Mixing
Find the min and max distance from origin of the curve $vert z+frac1zvert=a$
Clash Royale CLAN TAG#URR8PPP
up vote
2
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$z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation vert z + 1/z vert^2 = a^2. These parts yield the constraints:
$2xy(1+1/(x^2+y^2)^2)=0$ ;
$x^2-y^2+frac2(x^2-y^2)x^2+y^2+frac(x^2-y^2)(x^2+y^2)^2=a^2 $
I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.
What can I do? Any hints on how to approach this?
inequality complex-numbers geometric-inequalities
edited Aug 3 at 16:36
Batominovski
22.6k22776
asked Aug 23 '15 at 1:40
DLV
767820
add a comment |Â
up vote
2
down vote
favorite
$z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation vert z + 1/z vert^2 = a^2. These parts yield the constraints:
$2xy(1+1/(x^2+y^2)^2)=0$ ;
$x^2-y^2+frac2(x^2-y^2)x^2+y^2+frac(x^2-y^2)(x^2+y^2)^2=a^2 $
I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.
What can I do? Any hints on how to approach this?
inequality complex-numbers geometric-inequalities
edited Aug 3 at 16:36
Batominovski
22.6k22776
asked Aug 23 '15 at 1:40
DLV
767820
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation vert z + 1/z vert^2 = a^2. These parts yield the constraints:
$2xy(1+1/(x^2+y^2)^2)=0$ ;
$x^2-y^2+frac2(x^2-y^2)x^2+y^2+frac(x^2-y^2)(x^2+y^2)^2=a^2 $
I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.
What can I do? Any hints on how to approach this?
inequality complex-numbers geometric-inequalities
edited Aug 3 at 16:36
Batominovski
22.6k22776
asked Aug 23 '15 at 1:40
DLV
767820
$z$ is a complex number, by the way.
I've tried a lot of things and always end up with a huge algebraic mess and I've wondered if anyone of you has any idea on how to approach this problem.
One of the most "satisfactory" things I've done so far is to expand both the real and imaginary parts for the equation vert z + 1/z vert^2 = a^2. These parts yield the constraints:
$2xy(1+1/(x^2+y^2)^2)=0$ ;
$x^2-y^2+frac2(x^2-y^2)x^2+y^2+frac(x^2-y^2)(x^2+y^2)^2=a^2 $
I've graphed the equations and supposedly the intersection should be my set of points, but this method is far from analytical.
What can I do? Any hints on how to approach this?
inequality complex-numbers geometric-inequalities
edited Aug 3 at 16:36
Batominovski
22.6k22776
asked Aug 23 '15 at 1:40
DLV
767820
edited Aug 3 at 16:36
Batominovski
22.6k22776
edited Aug 3 at 16:36
Batominovski
22.6k22776
edited Aug 3 at 16:36
Batominovski
22.6k22776
22.6k22776
asked Aug 23 '15 at 1:40
DLV
767820
asked Aug 23 '15 at 1:40
DLV
767820
asked Aug 23 '15 at 1:40
DLV
767820
767820
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Short Solution:
By the Triangle Inequality,
$$a=left|z+frac1zright|geq left||z|-frac1zright|,,$$
giving
$$-aleq |z|-frac1zleq +a,,$$
which means
$$fracsqrta^2+4-a2 leq |z| leq fracsqrta^2+4+a2,.$$
Since $z=pm fracsqrta^2+4+a2$ and $z=pmfracsqrta^2+4-a2$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=rexp(textitheta)$ with $r>0$ and $thetainmathbbR$ is a solution to $left|z+frac1zright|=a$. Then, $$a^2=left|z+frac1zright|^2=r^2+frac1r^2+2cos(2theta),.$$
Hence,
$$a^2-2leq r^2+frac1r^2 leq a^2+2,.tag*$$
If $a>2$, then, $r$ must satisfy $left(r+frac1rright)^2geq a^2$, or $r+frac1rgeq a$, as well as$left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. This means
$$0leq rleq fraca-sqrta^2-42text or rgeq fraca+sqrta^2-42,,$$ along with $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Observe that $fraca-sqrta^2-42> fracsqrta^2+4-a2$ for all $a>2$. Thence,
$$fracsqrta^2+4-a2leq r leq fraca-sqrta^2-42text or fraca+sqrta^2-42leq rleq fracsqrta^2+4+a2,.$$
If $0 leq aleq 2$, then we have by AM-GM that $ r^2+frac1r^2geq 2geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. Therefore, $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Consequently, in both cases, we have $fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2$. The equality on the left side is satisfied when $z=pmfracsqrta^2+4-a2$, and the equality on the right is satisfied iff $z=pmfracsqrta^2+4+a2$. Ergo, the maximum value of $|z|=r$ is $fracsqrta^2+4+a2$, whereas the minimum value is $fracsqrta^2+4-a2$.
