Solving advanced complex inequalities
$begingroup$
$|z − 2| + |z + 2| ≤ 2$
I've tried solving it like this:
$|x+iy - 2| + |x+iy + 2| le 2\
|x - 2 + iy| + |x + 2 + iy| le 2\
sqrt{(x-2)^2+(y)^2} + sqrt{(x+2)^2+(y)^2} le 2\
sqrt{(x-2)^2+(y)^2} le 2 - sqrt{(x+2)^2+(y)^2}\
sqrt{x^2 - 4x + 4 + y^2} le 2 - sqrt{x^2 + 4x + 4 + y^2}\
x^2 - 4x +4 + y^2 le 4 - 4sqrt{x^2 + 4x + 4 + y^2} + x^2 + 4x + 4 + y^2\
-8x le 4 - 4sqrt{x^2 + 4x + 4 + y^2}\
-2x le 1 - sqrt{x^2 + 4x + 4 + y^2}\
-2x - 1lesqrt{x^2 + 4x + 4 + y^2}\
4x^2+4x + 1 le x^2 + 4x + 4 + y^2\
3x^2-y^2 le 3$
Which seems to be incorrect, what am I doing incorrectly in my procedure? Is my logic correct?
complex-numbers
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
$endgroup$
add a comment |
$begingroup$
$|z − 2| + |z + 2| ≤ 2$
I've tried solving it like this:
$|x+iy - 2| + |x+iy + 2| le 2\
|x - 2 + iy| + |x + 2 + iy| le 2\
sqrt{(x-2)^2+(y)^2} + sqrt{(x+2)^2+(y)^2} le 2\
sqrt{(x-2)^2+(y)^2} le 2 - sqrt{(x+2)^2+(y)^2}\
sqrt{x^2 - 4x + 4 + y^2} le 2 - sqrt{x^2 + 4x + 4 + y^2}\
x^2 - 4x +4 + y^2 le 4 - 4sqrt{x^2 + 4x + 4 + y^2} + x^2 + 4x + 4 + y^2\
-8x le 4 - 4sqrt{x^2 + 4x + 4 + y^2}\
-2x le 1 - sqrt{x^2 + 4x + 4 + y^2}\
-2x - 1lesqrt{x^2 + 4x + 4 + y^2}\
4x^2+4x + 1 le x^2 + 4x + 4 + y^2\
3x^2-y^2 le 3$
Which seems to be incorrect, what am I doing incorrectly in my procedure? Is my logic correct?
complex-numbers
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
$endgroup$
$begingroup$
I can see one wrong step here: From $-2x-1leq sqrt{x^2+4x+4+y^2}$ we cannot deduce the inequality for their squares, that you have written. An example: $-10 leq 3$ but $100notleq 9.$
$endgroup$
– user376343
Dec 5 '18 at 11:37
add a comment |
$begingroup$
$|z − 2| + |z + 2| ≤ 2$
I've tried solving it like this:
$|x+iy - 2| + |x+iy + 2| le 2\
|x - 2 + iy| + |x + 2 + iy| le 2\
sqrt{(x-2)^2+(y)^2} + sqrt{(x+2)^2+(y)^2} le 2\
sqrt{(x-2)^2+(y)^2} le 2 - sqrt{(x+2)^2+(y)^2}\
sqrt{x^2 - 4x + 4 + y^2} le 2 - sqrt{x^2 + 4x + 4 + y^2}\
x^2 - 4x +4 + y^2 le 4 - 4sqrt{x^2 + 4x + 4 + y^2} + x^2 + 4x + 4 + y^2\
-8x le 4 - 4sqrt{x^2 + 4x + 4 + y^2}\
-2x le 1 - sqrt{x^2 + 4x + 4 + y^2}\
-2x - 1lesqrt{x^2 + 4x + 4 + y^2}\
4x^2+4x + 1 le x^2 + 4x + 4 + y^2\
3x^2-y^2 le 3$
Which seems to be incorrect, what am I doing incorrectly in my procedure? Is my logic correct?
complex-numbers
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
$endgroup$
$|z − 2| + |z + 2| ≤ 2$
I've tried solving it like this:
$|x+iy - 2| + |x+iy + 2| le 2\
|x - 2 + iy| + |x + 2 + iy| le 2\
sqrt{(x-2)^2+(y)^2} + sqrt{(x+2)^2+(y)^2} le 2\
sqrt{(x-2)^2+(y)^2} le 2 - sqrt{(x+2)^2+(y)^2}\
sqrt{x^2 - 4x + 4 + y^2} le 2 - sqrt{x^2 + 4x + 4 + y^2}\
x^2 - 4x +4 + y^2 le 4 - 4sqrt{x^2 + 4x + 4 + y^2} + x^2 + 4x + 4 + y^2\
-8x le 4 - 4sqrt{x^2 + 4x + 4 + y^2}\
-2x le 1 - sqrt{x^2 + 4x + 4 + y^2}\
-2x - 1lesqrt{x^2 + 4x + 4 + y^2}\
4x^2+4x + 1 le x^2 + 4x + 4 + y^2\
3x^2-y^2 le 3$
Which seems to be incorrect, what am I doing incorrectly in my procedure? Is my logic correct?
