at 7:29
@Did Thanks for the hint. I knew that, but when you write it, I thought I could use $epsilon - delta$ definition, which directly leads to the result that I'm looking for.
â onurcanbektas
Aug 18 at 7:30
1
What is the reason for the down vote ?
â onurcanbektas
Aug 18 at 7:35
2
2
Hint: $$|z^n|=|z|^n$$
â Did
Aug 18 at 7:23
Hint: $$|z^n|=|z|^n$$
â Did
Aug 18 at 7:23
Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
â drhab
Aug 18 at 7:28
Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
â drhab
Aug 18 at 7:28
@drhab Yes, that is correct; it is just a typo.
â onurcanbektas
Aug 18 at 7:29
@drhab Yes, that is correct; it is just a typo.
â onurcanbektas
Aug 18 at 7:29
@Did Thanks for the hint. I knew that, but when you write it, I thought I could use $epsilon - delta$ definition, which directly leads to the result that I'm looking for.
â onurcanbektas
Aug 18 at 7:30
@Did Thanks for the hint. I knew that, but when you write it, I thought I could use $epsilon - delta$ definition, which directly leads to the result that I'm looking for.
â onurcanbektas
Aug 18 at 7:30
1
1
What is the reason for the down vote ?
â onurcanbektas
Aug 18 at 7:35
What is the reason for the down vote ?
â onurcanbektas
Aug 18 at 7:35
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
$z^n+1 to 0 iff lim |z^n+1 - 0| = 0 iff |z|^n+1 to 0.$
First iff: let $z^n = x_n +mathrm i y_n$, then by
$$
max(|x_n|, |y_n|) leqslant |z^n| leqslant |x_n| + |y_n|,
$$
we conclude the first iff.
Example: $z^2 = (a^2 - b^2) + mathrm i 2ab$, then $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
add a comment |Â
up vote
1
down vote
Just write your complex numbers in polar form â are you familiar with the polar representation of complex numbers?
Write $z=re^itheta=r(costheta+isintheta)$ with $r=|z|$. Then $z^n=r^ne^intheta$; $|e^intheta| = 1$ for all real $n,theta$ thus
$$|z^n|=|r^n|to0$$
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$z^n+1 to 0 iff lim |z^n+1 - 0| = 0 iff |z|^n+1 to 0.$
First iff: let $z^n = x_n +mathrm i y_n$, then by
$$
max(|x_n|, |y_n|) leqslant |z^n| leqslant |x_n| + |y_n|,
$$
we conclude the first iff.
Example: $z^2 = (a^2 - b^2) + mathrm i 2ab$, then $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
add a comment |Â
up vote
2
down vote
accepted
$z^n+1 to 0 iff lim |z^n+1 - 0| = 0 iff |z|^n+1 to 0.$
First iff: let $z^n = x_n +mathrm i y_n$, then by
$$
max(|x_n|, |y_n|) leqslant |z^n| leqslant |x_n| + |y_n|,
$$
we conclude the first iff.
Example: $z^2 = (a^2 - b^2) + mathrm i 2ab$, then $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$z^n+1 to 0 iff lim |z^n+1 - 0| = 0 iff |z|^n+1 to 0.$
First iff: let $z^n = x_n +mathrm i y_n$, then by
$$
max(|x_n|, |y_n|) leqslant |z^n| leqslant |x_n| + |y_n|,
$$
we conclude the first iff.
Example: $z^2 = (a^2 - b^2) + mathrm i 2ab$, then $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.
$z^n+1 to 0 iff lim |z^n+1 - 0| = 0 iff |z|^n+1 to 0.$
First iff: let $z^n = x_n +mathrm i y_n$, then by
$$
max(|x_n|, |y_n|) leqslant |z^n| leqslant |x_n| + |y_n|,
$$
we conclude the first iff.
Example: $z^2 = (a^2 - b^2) + mathrm i 2ab$, then $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.
edited Aug 18 at 7:28
answered Aug 18 at 7:22
xbh
2,402114
2,402114
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
add a comment |Â
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
Yes, I'm asking how to prove the first "iff" part.
â onurcanbektas
Aug 18 at 7:23
add a comment |Â
up vote
1
down vote
Just write your complex numbers in polar form â are you familiar with the polar representation of complex numbers?
Write $z=re^itheta=r(costheta+isintheta)$ with $r=|z|$. Then $z^n=r^ne^intheta$; $|e^intheta| = 1$ for all real $n,theta$ thus
$$|z^n|=|r^n|to0$$
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
add a comment |Â
up vote
1
down vote
Just write your complex numbers in polar form â are you familiar with the polar representation of complex numbers?
Write $z=re^itheta=r(costheta+isintheta)$ with $r=|z|$. Then $z^n=r^ne^intheta$; $|e^intheta| = 1$ for all real $n,theta$ thus
$$|z^n|=|r^n|to0$$
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just write your complex numbers in polar form â are you familiar with the polar representation of complex numbers?
Write $z=re^itheta=r(costheta+isintheta)$ with $r=|z|$. Then $z^n=r^ne^intheta$; $|e^intheta| = 1$ for all real $n,theta$ thus
$$|z^n|=|r^n|to0$$
Just write your complex numbers in polar form â are you familiar with the polar representation of complex numbers?
Write $z=re^itheta=r(costheta+isintheta)$ with $r=|z|$. Then $z^n=r^ne^intheta$; $|e^intheta| = 1$ for all real $n,theta$ thus
$$|z^n|=|r^n|to0$$
answered Aug 18 at 18:24
what a disgrace
33316
33316
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
add a comment |Â
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
Thanks for the answer; that is also a clear picture of what is happening, but I haven't examined the polar representation thoroughly, so I did not consent myself to use it.
â onurcanbektas
Aug 19 at 4:53
add a comment |Â
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2
Hint: $$|z^n|=|z|^n$$
â Did
Aug 18 at 7:23
Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
â drhab
Aug 18 at 7:28
@drhab Yes, that is correct; it is just a typo.
â onurcanbektas
Aug 18 at 7:29
@Did Thanks for the hint. I knew that, but when you write it, I thought I could use $epsilon - delta$ definition, which directly leads to the result that I'm looking for.
â onurcanbektas
Aug 18 at 7:30
1
What is the reason for the down vote ?
â onurcanbektas
Aug 18 at 7:35