Prove 2 segments in complex plane have the same length
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I saw a solution which stated that we simultaneously show that two segments in the complex plane $A$ and $B$ have the same length and are perpendicular if we can show that $B=iA$.
I get that the rotation by $i$ implies they're perpendicular but why does showing this imply they're of the same length?
complex-analysis geometry
edited Dec 4 '18 at 10:59
user376343
3,4133826
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
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add a comment |
$begingroup$
I saw a solution which stated that we simultaneously show that two segments in the complex plane $A$ and $B$ have the same length and are perpendicular if we can show that $B=iA$.
I get that the rotation by $i$ implies they're perpendicular but why does showing this imply they're of the same length?
complex-analysis geometry
edited Dec 4 '18 at 10:59
user376343
3,4133826
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
$endgroup$
2
$begingroup$
One of my rules in complex analysis is if I have a problem that I don't have a better approach to, I write $z=x+iy$ and see where that leads. That would work here.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:14
add a comment |
$begingroup$
I saw a solution which stated that we simultaneously show that two segments in the complex plane $A$ and $B$ have the same length and are perpendicular if we can show that $B=iA$.
I get that the rotation by $i$ implies they're perpendicular but why does showing this imply they're of the same length?
complex-analysis geometry
edited Dec 4 '18 at 10:59
user376343
3,4133826
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
$endgroup$
I saw a solution which stated that we simultaneously show that two segments in the complex plane $A$ and $B$ have the same length and are perpendicular if we can show that $B=iA$.
I get that the rotation by $i$ implies they're perpendicular but why does showing this imply they're of the same length?
complex-analysis geometry
complex-analysis geometry
edited Dec 4 '18 at 10:59
user376343
3,4133826
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
edited Dec 4 '18 at 10:59
user376343
3,4133826
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
edited Dec 4 '18 at 10:59
user376343
3,4133826
edited Dec 4 '18 at 10:59
user376343
3,4133826
edited Dec 4 '18 at 10:59
user376343
3,4133826
3,4133826
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
asked Dec 4 '18 at 3:05
john fowlesjohn fowles
1,203817
1,203817
2
$begingroup$
One of my rules in complex analysis is if I have a problem that I don't have a better approach to, I write $z=x+iy$ and see where that leads. That would work here.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:14
add a comment |
2
$begingroup$
One of my rules in complex analysis is if I have a problem that I don't have a better approach to, I write $z=x+iy$ and see where that leads. That would work here.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:14
2
2
$begingroup$
One of my rules in complex analysis is if I have a problem that I don't have a better approach to, I write $z=x+iy$ and see where that leads. That would work here.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:14
$begingroup$
One of my rules in complex analysis is if I have a problem that I don't have a better approach to, I write $z=x+iy$ and see where that leads. That would work here.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:14
add a comment |
1 Answer
1
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Let $x,y$ be the endpoints of $A$; so $ix,iy$ are the endpoints of $B$ (here, $x,y in mathbb{C}$). In the complex plane, the length of $A$ is just $|y - x|$; the length of $B$ is then:
$$|iy - ix| = |i(y - x)| = |i| |y - x| = |y-x|.$$
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
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1 Answer
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active
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Let $x,y$ be the endpoints of $A$; so $ix,iy$ are the endpoints of $B$ (here, $x,y in mathbb{C}$). In the complex plane, the length of $A$ is just $|y - x|$; the length of $B$ is then:
$$|iy - ix| = |i(y - x)| = |i| |y - x| = |y-x|.$$
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Let $x,y$ be the endpoints of $A$; so $ix,iy$ are the endpoints of $B$ (here, $x,y in mathbb{C}$). In the complex plane, the length of $A$ is just $|y - x|$; the length of $B$ is then:
$$|iy - ix| = |i(y - x)| = |i| |y - x| = |y-x|.$$
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Let $x,y$ be the endpoints of $A$; so $ix,iy$ are the endpoints of $B$ (here, $x,y in mathbb{C}$). In the complex plane, the length of $A$ is just $|y - x|$; the length of $B$ is then:
$$|iy - ix| = |i(y - x)| = |i| |y - x| = |y-x|.$$
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
$endgroup$
Let $x,y$ be the endpoints of $A$; so $ix,iy$ are the endpoints of $B$ (here, $x,y in mathbb{C}$). In the complex plane, the length of $A$ is just $|y - x|$; the length of $B$ is then:
$$|iy - ix| = |i(y - x)| = |i| |y - x| = |y-x|.$$
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
answered Dec 4 '18 at 3:10
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
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$begingroup$
One of my rules in complex analysis is if I have a problem that I don't have a better approach to, I write $z=x+iy$ and see where that leads. That would work here.
$endgroup$
– Ross Millikan
Dec 4 '18 at 3:14