How can we prove that $z^n+1 to 0_mathbbC$ as $nto infty$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-1
down vote

favorite












In the book of Function of One Complex Variable by Conway, at page 31, it is given that



enter image description here



However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $n to infty$ 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 $nto infty$ ?







share|cite|improve this question


















  • 2




    Hint: $$|z^n|=|z|^n$$
    – Did
    Aug 18 at 7:23










  • Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
    – drhab
    Aug 18 at 7:28










  • @drhab Yes, that is correct; it is just a typo.
    – 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.
    – onurcanbektas
    Aug 18 at 7:30






  • 1




    What is the reason for the down vote ?
    – onurcanbektas
    Aug 18 at 7:35














up vote
-1
down vote

favorite












In the book of Function of One Complex Variable by Conway, at page 31, it is given that



enter image description here



However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $n to infty$ 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 $nto infty$ ?







share|cite|improve this question


















  • 2




    Hint: $$|z^n|=|z|^n$$
    – Did
    Aug 18 at 7:23










  • Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
    – drhab
    Aug 18 at 7:28










  • @drhab Yes, that is correct; it is just a typo.
    – 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.
    – onurcanbektas
    Aug 18 at 7:30






  • 1




    What is the reason for the down vote ?
    – onurcanbektas
    Aug 18 at 7:35












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



enter image description here



However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $n to infty$ 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 $nto infty$ ?







share|cite|improve this question














In the book of Function of One Complex Variable by Conway, at page 31, it is given that



enter image description here



However, normally, if $z$ was a real number, we could argue that $z^n+1$ goes to zero as $n to infty$ 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 $nto infty$ ?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 7:39

























asked Aug 18 at 7:20









onurcanbektas

3,1221834




3,1221834







  • 2




    Hint: $$|z^n|=|z|^n$$
    – Did
    Aug 18 at 7:23










  • Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
    – drhab
    Aug 18 at 7:28










  • @drhab Yes, that is correct; it is just a typo.
    – 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.
    – onurcanbektas
    Aug 18 at 7:30






  • 1




    What is the reason for the down vote ?
    – onurcanbektas
    Aug 18 at 7:35












  • 2




    Hint: $$|z^n|=|z|^n$$
    – Did
    Aug 18 at 7:23










  • Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
    – drhab
    Aug 18 at 7:28










  • @drhab Yes, that is correct; it is just a typo.
    – 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.
    – onurcanbektas
    Aug 18 at 7:30






  • 1




    What is the reason for the down vote ?
    – onurcanbektas
    Aug 18 at 7:35







2




2




Hint: $$|z^n|=|z|^n$$
– Did
Aug 18 at 7:23




Hint: $$|z^n|=|z|^n$$
– Did
Aug 18 at 7:23












Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
– drhab
Aug 18 at 7:28




Correction:for $z^nto0$ you need $|z|<1$ (so $|z|leq1$ is not enough).
– drhab
Aug 18 at 7:28












@drhab Yes, that is correct; it is just a typo.
– onurcanbektas
Aug 18 at 7:29




@drhab Yes, that is correct; it is just a typo.
– 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.
– 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.
– onurcanbektas
Aug 18 at 7:30




1




1




What is the reason for the down vote ?
– onurcanbektas
Aug 18 at 7:35




What is the reason for the down vote ?
– onurcanbektas
Aug 18 at 7:35










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 $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.






share|cite|improve this answer






















  • Yes, I'm asking how to prove the first "iff" part.
    – onurcanbektas
    Aug 18 at 7:23

















up vote
1
down vote













Just write your complex numbers in polar form — 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



$$|z^n|=|r^n|to0$$






share|cite|improve this answer




















  • 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.
    – onurcanbektas
    Aug 19 at 4:53










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2886502%2fhow-can-we-prove-that-zn1-to-0-mathbbc-as-n-to-infty%23new-answer', 'question_page');

);

Post as a guest






























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 $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.






share|cite|improve this answer






















  • Yes, I'm asking how to prove the first "iff" part.
    – onurcanbektas
    Aug 18 at 7:23














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 $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.






share|cite|improve this answer






















  • Yes, I'm asking how to prove the first "iff" part.
    – onurcanbektas
    Aug 18 at 7:23












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 $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.






share|cite|improve this answer














$z^n+1 to 0 iff lim |z^n+1 - 0| = 0 iff |z|^n+1 to 0.$



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 $a^2 -b^2 leqslant a^2 + b^2$, $2ab leqslant 2|ab| leqslant a^2+b^2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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.
    – onurcanbektas
    Aug 18 at 7:23
















  • Yes, I'm asking how to prove the first "iff" part.
    – onurcanbektas
    Aug 18 at 7:23















Yes, I'm asking how to prove the first "iff" part.
– onurcanbektas
Aug 18 at 7:23




Yes, I'm asking how to prove the first "iff" part.
– onurcanbektas
Aug 18 at 7:23










up vote
1
down vote













Just write your complex numbers in polar form — 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



$$|z^n|=|r^n|to0$$






share|cite|improve this answer




















  • 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.
    – onurcanbektas
    Aug 19 at 4:53














up vote
1
down vote













Just write your complex numbers in polar form — 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



$$|z^n|=|r^n|to0$$






share|cite|improve this answer




















  • 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.
    – onurcanbektas
    Aug 19 at 4:53












up vote
1
down vote










up vote
1
down vote









Just write your complex numbers in polar form — 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



$$|z^n|=|r^n|to0$$






share|cite|improve this answer












Just write your complex numbers in polar form — 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



$$|z^n|=|r^n|to0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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.
    – 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.
    – 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.
– 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.
– onurcanbektas
Aug 19 at 4:53












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2886502%2fhow-can-we-prove-that-zn1-to-0-mathbbc-as-n-to-infty%23new-answer', 'question_page');

);

Post as a guest













































































這個網誌中的熱門文章

How to combine Bézier curves to a surface?

Mutual Information Always Non-negative

Why am i infinitely getting the same tweet with the Twitter Search API?