How can I locate this complex number without actually plotting it on a graph.












2














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







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  • 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
















2














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







share|cite|improve this question




















  • 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














2












2








2







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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




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edited Nov 27 '18 at 4:04









pointguard0

1,310821




1,310821










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














  • 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










2 Answers
2






active

oldest

votes


















3














Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$






share|cite|improve this answer





















  • 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



















0














$|z|$ 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.






share|cite|improve this answer





















  • Thanks for your help
    – priyanka kumari
    Nov 30 '18 at 0:58











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$






share|cite|improve this answer





















  • 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
















3














Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$






share|cite|improve this answer





















  • 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














3












3








3






Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$






share|cite|improve this answer












Hint: $|z|ge 0$, and
$$|z|^2-4|z|-5 = (|z|-2)^2-9<0 implies -3<|z|-2<3 implies ?$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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











0














$|z|$ 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.






share|cite|improve this answer





















  • Thanks for your help
    – priyanka kumari
    Nov 30 '18 at 0:58
















0














$|z|$ 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.






share|cite|improve this answer





















  • Thanks for your help
    – priyanka kumari
    Nov 30 '18 at 0:58














0












0








0






$|z|$ 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.






share|cite|improve this answer












$|z|$ 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 21:21









user376343

2,8932823




2,8932823












  • 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




Thanks for your help
– priyanka kumari
Nov 30 '18 at 0:58


















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