complex-numbers
complex-numbers
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
asked Dec 1 '18 at 15:43
ChrisKChrisK
1
1
$begingroup$
I can see one wrong step here: From $-2x-1leq sqrt{x^2+4x+4+y^2}$ we cannot deduce the inequality for their squares, that you have written. An example: $-10 leq 3$ but $100notleq 9.$
$endgroup$
– user376343
Dec 5 '18 at 11:37
add a comment |
$begingroup$
I can see one wrong step here: From $-2x-1leq sqrt{x^2+4x+4+y^2}$ we cannot deduce the inequality for their squares, that you have written. An example: $-10 leq 3$ but $100notleq 9.$
$endgroup$
– user376343
Dec 5 '18 at 11:37
$begingroup$
I can see one wrong step here: From $-2x-1leq sqrt{x^2+4x+4+y^2}$ we cannot deduce the inequality for their squares, that you have written. An example: $-10 leq 3$ but $100notleq 9.$
$endgroup$
– user376343
Dec 5 '18 at 11:37
$begingroup$
I can see one wrong step here: From $-2x-1leq sqrt{x^2+4x+4+y^2}$ we cannot deduce the inequality for their squares, that you have written. An example: $-10 leq 3$ but $100notleq 9.$
$endgroup$
– user376343
Dec 5 '18 at 11:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Solve the inequality geometrically.
Since $|z-w|$ is the distance between the points representing $z$ and $w$, the inequality $$|z − 2| + |z + 2| ≤ 2$$ can be interpreted as:
The sum of distances from $z$ to $2$ and from $z$ to $-2$ is smaller than $2.$
Put on $x$-axis $-2$ and $2$ and various points $z$ in the plane .... wherever is $zin mathbb{C},$ the sum of distances to $2$ and $-2$ is at least $4,$ obtained when $zin [-2,2].$
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
$endgroup$
add a comment |
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$begingroup$
Solve the inequality geometrically.
Since $|z-w|$ is the distance between the points representing $z$ and $w$, the inequality $$|z − 2| + |z + 2| ≤ 2$$ can be interpreted as:
The sum of distances from $z$ to $2$ and from $z$ to $-2$ is smaller than $2.$
Put on $x$-axis $-2$ and $2$ and various points $z$ in the plane .... wherever is $zin mathbb{C},$ the sum of distances to $2$ and $-2$ is at least $4,$ obtained when $zin [-2,2].$
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
$endgroup$
add a comment |
$begingroup$
Solve the inequality geometrically.
Since $|z-w|$ is the distance between the points representing $z$ and $w$, the inequality $$|z − 2| + |z + 2| ≤ 2$$ can be interpreted as:
The sum of distances from $z$ to $2$ and from $z$ to $-2$ is smaller than $2.$
Put on $x$-axis $-2$ and $2$ and various points $z$ in the plane .... wherever is $zin mathbb{C},$ the sum of distances to $2$ and $-2$ is at least $4,$ obtained when $zin [-2,2].$
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
$endgroup$
add a comment |
$begingroup$
Solve the inequality geometrically.
Since $|z-w|$ is the distance between the points representing $z$ and $w$, the inequality $$|z − 2| + |z + 2| ≤ 2$$ can be interpreted as:
The sum of distances from $z$ to $2$ and from $z$ to $-2$ is smaller than $2.$
Put on $x$-axis $-2$ and $2$ and various points $z$ in the plane .... wherever is $zin mathbb{C},$ the sum of distances to $2$ and $-2$ is at least $4,$ obtained when $zin [-2,2].$
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
$endgroup$
Solve the inequality geometrically.
Since $|z-w|$ is the distance between the points representing $z$ and $w$, the inequality $$|z − 2| + |z + 2| ≤ 2$$ can be interpreted as:
The sum of distances from $z$ to $2$ and from $z$ to $-2$ is smaller than $2.$
Put on $x$-axis $-2$ and $2$ and various points $z$ in the plane .... wherever is $zin mathbb{C},$ the sum of distances to $2$ and $-2$ is at least $4,$ obtained when $zin [-2,2].$
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
answered Dec 5 '18 at 11:48
user376343user376343
3,0332825
3,0332825
add a comment |
add a comment |
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$begingroup$
I can see one wrong step here: From $-2x-1leq sqrt{x^2+4x+4+y^2}$ we cannot deduce the inequality for their squares, that you have written. An example: $-10 leq 3$ but $100notleq 9.$
$endgroup$
– user376343
Dec 5 '18 at 11:37