Mixing
Complex numbers, how cand I show that |z1|=|z2|=|z3|?
Clash Royale CLAN TAG#URR8PPP
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Show that if $z_1,z_2,z_3$ are complex numbers, $z_1+z_2+z_3=0$ and $z_1^2+z_2^2+z_3^2=0$ then: $|z_1|=|z_2|=|z_3|$
complex-analysis complex-numbers
edited 32 mins ago
Olof Rubin
1007
asked 40 mins ago
anonimousfifiha
356
add a comment |Â
up vote
1
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favorite
Show that if $z_1,z_2,z_3$ are complex numbers, $z_1+z_2+z_3=0$ and $z_1^2+z_2^2+z_3^2=0$ then: $|z_1|=|z_2|=|z_3|$
complex-analysis complex-numbers
edited 32 mins ago
Olof Rubin
1007
asked 40 mins ago
anonimousfifiha
356
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that if $z_1,z_2,z_3$ are complex numbers, $z_1+z_2+z_3=0$ and $z_1^2+z_2^2+z_3^2=0$ then: $|z_1|=|z_2|=|z_3|$
complex-analysis complex-numbers
edited 32 mins ago
Olof Rubin
1007
asked 40 mins ago
anonimousfifiha
356
Show that if $z_1,z_2,z_3$ are complex numbers, $z_1+z_2+z_3=0$ and $z_1^2+z_2^2+z_3^2=0$ then: $|z_1|=|z_2|=|z_3|$
complex-analysis complex-numbers
complex-analysis complex-numbers
edited 32 mins ago
Olof Rubin
1007
asked 40 mins ago
anonimousfifiha
356
edited 32 mins ago
Olof Rubin
1007
asked 40 mins ago
anonimousfifiha
356
edited 32 mins ago
Olof Rubin
1007
edited 32 mins ago
Olof Rubin
1007
edited 32 mins ago
Olof Rubin
1007
1007
asked 40 mins ago
anonimousfifiha
356
asked 40 mins ago
anonimousfifiha
356
asked 40 mins ago
anonimousfifiha
356
356
add a comment |Â
add a comment |Â
3 Answers
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$$z_3=-z_1-z_2\
z_3^2=z_1^2+2z_1z_2+z_2^2$$
Since $z_3^2=-z_1^2-z_2^2$ you get
$$z_1^2+z_1z_2+z_2^2=0$$
Multiplying by $z_1-z_2$ you get
$$z_1^3=z_2^3$$
Applying absolute values you get
$$|z_1|^3=|z_2|^3$$
and hence
$$|z_1|=|z_2|$$
The equality $|z_1|=|z_3|$ can be obtained same way.
answered 24 mins ago
N. S.
100k5106200
(+1) Nicely done.
– Mark Viola
8 mins ago
add a comment |Â
up vote
4
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$$z_1 z_2 + z_2 z_3 + z_3 z_1 = big((z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)big)/2 = 0.$$
So $(z-z_1)(z-z_2)(z-z_3)=z^3-z_1 z_2 z_3$ for any $z$, and $z_1^3=z_2^3=z_3^3(=z_1 z_2 z_3)$.
answered 33 mins ago
metamorphy
1,9361313
(+1) Well done!
– Mark Viola
9 mins ago
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Let $f=(t-z_1)(t-z_2)(t-z_3)$.
Then $f$ is a cubic polynomial in $t$, with roots $z_1,z_2,z_3$.
In expanded form, $f$ can be expressed as
$$f=t^3-at^2+bt-c$$
where
beginalign*
a&=z_1+z_2+z_3\[4pt]
b&=z_1z_2+z_2z_3+z_3z_1\[4pt]
c&=z_1z_2z_3\[4pt]
endalign*
From $z_1+z_2+z_3=0$, we get $a=0$, and since we also have $z_1^2+z_2^2+z_3^2=0$, the identity
$$(z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)$$
yields $b=0$.
