Show that $(a-b)(c-d)(bara-bard)(barc-barb)+i(cbarc-dbard)textIm(cbarb-cbara-abarb)$ is real
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This is an exercise in Remmert's Theory of Complex Functions, GTM 122, page 17.
Show that for $a,b,c,dinmathbbC$ with $lvert arvert=lvert brvert=lvert crvert$ the complex number $$(a-b)(c-d)(bara-bard)(barc-barb)+i(cbarc-dbard)textIm(cbarb-cbara-abarb)$$ is real.
I let the above expression $w$ and tried showing the $w-barw=0$, but the expression is too complicated to handle and it seems like $w-barw$ depends on $d$, also. Is there any simpler way of showing $w$ is real? Please enlighten me.
complex-analysis complex-numbers
asked Sep 4 at 12:15
Dilemian
382512
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1
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This is an exercise in Remmert's Theory of Complex Functions, GTM 122, page 17.
Show that for $a,b,c,dinmathbbC$ with $lvert arvert=lvert brvert=lvert crvert$ the complex number $$(a-b)(c-d)(bara-bard)(barc-barb)+i(cbarc-dbard)textIm(cbarb-cbara-abarb)$$ is real.
I let the above expression $w$ and tried showing the $w-barw=0$, but the expression is too complicated to handle and it seems like $w-barw$ depends on $d$, also. Is there any simpler way of showing $w$ is real? Please enlighten me.
complex-analysis complex-numbers
asked Sep 4 at 12:15
Dilemian
382512
is not $lvert arvert=lvert brvert=lvert crvert=|d|$ ?
– Nosrati
Sep 4 at 12:18
No, only $lvert arvert=lvert brvert=lvert crvert$ :(
– Dilemian
Sep 4 at 12:21
So with $a=b=c=1$ and $dneq1$ the expression is not real.
– Nosrati
Sep 4 at 12:24
1
@Nosrati For $a=b=c=1$ the expression seems to be $0$.
– Delta-u
Sep 4 at 12:27
Hint: since it is supposed to be valid for all $d$, you can interpret it as a polynomial of second order in the absolute value of $d$. Hence collect terms of the same order in $|d|$ and check the three imaginary components independently and confirm the statement.
– Ronald Blaak
Sep 4 at 16:34
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is an exercise in Remmert's Theory of Complex Functions, GTM 122, page 17.
Show that for $a,b,c,dinmathbbC$ with $lvert arvert=lvert brvert=lvert crvert$ the complex number $$(a-b)(c-d)(bara-bard)(barc-barb)+i(cbarc-dbard)textIm(cbarb-cbara-abarb)$$ is real.
I let the above expression $w$ and tried showing the $w-barw=0$, but the expression is too complicated to handle and it seems like $w-barw$ depends on $d$, also. Is there any simpler way of showing $w$ is real? Please enlighten me.
complex-analysis complex-numbers
asked Sep 4 at 12:15
Dilemian
382512
This is an exercise in Remmert's Theory of Complex Functions, GTM 122, page 17.
Show that for $a,b,c,dinmathbbC$ with $lvert arvert=lvert brvert=lvert crvert$ the complex number $$(a-b)(c-d)(bara-bard)(barc-barb)+i(cbarc-dbard)textIm(cbarb-cbara-abarb)$$ is real.
I let the above expression $w$ and tried showing the $w-barw=0$, but the expression is too complicated to handle and it seems like $w-barw$ depends on $d$, also. Is there any simpler way of showing $w$ is real? Please enlighten me.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Sep 4 at 12:15
Dilemian
382512
asked Sep 4 at 12:15
Dilemian
382512
asked Sep 4 at 12:15
Dilemian
382512
asked Sep 4 at 12:15
Dilemian
382512
asked Sep 4 at 12:15
Dilemian
382512
382512
is not $lvert arvert=lvert brvert=lvert crvert=|d|$ ?
– Nosrati
Sep 4 at 12:18
No, only $lvert arvert=lvert brvert=lvert crvert$ :(
– Dilemian
Sep 4 at 12:21
So with $a=b=c=1$ and $dneq1$ the expression is not real.
– Nosrati
Sep 4 at 12:24
1
@Nosrati For $a=b=c=1$ the expression seems to be $0$.
