Mixing
Solve $| frac2+z2-z | < 1$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
How do I solve this equation $$left| frac2+z2-z right| < 1$$ for complex $z in Bbb C$?
I know the answer is $textRe(z) lt 0$ but I can not understand how to get there. I tried making the denominator real but how do I proceed from here?
$$left| frac2+z2-z right| = left| frac(2+z)^2(2-z)(2+z) right| = left| frac4+4z+z^24 - z^2 right|$$
algebra-precalculus inequality complex-numbers
edited Jul 25 at 9:37
greedoid
26.1k93473
asked Jul 24 at 20:03
philmcole
814418
add a comment |Â
up vote
1
down vote
favorite
How do I solve this equation $$left| frac2+z2-z right| < 1$$ for complex $z in Bbb C$?
I know the answer is $textRe(z) lt 0$ but I can not understand how to get there. I tried making the denominator real but how do I proceed from here?
$$left| frac2+z2-z right| = left| frac(2+z)^2(2-z)(2+z) right| = left| frac4+4z+z^24 - z^2 right|$$
algebra-precalculus inequality complex-numbers
edited Jul 25 at 9:37
greedoid
26.1k93473
asked Jul 24 at 20:03
philmcole
814418
solve the inequality by doing -1<z<1 and solving for z2
– ubuntu_noob
Jul 24 at 20:06
5
Hint: $left| frac2+z2-z right| = 1 iff |z-2|=|z+2|$ is the perpendicular bisector of the segment between $-2$ and $2$ i.e. the imaginary axis.
– dxiv
Jul 24 at 20:07
Interestingly, the solutions to $left|frac2+z2-zright|=r$, where $r<1$, form a circle.
– Batominovski
Jul 24 at 20:09
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How do I solve this equation $$left| frac2+z2-z right| < 1$$ for complex $z in Bbb C$?
I know the answer is $textRe(z) lt 0$ but I can not understand how to get there. I tried making the denominator real but how do I proceed from here?
$$left| frac2+z2-z right| = left| frac(2+z)^2(2-z)(2+z) right| = left| frac4+4z+z^24 - z^2 right|$$
algebra-precalculus inequality complex-numbers
edited Jul 25 at 9:37
greedoid
26.1k93473
asked Jul 24 at 20:03
philmcole
814418
How do I solve this equation $$left| frac2+z2-z right| < 1$$ for complex $z in Bbb C$?
I know the answer is $textRe(z) lt 0$ but I can not understand how to get there. I tried making the denominator real but how do I proceed from here?
$$left| frac2+z2-z right| = left| frac(2+z)^2(2-z)(2+z) right| = left| frac4+4z+z^24 - z^2 right|$$
algebra-precalculus inequality complex-numbers
edited Jul 25 at 9:37
greedoid
26.1k93473
asked Jul 24 at 20:03
philmcole
814418
edited Jul 25 at 9:37
greedoid
26.1k93473
edited Jul 25 at 9:37
greedoid
26.1k93473
edited Jul 25 at 9:37
greedoid
26.1k93473
26.1k93473
asked Jul 24 at 20:03
philmcole
814418
asked Jul 24 at 20:03
philmcole
814418
asked Jul 24 at 20:03
philmcole
814418
814418
solve the inequality by doing -1<z<1 and solving for z2
– ubuntu_noob
Jul 24 at 20:06
5
Hint: $left| frac2+z2-z right| = 1 iff |z-2|=|z+2|$ is the perpendicular bisector of the segment between $-2$ and $2$ i.e. the imaginary axis.
– dxiv
Jul 24 at 20:07
Interestingly, the solutions to $left|frac2+z2-zright|=r$, where $r<1$, form a circle.
– Batominovski
Jul 24 at 20:09
add a comment |Â
solve the inequality by doing -1<z<1 and solving for z2
– ubuntu_noob
Jul 24 at 20:06
5
Hint: $left| frac2+z2-z right| = 1 iff |z-2|=|z+2|$ is the perpendicular bisector of the segment between $-2$ and $2$ i.e. the imaginary axis.