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
1
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
1
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Short Solution:
By the Triangle Inequality,
$$a=left|z+frac1zright|geq left||z|-frac1zright|,,$$
giving
$$-aleq |z|-frac1zleq +a,,$$
which means
$$fracsqrta^2+4-a2 leq |z| leq fracsqrta^2+4+a2,.$$
Since $z=pm fracsqrta^2+4+a2$ and $z=pmfracsqrta^2+4-a2$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=rexp(textitheta)$ with $r>0$ and $thetainmathbbR$ is a solution to $left|z+frac1zright|=a$. Then, $$a^2=left|z+frac1zright|^2=r^2+frac1r^2+2cos(2theta),.$$
Hence,
$$a^2-2leq r^2+frac1r^2 leq a^2+2,.tag*$$
If $a>2$, then, $r$ must satisfy $left(r+frac1rright)^2geq a^2$, or $r+frac1rgeq a$, as well as$left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. This means
$$0leq rleq fraca-sqrta^2-42text or rgeq fraca+sqrta^2-42,,$$ along with $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Observe that $fraca-sqrta^2-42> fracsqrta^2+4-a2$ for all $a>2$. Thence,
$$fracsqrta^2+4-a2leq r leq fraca-sqrta^2-42text or fraca+sqrta^2-42leq rleq fracsqrta^2+4+a2,.$$
If $0 leq aleq 2$, then we have by AM-GM that $ r^2+frac1r^2geq 2geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. Therefore, $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Consequently, in both cases, we have $fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2$. The equality on the left side is satisfied when $z=pmfracsqrta^2+4-a2$, and the equality on the right is satisfied iff $z=pmfracsqrta^2+4+a2$. Ergo, the maximum value of $|z|=r$ is $fracsqrta^2+4+a2$, whereas the minimum value is $fracsqrta^2+4-a2$.
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
1
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
1
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
add a comment |Â
up vote
4
down vote
accepted
Short Solution:
By the Triangle Inequality,
$$a=left|z+frac1zright|geq left||z|-frac1zright|,,$$
giving
$$-aleq |z|-frac1zleq +a,,$$
which means
$$fracsqrta^2+4-a2 leq |z| leq fracsqrta^2+4+a2,.$$
Since $z=pm fracsqrta^2+4+a2$ and $z=pmfracsqrta^2+4-a2$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=rexp(textitheta)$ with $r>0$ and $thetainmathbbR$ is a solution to $left|z+frac1zright|=a$. Then, $$a^2=left|z+frac1zright|^2=r^2+frac1r^2+2cos(2theta),.$$
Hence,
$$a^2-2leq r^2+frac1r^2 leq a^2+2,.tag*$$
If $a>2$, then, $r$ must satisfy $left(r+frac1rright)^2geq a^2$, or $r+frac1rgeq a$, as well as$left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. This means
$$0leq rleq fraca-sqrta^2-42text or rgeq fraca+sqrta^2-42,,$$ along with $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Observe that $fraca-sqrta^2-42> fracsqrta^2+4-a2$ for all $a>2$. Thence,
$$fracsqrta^2+4-a2leq r leq fraca-sqrta^2-42text or fraca+sqrta^2-42leq rleq fracsqrta^2+4+a2,.$$
If $0 leq aleq 2$, then we have by AM-GM that $ r^2+frac1r^2geq 2geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. Therefore, $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Consequently, in both cases, we have $fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2$. The equality on the left side is satisfied when $z=pmfracsqrta^2+4-a2$, and the equality on the right is satisfied iff $z=pmfracsqrta^2+4+a2$. Ergo, the maximum value of $|z|=r$ is $fracsqrta^2+4+a2$, whereas the minimum value is $fracsqrta^2+4-a2$.