– Delta-u
Sep 4 at 12:27
Hint: since it is supposed to be valid for all $d$, you can interpret it as a polynomial of second order in the absolute value of $d$. Hence collect terms of the same order in $|d|$ and check the three imaginary components independently and confirm the statement.
– Ronald Blaak
Sep 4 at 16:34
 |Â
show 1 more comment
is not $lvert arvert=lvert brvert=lvert crvert=|d|$ ?
– Nosrati
Sep 4 at 12:18
No, only $lvert arvert=lvert brvert=lvert crvert$ :(
– Dilemian
Sep 4 at 12:21
So with $a=b=c=1$ and $dneq1$ the expression is not real.
– Nosrati
Sep 4 at 12:24
1
@Nosrati For $a=b=c=1$ the expression seems to be $0$.
– Delta-u
Sep 4 at 12:27
Hint: since it is supposed to be valid for all $d$, you can interpret it as a polynomial of second order in the absolute value of $d$. Hence collect terms of the same order in $|d|$ and check the three imaginary components independently and confirm the statement.
– Ronald Blaak
Sep 4 at 16:34
is not $lvert arvert=lvert brvert=lvert crvert=|d|$ ?
– Nosrati
Sep 4 at 12:18
is not $lvert arvert=lvert brvert=lvert crvert=|d|$ ?
– Nosrati
Sep 4 at 12:18
No, only $lvert arvert=lvert brvert=lvert crvert$ :(
– Dilemian
Sep 4 at 12:21
No, only $lvert arvert=lvert brvert=lvert crvert$ :(
– Dilemian
Sep 4 at 12:21
So with $a=b=c=1$ and $dneq1$ the expression is not real.
– Nosrati
Sep 4 at 12:24
So with $a=b=c=1$ and $dneq1$ the expression is not real.
– Nosrati
Sep 4 at 12:24
1
1
@Nosrati For $a=b=c=1$ the expression seems to be $0$.
– Delta-u
Sep 4 at 12:27
@Nosrati For $a=b=c=1$ the expression seems to be $0$.
– Delta-u
Sep 4 at 12:27
Hint: since it is supposed to be valid for all $d$, you can interpret it as a polynomial of second order in the absolute value of $d$. Hence collect terms of the same order in $|d|$ and check the three imaginary components independently and confirm the statement.
– Ronald Blaak
Sep 4 at 16:34
Hint: since it is supposed to be valid for all $d$, you can interpret it as a polynomial of second order in the absolute value of $d$. Hence collect terms of the same order in $|d|$ and check the three imaginary components independently and confirm the statement.
– Ronald Blaak
Sep 4 at 16:34
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If $b=0$ or $c=0$, then the complex number equals $0$ which is real.
In the following, $bnot =0$ and $cnot=0$.
In order to prove that the claim is true, it is sufficient to prove the following two lemmas :
Lemma 1 : $$textIm(cbarb-cbara-abarb)=-fracbar a(a-b)(b-c)(c-a)2bc$$
Lemma 2 : $$textIm((a-b)(c-d)(bara-bard)(barc-barb))=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bc$$
From the two lemmas, we see that the complex number equals $textRe((a-b)(c-d)(bara-bard)(barc-barb))$ which is real, so the claim is true.
Proof for lemma 1 :
Using $abar a=bbar b=cbar c$, we have
$$beginaligntextIm(cbarb-cbara-abarb)
&=frac 12left((cbarb-cbara-abarb)-overline(cbarb-cbara-abarb)right)
\\&=frac 12left((cbarb-cbara-abarb)-(bar c b-bar ca-bar a b)right)
\\&=frac 12left(cfracabar ab-cbara-afracabar ab-fracabar ac b+fracabar aca+bar a bright)
\\&=fracbar a2bcleft(c^2a-bc^2-ca^2-ab^2+a^2b+b^2cright)
\\&=fracbar a2bcleft(c^2(a-b)-c(a-b)(a+b)+ab(a-b)right)
\\&=fracbar a(a-b)2bcleft(c^2-c(a+b)+abright)
\\&=fracbar a(a-b)2bc(c-a)(c-b)
\\&=-fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
Proof for lemma 2 :
Using $abar a=bbar b=cbar c$, we have
$$beginalign&textIm((a-b)(c-d)(bara-bard)(barc-barb))
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-bar b)-frac 12overline(a-b)(c-d)(bara-bard)(barc-barb)
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-barb)-frac 12(bar a-bar b)(bar c-bar d)(a-d)(c-b)
\\&=frac 12(a-b)(c-d)(bara-bard)left(fracabar ac-fracabar abright)-frac 12left(bar a-fracabar abright)left(fracabar ac-bar dright)(a-d)(c-b)
\\&=fracabar a2bc(a-b)(c-d)(bara-bard)left(b-cright)-fracbar a2bcleft(b-aright)left(abar a-cbar dright)(a-d)(c-b)
\\&=fracbar a(a-b)(b-c)2bcleft(a(c-d)(bara-bard)-c(bar c-bar d)(a-d)right)
\\&=fracbar a(a-b)(b-c)2bcleft((c-a)cbar c-(c-a)dbar dright)
\\&=fracbar a(a-b)(b-c)(c-a)(cbar c-dbar d)2bc
\\&=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
answered Sep 10 at 5:55
mathlove
87.7k877209
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $b=0$ or $c=0$, then the complex number equals $0$ which is real.