Thus, $f=t^3-c$, hence, since
$$f(z_1)=f(z_2)=f(z_3)=0$$
we get
$$z_1^3=z_2^3=z_3^3=c$$
so $|z_1|^3=|z_2|^3=|z_3|^3$, which yields $|z_1|=|z_2|=|z_3|$.
answered 12 mins ago
quasi
34.5k22561
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
$$z_3=-z_1-z_2\
z_3^2=z_1^2+2z_1z_2+z_2^2$$
Since $z_3^2=-z_1^2-z_2^2$ you get
$$z_1^2+z_1z_2+z_2^2=0$$
Multiplying by $z_1-z_2$ you get
$$z_1^3=z_2^3$$
Applying absolute values you get
$$|z_1|^3=|z_2|^3$$
and hence
$$|z_1|=|z_2|$$
The equality $|z_1|=|z_3|$ can be obtained same way.
answered 24 mins ago
N. S.
100k5106200
(+1) Nicely done.
– Mark Viola
8 mins ago
add a comment |Â
up vote
5
down vote
$$z_3=-z_1-z_2\
z_3^2=z_1^2+2z_1z_2+z_2^2$$
Since $z_3^2=-z_1^2-z_2^2$ you get
$$z_1^2+z_1z_2+z_2^2=0$$
Multiplying by $z_1-z_2$ you get
$$z_1^3=z_2^3$$
Applying absolute values you get
$$|z_1|^3=|z_2|^3$$
and hence
$$|z_1|=|z_2|$$
The equality $|z_1|=|z_3|$ can be obtained same way.
answered 24 mins ago
N. S.
100k5106200
(+1) Nicely done.
– Mark Viola
8 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$z_3=-z_1-z_2\
z_3^2=z_1^2+2z_1z_2+z_2^2$$
Since $z_3^2=-z_1^2-z_2^2$ you get
$$z_1^2+z_1z_2+z_2^2=0$$
Multiplying by $z_1-z_2$ you get
$$z_1^3=z_2^3$$
Applying absolute values you get
$$|z_1|^3=|z_2|^3$$
and hence
$$|z_1|=|z_2|$$
The equality $|z_1|=|z_3|$ can be obtained same way.
answered 24 mins ago
N. S.
100k5106200
$$z_3=-z_1-z_2\
z_3^2=z_1^2+2z_1z_2+z_2^2$$
Since $z_3^2=-z_1^2-z_2^2$ you get
$$z_1^2+z_1z_2+z_2^2=0$$
Multiplying by $z_1-z_2$ you get
$$z_1^3=z_2^3$$
Applying absolute values you get
$$|z_1|^3=|z_2|^3$$
and hence
$$|z_1|=|z_2|$$
The equality $|z_1|=|z_3|$ can be obtained same way.
answered 24 mins ago
N. S.
100k5106200
answered 24 mins ago
N. S.
100k5106200
answered 24 mins ago
N. S.
100k5106200
answered 24 mins ago
N. S.
100k5106200
100k5106200
(+1) Nicely done.
– Mark Viola
8 mins ago
add a comment |Â
(+1) Nicely done.
– Mark Viola
8 mins ago
(+1) Nicely done.
– Mark Viola
8 mins ago
(+1) Nicely done.
– Mark Viola
8 mins ago
add a comment |Â
up vote
4
down vote
$$z_1 z_2 + z_2 z_3 + z_3 z_1 = big((z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)big)/2 = 0.$$
So $(z-z_1)(z-z_2)(z-z_3)=z^3-z_1 z_2 z_3$ for any $z$, and $z_1^3=z_2^3=z_3^3(=z_1 z_2 z_3)$.
answered 33 mins ago
metamorphy
1,9361313
(+1) Well done!
– Mark Viola
9 mins ago
add a comment |Â
up vote
4
down vote
$$z_1 z_2 + z_2 z_3 + z_3 z_1 = big((z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)big)/2 = 0.$$
So $(z-z_1)(z-z_2)(z-z_3)=z^3-z_1 z_2 z_3$ for any $z$, and $z_1^3=z_2^3=z_3^3(=z_1 z_2 z_3)$.
answered 33 mins ago
metamorphy
1,9361313
(+1) Well done!
– Mark Viola
9 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$$z_1 z_2 + z_2 z_3 + z_3 z_1 = big((z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)big)/2 = 0.$$
So $(z-z_1)(z-z_2)(z-z_3)=z^3-z_1 z_2 z_3$ for any $z$, and $z_1^3=z_2^3=z_3^3(=z_1 z_2 z_3)$.
answered 33 mins ago
metamorphy
1,9361313
$$z_1 z_2 + z_2 z_3 + z_3 z_1 = big((z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)big)/2 = 0.$$
So $(z-z_1)(z-z_2)(z-z_3)=z^3-z_1 z_2 z_3$ for any $z$, and $z_1^3=z_2^3=z_3^3(=z_1 z_2 z_3)$.
answered 33 mins ago
metamorphy
1,9361313
edited 13 mins ago
answered 33 mins ago
metamorphy
1,9361313
answered 33 mins ago
metamorphy
1,9361313
answered 33 mins ago
metamorphy
1,9361313
1,9361313
(+1) Well done!
