Mixing
Linking complex loci to vectors
$begingroup$
Find the locus of $z$ such that
$$argleft( frac{ z^2 - 1}{ z^2 + 1} right) = 0~, qquad z neq pm i$$
I am able to solve this by substituting $z$ as $x+iy$ and proceeding algebraically.
My question is
How can I solve for this loci in terms of vectors (maybe with a bit plane geometry)?
For example, if I see an expression of the form |$z$|=$r$, I know that it's a circle.
complex-numbers vectors locus
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
$endgroup$
add a comment |
$begingroup$
Find the locus of $z$ such that
$$argleft( frac{ z^2 - 1}{ z^2 + 1} right) = 0~, qquad z neq pm i$$
I am able to solve this by substituting $z$ as $x+iy$ and proceeding algebraically.
My question is
How can I solve for this loci in terms of vectors (maybe with a bit plane geometry)?
For example, if I see an expression of the form |$z$|=$r$, I know that it's a circle.
complex-numbers vectors locus
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
$endgroup$
$begingroup$
Let's see if my edit is deemed an improvement that can prevent further downvoting and vote-to-close. Sameer Thakur please feel free to overwrite my edit.
$endgroup$
– Lee David Chung Lin
Jan 10 at 12:59
add a comment |
$begingroup$
Find the locus of $z$ such that
$$argleft( frac{ z^2 - 1}{ z^2 + 1} right) = 0~, qquad z neq pm i$$
I am able to solve this by substituting $z$ as $x+iy$ and proceeding algebraically.
My question is
How can I solve for this loci in terms of vectors (maybe with a bit plane geometry)?
For example, if I see an expression of the form |$z$|=$r$, I know that it's a circle.
complex-numbers vectors locus
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
$endgroup$
Find the locus of $z$ such that
$$argleft( frac{ z^2 - 1}{ z^2 + 1} right) = 0~, qquad z neq pm i$$
I am able to solve this by substituting $z$ as $x+iy$ and proceeding algebraically.
My question is
How can I solve for this loci in terms of vectors (maybe with a bit plane geometry)?
For example, if I see an expression of the form |$z$|=$r$, I know that it's a circle.
complex-numbers vectors locus
complex-numbers vectors locus
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
edited Jan 10 at 13:00
Lee David Chung Lin
3,88531140
3,88531140
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
asked Jan 4 at 18:10
Sameer ThakurSameer Thakur
285
285
$begingroup$
Let's see if my edit is deemed an improvement that can prevent further downvoting and vote-to-close. Sameer Thakur please feel free to overwrite my edit.
$endgroup$
– Lee David Chung Lin
Jan 10 at 12:59
add a comment |
$begingroup$
Let's see if my edit is deemed an improvement that can prevent further downvoting and vote-to-close. Sameer Thakur please feel free to overwrite my edit.
$endgroup$
– Lee David Chung Lin
Jan 10 at 12:59
$begingroup$
Let's see if my edit is deemed an improvement that can prevent further downvoting and vote-to-close. Sameer Thakur please feel free to overwrite my edit.
$endgroup$
– Lee David Chung Lin
Jan 10 at 12:59
$begingroup$
Let's see if my edit is deemed an improvement that can prevent further downvoting and vote-to-close. Sameer Thakur please feel free to overwrite my edit.
$endgroup$
– Lee David Chung Lin
Jan 10 at 12:59
add a comment |
1 Answer
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$begingroup$
$argleft( frac{ z - z_A}{ z - z_B} right)$ expresses the oriented angle between the vectors $vec{AM}, vec{BM},$ where the points $A,B,M$ represent the complex numbers $z_A, z_B, z$ respectively.
In particular, $argleft( frac{ z - z_A}{ z - z_B} right)=0$ says that the points $A,B$ lie on the same half-line with the bound $M.$ Moreover, $Mneq A$ because $arg$ is not defined for $0,$ and $Mneq B$ for a trivial reason.
Return to the given equation.
The points $Q(z^2)$ such that $0=argleft( frac{ z^2 - 1}{ z^2 + 1} right) = argleft( frac{ z^2 - 1}{ z^2 - (-1)} right)$ are collinear with the points representing $1$ or $-1,$ where $Q$ is the limit point of a convenient half-line. This gives $z^2in (-infty,-1)cup (1,infty).$
From this one concludes easily for $z:$
The locus is the union of four half-lines which are parts of $x-$axis or $y-$axis, and do not contain the segments $[-1,1]$ nor $[-i,i].$
answered Jan 10 at 13:50
user376343user376343
3,3833826
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$argleft( frac{ z - z_A}{ z - z_B} right)$ expresses the oriented angle between the vectors $vec{AM}, vec{BM},$ where the points $A,B,M$ represent the complex numbers $z_A, z_B, z$ respectively.
