Proving two lines are perpendicular
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Given $z_1,z_2,z_3 $ and $z_4$ are complex numbers, prove that the line joining $z_1,z_2$ and the line joining from $z_3,z_4$ are perpendicular iff $Re{(z_1-z_2)(bar z_3-bar z_4)}=0$. Try not to use polar form.
I try to start with writing $Re{(z_1-z_2)(bar z_3-bar z_4)}=Re{z_1bar z_3}-Re{z_1bar z_4}-Re{z_2bar z_3}+Re{z_2bar z_4}$ (I'm not sure if it's right)
Then how can I make use of the perpendicular condition? Any hints for the reverse direction, or I just have to reverse the argument?
Thank you!
complex-numbers
asked Oct 21 '16 at 7:50
mathshungry
425312
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up vote
0
down vote
favorite
Given $z_1,z_2,z_3 $ and $z_4$ are complex numbers, prove that the line joining $z_1,z_2$ and the line joining from $z_3,z_4$ are perpendicular iff $Re{(z_1-z_2)(bar z_3-bar z_4)}=0$. Try not to use polar form.
I try to start with writing $Re{(z_1-z_2)(bar z_3-bar z_4)}=Re{z_1bar z_3}-Re{z_1bar z_4}-Re{z_2bar z_3}+Re{z_2bar z_4}$ (I'm not sure if it's right)
Then how can I make use of the perpendicular condition? Any hints for the reverse direction, or I just have to reverse the argument?
Thank you!
complex-numbers
asked Oct 21 '16 at 7:50
mathshungry
425312
If $v$ and $w$ are orthogonal vectors then $vcdot w=0$. This is the definition of orthoganility.
– Masacroso
Oct 21 '16 at 8:08
how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$?
– mathshungry
Oct 21 '16 at 9:28
1
The euclidean dot product on $Bbb C^n$ is defined as $$vcdot w:=sum_{k=1}^n v_k bar{w}_k$$ See the definition and characteristics of any dot product here.
– Masacroso
Oct 21 '16 at 9:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $z_1,z_2,z_3 $ and $z_4$ are complex numbers, prove that the line joining $z_1,z_2$ and the line joining from $z_3,z_4$ are perpendicular iff $Re{(z_1-z_2)(bar z_3-bar z_4)}=0$. Try not to use polar form.
I try to start with writing $Re{(z_1-z_2)(bar z_3-bar z_4)}=Re{z_1bar z_3}-Re{z_1bar z_4}-Re{z_2bar z_3}+Re{z_2bar z_4}$ (I'm not sure if it's right)
Then how can I make use of the perpendicular condition? Any hints for the reverse direction, or I just have to reverse the argument?
Thank you!
complex-numbers
asked Oct 21 '16 at 7:50
mathshungry
425312
Given $z_1,z_2,z_3 $ and $z_4$ are complex numbers, prove that the line joining $z_1,z_2$ and the line joining from $z_3,z_4$ are perpendicular iff $Re{(z_1-z_2)(bar z_3-bar z_4)}=0$. Try not to use polar form.
I try to start with writing $Re{(z_1-z_2)(bar z_3-bar z_4)}=Re{z_1bar z_3}-Re{z_1bar z_4}-Re{z_2bar z_3}+Re{z_2bar z_4}$ (I'm not sure if it's right)
Then how can I make use of the perpendicular condition? Any hints for the reverse direction, or I just have to reverse the argument?
Thank you!
complex-numbers
complex-numbers
asked Oct 21 '16 at 7:50
mathshungry
425312
asked Oct 21 '16 at 7:50
mathshungry
425312
asked Oct 21 '16 at 7:50
mathshungry
425312
asked Oct 21 '16 at 7:50
mathshungry
425312
asked Oct 21 '16 at 7:50
mathshungry
425312
425312
If $v$ and $w$ are orthogonal vectors then $vcdot w=0$. This is the definition of orthoganility.
– Masacroso
Oct 21 '16 at 8:08
how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$?
– mathshungry
Oct 21 '16 at 9:28
1
The euclidean dot product on $Bbb C^n$ is defined as $$vcdot w:=sum_{k=1}^n v_k bar{w}_k$$ See the definition and characteristics of any dot product here.
– Masacroso
Oct 21 '16 at 9:50
add a comment |
If $v$ and $w$ are orthogonal vectors then $vcdot w=0$. This is the definition of orthoganility.
– Masacroso
Oct 21 '16 at 8:08
how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$?
