How can I locate this complex number without actually plotting it on a graph.
The question is $$log_{frac{1}{√3}}frac{|z|^2–|z|+1}{2+|z|}>–2.$$ I tried to solve this and simplified it to $$|z|^2-4|z|-5<0.$$
But I wasn't able to locate this number, please tell me the procedure as well as the answer how to do it without actually plotting it on the graph as use of graph is not allowed. Assume $z=x+iy$.
complex-numbers complex-geometry
edited Nov 27 '18 at 4:04
pointguard0
1,310821
asked Nov 27 '18 at 3:02
priyanka kumari
1297
add a comment |
The question is $$log_{frac{1}{√3}}frac{|z|^2–|z|+1}{2+|z|}>–2.$$ I tried to solve this and simplified it to $$|z|^2-4|z|-5<0.$$
But I wasn't able to locate this number, please tell me the procedure as well as the answer how to do it without actually plotting it on the graph as use of graph is not allowed. Assume $z=x+iy$.
complex-numbers complex-geometry
edited Nov 27 '18 at 4:04
pointguard0
1,310821
asked Nov 27 '18 at 3:02
priyanka kumari
1297
1
use polar coordinates. In polar coordinates we have $z=|z| e^{ialpha}$ for some $alphain(-pi, pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution
– Masacroso
Nov 27 '18 at 3:09
add a comment |
The question is $$log_{frac{1}{√3}}frac{|z|^2–|z|+1}{2+|z|}>–2.$$ I tried to solve this and simplified it to $$|z|^2-4|z|-5<0.$$
But I wasn't able to locate this number, please tell me the procedure as well as the answer how to do it without actually plotting it on the graph as use of graph is not allowed. Assume $z=x+iy$.
complex-numbers complex-geometry
edited Nov 27 '18 at 4:04
pointguard0
1,310821
asked Nov 27 '18 at 3:02
priyanka kumari
1297
The question is $$log_{frac{1}{√3}}frac{|z|^2–|z|+1}{2+|z|}>–2.$$ I tried to solve this and simplified it to $$|z|^2-4|z|-5<0.$$
But I wasn't able to locate this number, please tell me the procedure as well as the answer how to do it without actually plotting it on the graph as use of graph is not allowed. Assume $z=x+iy$.
complex-numbers complex-geometry
complex-numbers complex-geometry
edited Nov 27 '18 at 4:04
pointguard0
1,310821
asked Nov 27 '18 at 3:02
priyanka kumari
1297
edited Nov 27 '18 at 4:04
pointguard0
1,310821
asked Nov 27 '18 at 3:02
priyanka kumari
1297
edited Nov 27 '18 at 4:04
pointguard0
1,310821
edited Nov 27 '18 at 4:04
pointguard0
1,310821
edited Nov 27 '18 at 4:04
pointguard0
1,310821
1,310821
asked Nov 27 '18 at 3:02
priyanka kumari
1297
asked Nov 27 '18 at 3:02
priyanka kumari
1297
asked Nov 27 '18 at 3:02
priyanka kumari
1297
1297
1
use polar coordinates. In polar coordinates we have $z=|z| e^{ialpha}$ for some $alphain(-pi, pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution
– Masacroso
Nov 27 '18 at 3:09
add a comment |
1
use polar coordinates. In polar coordinates we have $z=|z| e^{ialpha}$ for some $alphain(-pi, pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution
– Masacroso
Nov 27 '18 at 3:09
1
1
use polar coordinates. In polar coordinates we have $z=|z| e^{ialpha}$ for some $alphain(-pi, pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution
– Masacroso
Nov 27 '18 at 3:09
use polar coordinates. In polar coordinates we have $z=|z| e^{ialpha}$ for some $alphain(-pi, pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution
– Masacroso
Nov 27 '18 at 3:09
add a comment |
2 Answers
2
active
oldest
votes
Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
add a comment |
$|z|$ is the distance from $0$ to the point $z,$ denote it $r.$
Then
$|z|^2-4|z|-5<0$ rewrites $r^2-4r-5<0$ and is equivalent to $$(r-5)(r+1)<0.$$ Since $(r+1)$ is clearly positive (it is a distance + $1$), we solve $$r-5<0$$ or, in terms of $z,$ $$|z|<5$$
Geometrically: the solutions form the circle centered in $0$, with radius $5,$ without the boundary circumference.
answered Nov 29 '18 at 21:21
user376343
2,8932823
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
add a comment |
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015270%2fhow-can-i-locate-this-complex-number-without-actually-plotting-it-on-a-graph%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
add a comment |
Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
add a comment |
Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
answered Nov 27 '18 at 3:38
Math Lover
13.7k31435
13.7k31435
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
add a comment |
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
I got some hint it will be a circle with radius 2 units but what after that
– priyanka kumari
Nov 27 '18 at 4:49
add a comment |
$|z|$ is the distance from $0$ to the point $z,$ denote it $r.$
Then
$|z|^2-4|z|-5<0$ rewrites $r^2-4r-5<0$ and is equivalent to $$(r-5)(r+1)<0.$$ Since $(r+1)$ is clearly positive (it is a distance + $1$), we solve $$r-5<0$$ or, in terms of $z,$ $$|z|<5$$
Geometrically: the solutions form the circle centered in $0$, with radius $5,$ without the boundary circumference.
answered Nov 29 '18 at 21:21
user376343
2,8932823
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
add a comment |
$|z|$ is the distance from $0$ to the point $z,$ denote it $r.$
Then
$|z|^2-4|z|-5<0$ rewrites $r^2-4r-5<0$ and is equivalent to $$(r-5)(r+1)<0.$$ Since $(r+1)$ is clearly positive (it is a distance + $1$), we solve $$r-5<0$$ or, in terms of $z,$ $$|z|<5$$
Geometrically: the solutions form the circle centered in $0$, with radius $5,$ without the boundary circumference.
answered Nov 29 '18 at 21:21
user376343
2,8932823
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
add a comment |
$|z|$ is the distance from $0$ to the point $z,$ denote it $r.$
Then
$|z|^2-4|z|-5<0$ rewrites $r^2-4r-5<0$ and is equivalent to $$(r-5)(r+1)<0.$$ Since $(r+1)$ is clearly positive (it is a distance + $1$), we solve $$r-5<0$$ or, in terms of $z,$ $$|z|<5$$
Geometrically: the solutions form the circle centered in $0$, with radius $5,$ without the boundary circumference.
answered Nov 29 '18 at 21:21
user376343
2,8932823
$|z|$ is the distance from $0$ to the point $z,$ denote it $r.$
Then
$|z|^2-4|z|-5<0$ rewrites $r^2-4r-5<0$ and is equivalent to $$(r-5)(r+1)<0.$$ Since $(r+1)$ is clearly positive (it is a distance + $1$), we solve $$r-5<0$$ or, in terms of $z,$ $$|z|<5$$
Geometrically: the solutions form the circle centered in $0$, with radius $5,$ without the boundary circumference.
answered Nov 29 '18 at 21:21
user376343
2,8932823
answered Nov 29 '18 at 21:21
user376343
2,8932823
answered Nov 29 '18 at 21:21
user376343
2,8932823
answered Nov 29 '18 at 21:21
user376343
2,8932823
2,8932823
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
add a comment |
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015270%2fhow-can-i-locate-this-complex-number-without-actually-plotting-it-on-a-graph%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
use polar coordinates. In polar coordinates we have $z=|z| e^{ialpha}$ for some $alphain(-pi, pi]$. Then, from your inequality, you can estimate $r$. However the inequality doesn't defines a unique possible solution
– Masacroso
Nov 27 '18 at 3:09