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
1
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
1
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Short Solution:
By the Triangle Inequality,
$$a=left|z+frac1zright|geq left||z|-frac1zright|,,$$
giving
$$-aleq |z|-frac1zleq +a,,$$
which means
$$fracsqrta^2+4-a2 leq |z| leq fracsqrta^2+4+a2,.$$
Since $z=pm fracsqrta^2+4+a2$ and $z=pmfracsqrta^2+4-a2$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=rexp(textitheta)$ with $r>0$ and $thetainmathbbR$ is a solution to $left|z+frac1zright|=a$. Then, $$a^2=left|z+frac1zright|^2=r^2+frac1r^2+2cos(2theta),.$$
Hence,
$$a^2-2leq r^2+frac1r^2 leq a^2+2,.tag*$$
If $a>2$, then, $r$ must satisfy $left(r+frac1rright)^2geq a^2$, or $r+frac1rgeq a$, as well as$left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. This means
$$0leq rleq fraca-sqrta^2-42text or rgeq fraca+sqrta^2-42,,$$ along with $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Observe that $fraca-sqrta^2-42> fracsqrta^2+4-a2$ for all $a>2$. Thence,
$$fracsqrta^2+4-a2leq r leq fraca-sqrta^2-42text or fraca+sqrta^2-42leq rleq fracsqrta^2+4+a2,.$$
If $0 leq aleq 2$, then we have by AM-GM that $ r^2+frac1r^2geq 2geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. Therefore, $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Consequently, in both cases, we have $fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2$. The equality on the left side is satisfied when $z=pmfracsqrta^2+4-a2$, and the equality on the right is satisfied iff $z=pmfracsqrta^2+4+a2$. Ergo, the maximum value of $|z|=r$ is $fracsqrta^2+4+a2$, whereas the minimum value is $fracsqrta^2+4-a2$.
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
Short Solution:
By the Triangle Inequality,
$$a=left|z+frac1zright|geq left||z|-frac1zright|,,$$
giving
$$-aleq |z|-frac1zleq +a,,$$
which means
$$fracsqrta^2+4-a2 leq |z| leq fracsqrta^2+4+a2,.$$
Since $z=pm fracsqrta^2+4+a2$ and $z=pmfracsqrta^2+4-a2$ are solutions, both bounds are sharp.
Long Solution: I like this one better, though, because it tells you all possible values of $|z|$.
Suppose that $z=rexp(textitheta)$ with $r>0$ and $thetainmathbbR$ is a solution to $left|z+frac1zright|=a$. Then, $$a^2=left|z+frac1zright|^2=r^2+frac1r^2+2cos(2theta),.$$
Hence,
$$a^2-2leq r^2+frac1r^2 leq a^2+2,.tag*$$
If $a>2$, then, $r$ must satisfy $left(r+frac1rright)^2geq a^2$, or $r+frac1rgeq a$, as well as$left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. This means
$$0leq rleq fraca-sqrta^2-42text or rgeq fraca+sqrta^2-42,,$$ along with $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Observe that $fraca-sqrta^2-42> fracsqrta^2+4-a2$ for all $a>2$. Thence,
$$fracsqrta^2+4-a2leq r leq fraca-sqrta^2-42text or fraca+sqrta^2-42leq rleq fracsqrta^2+4+a2,.$$
If $0 leq aleq 2$, then we have by AM-GM that $ r^2+frac1r^2geq 2geq a^2-2$. Thus, only the inequality on the right-hand side of (*) is relevant. We have $left(r-frac1rright)^2leq a^2$, or $-aleq r-frac1rleq +a$. Therefore, $$fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2,.$$
Consequently, in both cases, we have $fracsqrta^2+4-a2 leq r leq fracsqrta^2+4+a2$. The equality on the left side is satisfied when $z=pmfracsqrta^2+4-a2$, and the equality on the right is satisfied iff $z=pmfracsqrta^2+4+a2$. Ergo, the maximum value of $|z|=r$ is $fracsqrta^2+4+a2$, whereas the minimum value is $fracsqrta^2+4-a2$.
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
edited May 25 '16 at 15:13
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
answered Aug 23 '15 at 2:32
Batominovski
22.6k22776
22.6k22776
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
1
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
1
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
add a comment |Â
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
1
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
1
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
Thanks. I'm studying your answer, I'll provide feedback a bit later.
– DLV
Aug 23 '15 at 2:44
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
I've stared at the equations from paragraph 3 onwards for a long time now. I don't know where they're coming from... What should I try to see where they're coming from? By paragraph 3 I mean after the conclusion that $-a leq r-1/r leq a$
– DLV
Aug 23 '15 at 3:04
1
1
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
If $x^2leq y^2$ and $ygeq 0$, then $|x|leq |y|=y$, so $-yleq xleq +y$. I have posted another solution that is a lot simpler.
– Batominovski
Aug 23 '15 at 3:10
1
1
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
I think I misunderstood your question. If $r-frac1rleq +a$ and $rgeq 0$, then you have $0 leq r leq fracsqrta^2+4+a2$, since it is equivalent to $r^2-2ra-1leq 0$. If $r-frac1rgeq-a$ and $rgeq 0$, then you have $rgeq fracsqrta^2+4-a2$, since it is equivalent to $r^2+2ra-1geq 0$.
– Batominovski
Aug 23 '15 at 3:19
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
Thanks! I'll keep studying your answer but I think I've got it now :D
– DLV
Aug 23 '15 at 3:46
add a comment |Â
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