In the following, $bnot =0$ and $cnot=0$.
In order to prove that the claim is true, it is sufficient to prove the following two lemmas :
Lemma 1 : $$textIm(cbarb-cbara-abarb)=-fracbar a(a-b)(b-c)(c-a)2bc$$
Lemma 2 : $$textIm((a-b)(c-d)(bara-bard)(barc-barb))=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bc$$
From the two lemmas, we see that the complex number equals $textRe((a-b)(c-d)(bara-bard)(barc-barb))$ which is real, so the claim is true.
Proof for lemma 1 :
Using $abar a=bbar b=cbar c$, we have
$$beginaligntextIm(cbarb-cbara-abarb)
&=frac 12left((cbarb-cbara-abarb)-overline(cbarb-cbara-abarb)right)
\\&=frac 12left((cbarb-cbara-abarb)-(bar c b-bar ca-bar a b)right)
\\&=frac 12left(cfracabar ab-cbara-afracabar ab-fracabar ac b+fracabar aca+bar a bright)
\\&=fracbar a2bcleft(c^2a-bc^2-ca^2-ab^2+a^2b+b^2cright)
\\&=fracbar a2bcleft(c^2(a-b)-c(a-b)(a+b)+ab(a-b)right)
\\&=fracbar a(a-b)2bcleft(c^2-c(a+b)+abright)
\\&=fracbar a(a-b)2bc(c-a)(c-b)
\\&=-fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
Proof for lemma 2 :
Using $abar a=bbar b=cbar c$, we have
$$beginalign&textIm((a-b)(c-d)(bara-bard)(barc-barb))
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-bar b)-frac 12overline(a-b)(c-d)(bara-bard)(barc-barb)
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-barb)-frac 12(bar a-bar b)(bar c-bar d)(a-d)(c-b)
\\&=frac 12(a-b)(c-d)(bara-bard)left(fracabar ac-fracabar abright)-frac 12left(bar a-fracabar abright)left(fracabar ac-bar dright)(a-d)(c-b)
\\&=fracabar a2bc(a-b)(c-d)(bara-bard)left(b-cright)-fracbar a2bcleft(b-aright)left(abar a-cbar dright)(a-d)(c-b)
\\&=fracbar a(a-b)(b-c)2bcleft(a(c-d)(bara-bard)-c(bar c-bar d)(a-d)right)
\\&=fracbar a(a-b)(b-c)2bcleft((c-a)cbar c-(c-a)dbar dright)
\\&=fracbar a(a-b)(b-c)(c-a)(cbar c-dbar d)2bc
\\&=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
answered Sep 10 at 5:55
mathlove
87.7k877209
add a comment |Â
up vote
2
down vote
accepted
If $b=0$ or $c=0$, then the complex number equals $0$ which is real.
In the following, $bnot =0$ and $cnot=0$.
In order to prove that the claim is true, it is sufficient to prove the following two lemmas :
Lemma 1 : $$textIm(cbarb-cbara-abarb)=-fracbar a(a-b)(b-c)(c-a)2bc$$
Lemma 2 : $$textIm((a-b)(c-d)(bara-bard)(barc-barb))=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bc$$
From the two lemmas, we see that the complex number equals $textRe((a-b)(c-d)(bara-bard)(barc-barb))$ which is real, so the claim is true.