– mathshungry
Oct 21 '16 at 9:28
1
The euclidean dot product on $Bbb C^n$ is defined as $$vcdot w:=sum_{k=1}^n v_k bar{w}_k$$ See the definition and characteristics of any dot product here.
– Masacroso
Oct 21 '16 at 9:50
If $v$ and $w$ are orthogonal vectors then $vcdot w=0$. This is the definition of orthoganility.
– Masacroso
Oct 21 '16 at 8:08
If $v$ and $w$ are orthogonal vectors then $vcdot w=0$. This is the definition of orthoganility.
– Masacroso
Oct 21 '16 at 8:08
how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$?
– mathshungry
Oct 21 '16 at 9:28
how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$?
– mathshungry
Oct 21 '16 at 9:28
1
1
The euclidean dot product on $Bbb C^n$ is defined as $$vcdot w:=sum_{k=1}^n v_k bar{w}_k$$ See the definition and characteristics of any dot product here.
– Masacroso
Oct 21 '16 at 9:50
The euclidean dot product on $Bbb C^n$ is defined as $$vcdot w:=sum_{k=1}^n v_k bar{w}_k$$ See the definition and characteristics of any dot product here.
– Masacroso
Oct 21 '16 at 9:50
add a comment |
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Let $z_1-z_2=ae^{itheta}$ and $z_3-z_4=be^{iphi}$, then
$$(z_1-z_2)overline{(z_3-z_4)} = abe^{i(theta-phi)}$$
$$operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}] = abcos (theta-phi)$$
begin{align*}
operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}]=0
& iff cos (theta-phi) =0 \
& iff theta-phi=left(n+frac{1}{2} right) pi \
& iff (z_1-z_2)perp (z_3-z_4)
end{align*}
where $a$, $b> 0$
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
add a comment |
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Let $z_1-z_2=ae^{itheta}$ and $z_3-z_4=be^{iphi}$, then
$$(z_1-z_2)overline{(z_3-z_4)} = abe^{i(theta-phi)}$$
$$operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}] = abcos (theta-phi)$$
begin{align*}
operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}]=0
& iff cos (theta-phi) =0 \
& iff theta-phi=left(n+frac{1}{2} right) pi \
& iff (z_1-z_2)perp (z_3-z_4)
end{align*}
where $a$, $b> 0$
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
add a comment |
up vote
0
down vote
Let $z_1-z_2=ae^{itheta}$ and $z_3-z_4=be^{iphi}$, then
$$(z_1-z_2)overline{(z_3-z_4)} = abe^{i(theta-phi)}$$
$$operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}] = abcos (theta-phi)$$
begin{align*}
operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}]=0
& iff cos (theta-phi) =0 \
& iff theta-phi=left(n+frac{1}{2} right) pi \
& iff (z_1-z_2)perp (z_3-z_4)
end{align*}
where $a$, $b> 0$
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $z_1-z_2=ae^{itheta}$ and $z_3-z_4=be^{iphi}$, then
$$(z_1-z_2)overline{(z_3-z_4)} = abe^{i(theta-phi)}$$
$$operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}] = abcos (theta-phi)$$
begin{align*}
operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}]=0
& iff cos (theta-phi) =0 \
& iff theta-phi=left(n+frac{1}{2} right) pi \
& iff (z_1-z_2)perp (z_3-z_4)
end{align*}
where $a$, $b> 0$
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
Let $z_1-z_2=ae^{itheta}$ and $z_3-z_4=be^{iphi}$, then
$$(z_1-z_2)overline{(z_3-z_4)} = abe^{i(theta-phi)}$$
$$operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}] = abcos (theta-phi)$$
begin{align*}
operatorname{Re}[(z_1-z_2)overline{(z_3-z_4)}]=0
& iff cos (theta-phi) =0 \
& iff theta-phi=left(n+frac{1}{2} right) pi \
& iff (z_1-z_2)perp (z_3-z_4)
end{align*}
where $a$, $b> 0$
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
edited Oct 21 '16 at 16:58
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
answered Oct 21 '16 at 16:50
Ng Chung Tak
13.8k31234
13.8k31234
add a comment |
add a comment |
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If $v$ and $w$ are orthogonal vectors then $vcdot w=0$. This is the definition of orthoganility.
– Masacroso
Oct 21 '16 at 8:08
how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$?
– mathshungry
Oct 21 '16 at 9:28
1
The euclidean dot product on $Bbb C^n$ is defined as $$vcdot w:=sum_{k=1}^n v_k bar{w}_k$$ See the definition and characteristics of any dot product here.
– Masacroso
Oct 21 '16 at 9:50