How to factorize $zz^*-4z-4z^*+12=0$ (where $z^*$ is the complex conjugate of $z$)
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I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.
I'm trying to factorise this using the completed square method but had no luck so far.
Could use some help.
complex-numbers quadratics
edited Dec 2 at 5:29
Blue
47.1k870148
asked Dec 2 at 5:15
Usama Abdul
82
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up vote
1
down vote
favorite
I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.
I'm trying to factorise this using the completed square method but had no luck so far.
Could use some help.
complex-numbers quadratics
edited Dec 2 at 5:29
Blue
47.1k870148
asked Dec 2 at 5:15
Usama Abdul
82
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
– Mohammad Zuhair Khan
Dec 2 at 5:25
Alright, got it now. Thanks a lot.
– Usama Abdul
Dec 2 at 5:33
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.
I'm trying to factorise this using the completed square method but had no luck so far.
Could use some help.
complex-numbers quadratics
edited Dec 2 at 5:29
Blue
47.1k870148
asked Dec 2 at 5:15
Usama Abdul
82
I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.
I'm trying to factorise this using the completed square method but had no luck so far.
Could use some help.
complex-numbers quadratics
complex-numbers quadratics
edited Dec 2 at 5:29
Blue
47.1k870148
asked Dec 2 at 5:15
Usama Abdul
82
edited Dec 2 at 5:29
Blue
47.1k870148
asked Dec 2 at 5:15
Usama Abdul
82
edited Dec 2 at 5:29
Blue
47.1k870148
edited Dec 2 at 5:29
Blue
47.1k870148
edited Dec 2 at 5:29
Blue
47.1k870148
47.1k870148
asked Dec 2 at 5:15
Usama Abdul
82
asked Dec 2 at 5:15
Usama Abdul
82
asked Dec 2 at 5:15
Usama Abdul
82
82
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
– Mohammad Zuhair Khan
Dec 2 at 5:25
Alright, got it now. Thanks a lot.
– Usama Abdul
Dec 2 at 5:33
add a comment |
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
– Mohammad Zuhair Khan
Dec 2 at 5:25
Alright, got it now. Thanks a lot.
– Usama Abdul
Dec 2 at 5:33
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
– Mohammad Zuhair Khan
Dec 2 at 5:25
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
– Mohammad Zuhair Khan
Dec 2 at 5:25
Alright, got it now. Thanks a lot.
– Usama Abdul
Dec 2 at 5:33
Alright, got it now. Thanks a lot.
– Usama Abdul
Dec 2 at 5:33
add a comment |
2 Answers
2
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oldest
votes
up vote
2
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accepted
begin{align}
zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
&= z(z^* -4) -4(z^* -4) -4\
&= (z-4)(z^* -4) -4\
&= (z-4)(z-4)^* -4\
&= |z-4|^2 -4.
end{align}
answered Dec 2 at 5:28
Rócherz
2,6962721
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Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write
$zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$
Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.
answered Dec 2 at 5:52
zwim
11.3k628
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
begin{align}
zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
&= z(z^* -4) -4(z^* -4) -4\
&= (z-4)(z^* -4) -4\
&= (z-4)(z-4)^* -4\
&= |z-4|^2 -4.
end{align}
answered Dec 2 at 5:28
Rócherz
2,6962721
add a comment |
up vote
2
down vote
accepted
begin{align}
zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
&= z(z^* -4) -4(z^* -4) -4\
&= (z-4)(z^* -4) -4\
&= (z-4)(z-4)^* -4\
&= |z-4|^2 -4.
end{align}
answered Dec 2 at 5:28
Rócherz
2,6962721
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
begin{align}
zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
&= z(z^* -4) -4(z^* -4) -4\
&= (z-4)(z^* -4) -4\
&= (z-4)(z-4)^* -4\
&= |z-4|^2 -4.
end{align}
answered Dec 2 at 5:28
Rócherz
2,6962721
begin{align}
zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
&= z(z^* -4) -4(z^* -4) -4\
&= (z-4)(z^* -4) -4\
&= (z-4)(z-4)^* -4\
&= |z-4|^2 -4.
end{align}
answered Dec 2 at 5:28
Rócherz
2,6962721
answered Dec 2 at 5:28
Rócherz
2,6962721
answered Dec 2 at 5:28
Rócherz
2,6962721
answered Dec 2 at 5:28
Rócherz
2,6962721
2,6962721
add a comment |
add a comment |
up vote
1
down vote
Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write
$zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$
Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.
answered Dec 2 at 5:52
zwim
11.3k628
add a comment |
up vote
1
down vote
Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write
$zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$
Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.
answered Dec 2 at 5:52
zwim
11.3k628
add a comment |
up vote
1
down vote
up vote
1
down vote
Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write
$zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$
Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.
answered Dec 2 at 5:52
zwim
11.3k628
Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write
$zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$
Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.
answered Dec 2 at 5:52
zwim
11.3k628
answered Dec 2 at 5:52
zwim
11.3k628
answered Dec 2 at 5:52
zwim
11.3k628
answered Dec 2 at 5:52
zwim
11.3k628
11.3k628
add a comment |
add a comment |
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Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
– Mohammad Zuhair Khan
Dec 2 at 5:25
Alright, got it now. Thanks a lot.
– Usama Abdul
Dec 2 at 5:33