Proof for lemma 1 :
Using $abar a=bbar b=cbar c$, we have
$$beginaligntextIm(cbarb-cbara-abarb)
&=frac 12left((cbarb-cbara-abarb)-overline(cbarb-cbara-abarb)right)
\\&=frac 12left((cbarb-cbara-abarb)-(bar c b-bar ca-bar a b)right)
\\&=frac 12left(cfracabar ab-cbara-afracabar ab-fracabar ac b+fracabar aca+bar a bright)
\\&=fracbar a2bcleft(c^2a-bc^2-ca^2-ab^2+a^2b+b^2cright)
\\&=fracbar a2bcleft(c^2(a-b)-c(a-b)(a+b)+ab(a-b)right)
\\&=fracbar a(a-b)2bcleft(c^2-c(a+b)+abright)
\\&=fracbar a(a-b)2bc(c-a)(c-b)
\\&=-fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
Proof for lemma 2 :
Using $abar a=bbar b=cbar c$, we have
$$beginalign&textIm((a-b)(c-d)(bara-bard)(barc-barb))
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-bar b)-frac 12overline(a-b)(c-d)(bara-bard)(barc-barb)
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-barb)-frac 12(bar a-bar b)(bar c-bar d)(a-d)(c-b)
\\&=frac 12(a-b)(c-d)(bara-bard)left(fracabar ac-fracabar abright)-frac 12left(bar a-fracabar abright)left(fracabar ac-bar dright)(a-d)(c-b)
\\&=fracabar a2bc(a-b)(c-d)(bara-bard)left(b-cright)-fracbar a2bcleft(b-aright)left(abar a-cbar dright)(a-d)(c-b)
\\&=fracbar a(a-b)(b-c)2bcleft(a(c-d)(bara-bard)-c(bar c-bar d)(a-d)right)
\\&=fracbar a(a-b)(b-c)2bcleft((c-a)cbar c-(c-a)dbar dright)
\\&=fracbar a(a-b)(b-c)(c-a)(cbar c-dbar d)2bc
\\&=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
answered Sep 10 at 5:55
mathlove
87.7k877209
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $b=0$ or $c=0$, then the complex number equals $0$ which is real.
In the following, $bnot =0$ and $cnot=0$.
In order to prove that the claim is true, it is sufficient to prove the following two lemmas :
Lemma 1 : $$textIm(cbarb-cbara-abarb)=-fracbar a(a-b)(b-c)(c-a)2bc$$
Lemma 2 : $$textIm((a-b)(c-d)(bara-bard)(barc-barb))=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bc$$
From the two lemmas, we see that the complex number equals $textRe((a-b)(c-d)(bara-bard)(barc-barb))$ which is real, so the claim is true.
Proof for lemma 1 :
Using $abar a=bbar b=cbar c$, we have
$$beginaligntextIm(cbarb-cbara-abarb)
&=frac 12left((cbarb-cbara-abarb)-overline(cbarb-cbara-abarb)right)
\\&=frac 12left((cbarb-cbara-abarb)-(bar c b-bar ca-bar a b)right)
\\&=frac 12left(cfracabar ab-cbara-afracabar ab-fracabar ac b+fracabar aca+bar a bright)
\\&=fracbar a2bcleft(c^2a-bc^2-ca^2-ab^2+a^2b+b^2cright)
\\&=fracbar a2bcleft(c^2(a-b)-c(a-b)(a+b)+ab(a-b)right)
\\&=fracbar a(a-b)2bcleft(c^2-c(a+b)+abright)
\\&=fracbar a(a-b)2bc(c-a)(c-b)
\\&=-fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
Proof for lemma 2 :
Using $abar a=bbar b=cbar c$, we have
$$beginalign&textIm((a-b)(c-d)(bara-bard)(barc-barb))
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-bar b)-frac 12overline(a-b)(c-d)(bara-bard)(barc-barb)
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-barb)-frac 12(bar a-bar b)(bar c-bar d)(a-d)(c-b)
\\&=frac 12(a-b)(c-d)(bara-bard)left(fracabar ac-fracabar abright)-frac 12left(bar a-fracabar abright)left(fracabar ac-bar dright)(a-d)(c-b)
\\&=fracabar a2bc(a-b)(c-d)(bara-bard)left(b-cright)-fracbar a2bcleft(b-aright)left(abar a-cbar dright)(a-d)(c-b)
\\&=fracbar a(a-b)(b-c)2bcleft(a(c-d)(bara-bard)-c(bar c-bar d)(a-d)right)
\\&=fracbar a(a-b)(b-c)2bcleft((c-a)cbar c-(c-a)dbar dright)
\\&=fracbar a(a-b)(b-c)(c-a)(cbar c-dbar d)2bc
\\&=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
answered Sep 10 at 5:55
mathlove
87.7k877209
If $b=0$ or $c=0$, then the complex number equals $0$ which is real.
