$z in mathbb{C}^-$ where $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$ identity
$begingroup$
Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,
where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.
Why is this true?
complex-numbers
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,
where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.
Why is this true?
complex-numbers
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
$endgroup$
1
$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54
$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56
4
$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58
$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27
add a comment |
$begingroup$
Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,
where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.
Why is this true?
complex-numbers
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
$endgroup$
Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,
where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.
Why is this true?
complex-numbers
complex-numbers
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
edited Dec 6 '18 at 22:36
pablo_mathscobar
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
asked Dec 6 '18 at 21:53
pablo_mathscobarpablo_mathscobar
996
996
1
$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54
$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56
4
$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58
$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27
add a comment |
1
$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54
$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56
4
$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58
$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27
1
1
$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54
$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54
$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56
$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56
4
4
$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58
$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58
$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27
$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
hint
$$|2-z|^2=(2-z)(2-bar{z})$$
$$=4+zbar{z}-2(z+bar{z})$$
$$=4+|z|^2-2(x+iy+x-iy)$$
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
This identity is true for any $zinmathbb C$
$|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint
$$|2-z|^2=(2-z)(2-bar{z})$$
$$=4+zbar{z}-2(z+bar{z})$$
$$=4+|z|^2-2(x+iy+x-iy)$$
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
hint
$$|2-z|^2=(2-z)(2-bar{z})$$
$$=4+zbar{z}-2(z+bar{z})$$
$$=4+|z|^2-2(x+iy+x-iy)$$
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
hint
$$|2-z|^2=(2-z)(2-bar{z})$$
$$=4+zbar{z}-2(z+bar{z})$$
$$=4+|z|^2-2(x+iy+x-iy)$$
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
hint
$$|2-z|^2=(2-z)(2-bar{z})$$
$$=4+zbar{z}-2(z+bar{z})$$
$$=4+|z|^2-2(x+iy+x-iy)$$
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 6 '18 at 21:58
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
$begingroup$
This identity is true for any $zinmathbb C$
$|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
This identity is true for any $zinmathbb C$
$|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
$endgroup$
add a comment |
$begingroup$
This identity is true for any $zinmathbb C$
$|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
$endgroup$
This identity is true for any $zinmathbb C$
$|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
answered Dec 6 '18 at 22:00
Shubham JohriShubham Johri
5,172717
5,172717
add a comment |
add a comment |
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1
$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54
$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56
4
$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58
$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27