In particular, $argleft( frac{ z - z_A}{ z - z_B} right)=0$ says that the points $A,B$ lie on the same half-line with the bound $M.$ Moreover, $Mneq A$ because $arg$ is not defined for $0,$ and $Mneq B$ for a trivial reason.
Return to the given equation.
The points $Q(z^2)$ such that $0=argleft( frac{ z^2 - 1}{ z^2 + 1} right) = argleft( frac{ z^2 - 1}{ z^2 - (-1)} right)$ are collinear with the points representing $1$ or $-1,$ where $Q$ is the limit point of a convenient half-line. This gives $z^2in (-infty,-1)cup (1,infty).$
From this one concludes easily for $z:$
The locus is the union of four half-lines which are parts of $x-$axis or $y-$axis, and do not contain the segments $[-1,1]$ nor $[-i,i].$
answered Jan 10 at 13:50
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
$argleft( frac{ z - z_A}{ z - z_B} right)$ expresses the oriented angle between the vectors $vec{AM}, vec{BM},$ where the points $A,B,M$ represent the complex numbers $z_A, z_B, z$ respectively.
In particular, $argleft( frac{ z - z_A}{ z - z_B} right)=0$ says that the points $A,B$ lie on the same half-line with the bound $M.$ Moreover, $Mneq A$ because $arg$ is not defined for $0,$ and $Mneq B$ for a trivial reason.
Return to the given equation.
The points $Q(z^2)$ such that $0=argleft( frac{ z^2 - 1}{ z^2 + 1} right) = argleft( frac{ z^2 - 1}{ z^2 - (-1)} right)$ are collinear with the points representing $1$ or $-1,$ where $Q$ is the limit point of a convenient half-line. This gives $z^2in (-infty,-1)cup (1,infty).$
From this one concludes easily for $z:$
The locus is the union of four half-lines which are parts of $x-$axis or $y-$axis, and do not contain the segments $[-1,1]$ nor $[-i,i].$
answered Jan 10 at 13:50
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
$argleft( frac{ z - z_A}{ z - z_B} right)$ expresses the oriented angle between the vectors $vec{AM}, vec{BM},$ where the points $A,B,M$ represent the complex numbers $z_A, z_B, z$ respectively.
In particular, $argleft( frac{ z - z_A}{ z - z_B} right)=0$ says that the points $A,B$ lie on the same half-line with the bound $M.$ Moreover, $Mneq A$ because $arg$ is not defined for $0,$ and $Mneq B$ for a trivial reason.
Return to the given equation.
The points $Q(z^2)$ such that $0=argleft( frac{ z^2 - 1}{ z^2 + 1} right) = argleft( frac{ z^2 - 1}{ z^2 - (-1)} right)$ are collinear with the points representing $1$ or $-1,$ where $Q$ is the limit point of a convenient half-line. This gives $z^2in (-infty,-1)cup (1,infty).$
From this one concludes easily for $z:$
The locus is the union of four half-lines which are parts of $x-$axis or $y-$axis, and do not contain the segments $[-1,1]$ nor $[-i,i].$
answered Jan 10 at 13:50
user376343user376343
3,3833826
$endgroup$
$argleft( frac{ z - z_A}{ z - z_B} right)$ expresses the oriented angle between the vectors $vec{AM}, vec{BM},$ where the points $A,B,M$ represent the complex numbers $z_A, z_B, z$ respectively.
In particular, $argleft( frac{ z - z_A}{ z - z_B} right)=0$ says that the points $A,B$ lie on the same half-line with the bound $M.$ Moreover, $Mneq A$ because $arg$ is not defined for $0,$ and $Mneq B$ for a trivial reason.
Return to the given equation.
The points $Q(z^2)$ such that $0=argleft( frac{ z^2 - 1}{ z^2 + 1} right) = argleft( frac{ z^2 - 1}{ z^2 - (-1)} right)$ are collinear with the points representing $1$ or $-1,$ where $Q$ is the limit point of a convenient half-line. This gives $z^2in (-infty,-1)cup (1,infty).$
From this one concludes easily for $z:$
The locus is the union of four half-lines which are parts of $x-$axis or $y-$axis, and do not contain the segments $[-1,1]$ nor $[-i,i].$
answered Jan 10 at 13:50
user376343user376343
3,3833826
answered Jan 10 at 13:50
user376343user376343
3,3833826
answered Jan 10 at 13:50
user376343user376343
3,3833826
answered Jan 10 at 13:50
user376343user376343
3,3833826
3,3833826
add a comment |
add a comment |
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$begingroup$
Let's see if my edit is deemed an improvement that can prevent further downvoting and vote-to-close. Sameer Thakur please feel free to overwrite my edit.
$endgroup$
– Lee David Chung Lin
Jan 10 at 12:59