In the following, $bnot =0$ and $cnot=0$.
In order to prove that the claim is true, it is sufficient to prove the following two lemmas :
Lemma 1 : $$textIm(cbarb-cbara-abarb)=-fracbar a(a-b)(b-c)(c-a)2bc$$
Lemma 2 : $$textIm((a-b)(c-d)(bara-bard)(barc-barb))=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bc$$
From the two lemmas, we see that the complex number equals $textRe((a-b)(c-d)(bara-bard)(barc-barb))$ which is real, so the claim is true.
Proof for lemma 1 :
Using $abar a=bbar b=cbar c$, we have
$$beginaligntextIm(cbarb-cbara-abarb)
&=frac 12left((cbarb-cbara-abarb)-overline(cbarb-cbara-abarb)right)
\\&=frac 12left((cbarb-cbara-abarb)-(bar c b-bar ca-bar a b)right)
\\&=frac 12left(cfracabar ab-cbara-afracabar ab-fracabar ac b+fracabar aca+bar a bright)
\\&=fracbar a2bcleft(c^2a-bc^2-ca^2-ab^2+a^2b+b^2cright)
\\&=fracbar a2bcleft(c^2(a-b)-c(a-b)(a+b)+ab(a-b)right)
\\&=fracbar a(a-b)2bcleft(c^2-c(a+b)+abright)
\\&=fracbar a(a-b)2bc(c-a)(c-b)
\\&=-fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
Proof for lemma 2 :
Using $abar a=bbar b=cbar c$, we have
$$beginalign&textIm((a-b)(c-d)(bara-bard)(barc-barb))
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-bar b)-frac 12overline(a-b)(c-d)(bara-bard)(barc-barb)
\\&=frac 12(a-b)(c-d)(bara-bard)(barc-barb)-frac 12(bar a-bar b)(bar c-bar d)(a-d)(c-b)
\\&=frac 12(a-b)(c-d)(bara-bard)left(fracabar ac-fracabar abright)-frac 12left(bar a-fracabar abright)left(fracabar ac-bar dright)(a-d)(c-b)
\\&=fracabar a2bc(a-b)(c-d)(bara-bard)left(b-cright)-fracbar a2bcleft(b-aright)left(abar a-cbar dright)(a-d)(c-b)
\\&=fracbar a(a-b)(b-c)2bcleft(a(c-d)(bara-bard)-c(bar c-bar d)(a-d)right)
\\&=fracbar a(a-b)(b-c)2bcleft((c-a)cbar c-(c-a)dbar dright)
\\&=fracbar a(a-b)(b-c)(c-a)(cbar c-dbar d)2bc
\\&=(cbar c-dbar d)times fracbar a(a-b)(b-c)(c-a)2bcqquadsquareendalign$$
answered Sep 10 at 5:55
mathlove
87.7k877209
answered Sep 10 at 5:55
mathlove
87.7k877209
answered Sep 10 at 5:55
mathlove
87.7k877209
answered Sep 10 at 5:55
mathlove
87.7k877209
87.7k877209
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is not $lvert arvert=lvert brvert=lvert crvert=|d|$ ?
– Nosrati
Sep 4 at 12:18
No, only $lvert arvert=lvert brvert=lvert crvert$ :(
– Dilemian
Sep 4 at 12:21
So with $a=b=c=1$ and $dneq1$ the expression is not real.
– Nosrati
Sep 4 at 12:24
1
@Nosrati For $a=b=c=1$ the expression seems to be $0$.
– Delta-u
Sep 4 at 12:27
Hint: since it is supposed to be valid for all $d$, you can interpret it as a polynomial of second order in the absolute value of $d$. Hence collect terms of the same order in $|d|$ and check the three imaginary components independently and confirm the statement.
– Ronald Blaak
Sep 4 at 16:34