– dxiv
Jul 24 at 20:07
Interestingly, the solutions to $left|frac2+z2-zright|=r$, where $r<1$, form a circle.
– Batominovski
Jul 24 at 20:09
solve the inequality by doing -1<z<1 and solving for z2
– ubuntu_noob
Jul 24 at 20:06
solve the inequality by doing -1<z<1 and solving for z2
– ubuntu_noob
Jul 24 at 20:06
5
5
Hint: $left| frac2+z2-z right| = 1 iff |z-2|=|z+2|$ is the perpendicular bisector of the segment between $-2$ and $2$ i.e. the imaginary axis.
– dxiv
Jul 24 at 20:07
Hint: $left| frac2+z2-z right| = 1 iff |z-2|=|z+2|$ is the perpendicular bisector of the segment between $-2$ and $2$ i.e. the imaginary axis.
– dxiv
Jul 24 at 20:07
Interestingly, the solutions to $left|frac2+z2-zright|=r$, where $r<1$, form a circle.
– Batominovski
Jul 24 at 20:09
Interestingly, the solutions to $left|frac2+z2-zright|=r$, where $r<1$, form a circle.
– Batominovski
Jul 24 at 20:09
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Rewrite like this $$|z-(-2)|<|z-2|$$
so $z$ is closer to $-2$ than to $2$. The set of points $z$ such that $z$ is equally apart from $2$ and $-2$ is $ operatornameRe(z)=0$ so in this case we have $ operatornameRe(z)<0$.
edited Jul 24 at 20:13
Bernard
110k635103
answered Jul 24 at 20:09
greedoid
26.1k93473
add a comment |Â
up vote
3
down vote
A more laborious algebraic approach is given here. The condition $left|frac2+z2-zright|<1$ is equivalent to
$$beginalign4+4,textRe(z)+|z|^2&=4+2z+2barz+zbarz=(2+z)(2+barz)=|2+z|^2
\&<|2-z|^2=(2-z)(2-barz)=4-2z-2barz+zbarz=4-4,textRe(z)+|z|^2,.endalign$$
Thus, $left|frac2+z2-zright|<1$ if and only if $textRe(z)<0$.
answered Jul 24 at 20:15
Batominovski
23.1k22777
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Rewrite like this $$|z-(-2)|<|z-2|$$
so $z$ is closer to $-2$ than to $2$. The set of points $z$ such that $z$ is equally apart from $2$ and $-2$ is $ operatornameRe(z)=0$ so in this case we have $ operatornameRe(z)<0$.
edited Jul 24 at 20:13
Bernard
110k635103
answered Jul 24 at 20:09
greedoid
26.1k93473
add a comment |Â
up vote
3
down vote
accepted
Rewrite like this $$|z-(-2)|<|z-2|$$
so $z$ is closer to $-2$ than to $2$. The set of points $z$ such that $z$ is equally apart from $2$ and $-2$ is $ operatornameRe(z)=0$ so in this case we have $ operatornameRe(z)<0$.
edited Jul 24 at 20:13
Bernard
110k635103
answered Jul 24 at 20:09
greedoid
26.1k93473
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Rewrite like this $$|z-(-2)|<|z-2|$$
so $z$ is closer to $-2$ than to $2$. The set of points $z$ such that $z$ is equally apart from $2$ and $-2$ is $ operatornameRe(z)=0$ so in this case we have $ operatornameRe(z)<0$.
edited Jul 24 at 20:13
Bernard
110k635103
answered Jul 24 at 20:09
greedoid
26.1k93473
Rewrite like this $$|z-(-2)|<|z-2|$$
so $z$ is closer to $-2$ than to $2$. The set of points $z$ such that $z$ is equally apart from $2$ and $-2$ is $ operatornameRe(z)=0$ so in this case we have $ operatornameRe(z)<0$.
edited Jul 24 at 20:13
Bernard
110k635103
answered Jul 24 at 20:09
greedoid
26.1k93473
edited Jul 24 at 20:13
Bernard
110k635103
edited Jul 24 at 20:13
Bernard
110k635103
edited Jul 24 at 20:13
Bernard
110k635103
110k635103
answered Jul 24 at 20:09
greedoid
26.1k93473
answered Jul 24 at 20:09
greedoid
26.1k93473
answered Jul 24 at 20:09
greedoid
26.1k93473
26.1k93473
add a comment |Â
add a comment |Â
up vote
3
down vote
A more laborious algebraic approach is given here. The condition $left|frac2+z2-zright|<1$ is equivalent to
$$beginalign4+4,textRe(z)+|z|^2&=4+2z+2barz+zbarz=(2+z)(2+barz)=|2+z|^2
\&<|2-z|^2=(2-z)(2-barz)=4-2z-2barz+zbarz=4-4,textRe(z)+|z|^2,.endalign$$
Thus, $left|frac2+z2-zright|<1$ if and only if $textRe(z)<0$.
answered Jul 24 at 20:15
Batominovski
23.1k22777
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
add a comment |Â
up vote
3
down vote
A more laborious algebraic approach is given here. The condition $left|frac2+z2-zright|<1$ is equivalent to
$$beginalign4+4,textRe(z)+|z|^2&=4+2z+2barz+zbarz=(2+z)(2+barz)=|2+z|^2
\&<|2-z|^2=(2-z)(2-barz)=4-2z-2barz+zbarz=4-4,textRe(z)+|z|^2,.endalign$$
Thus, $left|frac2+z2-zright|<1$ if and only if $textRe(z)<0$.
answered Jul 24 at 20:15
Batominovski
23.1k22777
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A more laborious algebraic approach is given here. The condition $left|frac2+z2-zright|<1$ is equivalent to
$$beginalign4+4,textRe(z)+|z|^2&=4+2z+2barz+zbarz=(2+z)(2+barz)=|2+z|^2
\&<|2-z|^2=(2-z)(2-barz)=4-2z-2barz+zbarz=4-4,textRe(z)+|z|^2,.endalign$$
Thus, $left|frac2+z2-zright|<1$ if and only if $textRe(z)<0$.
answered Jul 24 at 20:15
Batominovski
23.1k22777
A more laborious algebraic approach is given here. The condition $left|frac2+z2-zright|<1$ is equivalent to
$$beginalign4+4,textRe(z)+|z|^2&=4+2z+2barz+zbarz=(2+z)(2+barz)=|2+z|^2
\&<|2-z|^2=(2-z)(2-barz)=4-2z-2barz+zbarz=4-4,textRe(z)+|z|^2,.endalign$$
Thus, $left|frac2+z2-zright|<1$ if and only if $textRe(z)<0$.
answered Jul 24 at 20:15
Batominovski
23.1k22777
edited Jul 25 at 0:33
answered Jul 24 at 20:15
Batominovski
23.1k22777
answered Jul 24 at 20:15
Batominovski
23.1k22777
answered Jul 24 at 20:15
Batominovski
23.1k22777
23.1k22777
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
add a comment |Â
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
Thanks, this was the trick I was searching for. I should try squaring the next time first!
– philmcole
Jul 24 at 20:34
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861723%2fsolve-frac2z2-z-1%23new-answer', 'question_page');
);
Post as a guest
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
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
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
solve the inequality by doing -1<z<1 and solving for z2
– ubuntu_noob
Jul 24 at 20:06
5
Hint: $left| frac2+z2-z right| = 1 iff |z-2|=|z+2|$ is the perpendicular bisector of the segment between $-2$ and $2$ i.e. the imaginary axis.
– dxiv
Jul 24 at 20:07
Interestingly, the solutions to $left|frac2+z2-zright|=r$, where $r<1$, form a circle.
– Batominovski
Jul 24 at 20:09