– Mark Viola
9 mins ago
add a comment |Â
(+1) Well done!
– Mark Viola
9 mins ago
(+1) Well done!
– Mark Viola
9 mins ago
(+1) Well done!
– Mark Viola
9 mins ago
add a comment |Â
up vote
0
down vote
Let $f=(t-z_1)(t-z_2)(t-z_3)$.
Then $f$ is a cubic polynomial in $t$, with roots $z_1,z_2,z_3$.
In expanded form, $f$ can be expressed as
$$f=t^3-at^2+bt-c$$
where
beginalign*
a&=z_1+z_2+z_3\[4pt]
b&=z_1z_2+z_2z_3+z_3z_1\[4pt]
c&=z_1z_2z_3\[4pt]
endalign*
From $z_1+z_2+z_3=0$, we get $a=0$, and since we also have $z_1^2+z_2^2+z_3^2=0$, the identity
$$(z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)$$
yields $b=0$.
Thus, $f=t^3-c$, hence, since
$$f(z_1)=f(z_2)=f(z_3)=0$$
we get
$$z_1^3=z_2^3=z_3^3=c$$
so $|z_1|^3=|z_2|^3=|z_3|^3$, which yields $|z_1|=|z_2|=|z_3|$.
answered 12 mins ago
quasi
34.5k22561
add a comment |Â
up vote
0
down vote
Let $f=(t-z_1)(t-z_2)(t-z_3)$.
Then $f$ is a cubic polynomial in $t$, with roots $z_1,z_2,z_3$.
In expanded form, $f$ can be expressed as
$$f=t^3-at^2+bt-c$$
where
beginalign*
a&=z_1+z_2+z_3\[4pt]
b&=z_1z_2+z_2z_3+z_3z_1\[4pt]
c&=z_1z_2z_3\[4pt]
endalign*
From $z_1+z_2+z_3=0$, we get $a=0$, and since we also have $z_1^2+z_2^2+z_3^2=0$, the identity
$$(z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)$$
yields $b=0$.
Thus, $f=t^3-c$, hence, since
$$f(z_1)=f(z_2)=f(z_3)=0$$
we get
$$z_1^3=z_2^3=z_3^3=c$$
so $|z_1|^3=|z_2|^3=|z_3|^3$, which yields $|z_1|=|z_2|=|z_3|$.
answered 12 mins ago
quasi
34.5k22561
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $f=(t-z_1)(t-z_2)(t-z_3)$.
Then $f$ is a cubic polynomial in $t$, with roots $z_1,z_2,z_3$.
In expanded form, $f$ can be expressed as
$$f=t^3-at^2+bt-c$$
where
beginalign*
a&=z_1+z_2+z_3\[4pt]
b&=z_1z_2+z_2z_3+z_3z_1\[4pt]
c&=z_1z_2z_3\[4pt]
endalign*
From $z_1+z_2+z_3=0$, we get $a=0$, and since we also have $z_1^2+z_2^2+z_3^2=0$, the identity
$$(z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)$$
yields $b=0$.
Thus, $f=t^3-c$, hence, since
$$f(z_1)=f(z_2)=f(z_3)=0$$
we get
$$z_1^3=z_2^3=z_3^3=c$$
so $|z_1|^3=|z_2|^3=|z_3|^3$, which yields $|z_1|=|z_2|=|z_3|$.
answered 12 mins ago
quasi
34.5k22561
Let $f=(t-z_1)(t-z_2)(t-z_3)$.
Then $f$ is a cubic polynomial in $t$, with roots $z_1,z_2,z_3$.
In expanded form, $f$ can be expressed as
$$f=t^3-at^2+bt-c$$
where
beginalign*
a&=z_1+z_2+z_3\[4pt]
b&=z_1z_2+z_2z_3+z_3z_1\[4pt]
c&=z_1z_2z_3\[4pt]
endalign*
From $z_1+z_2+z_3=0$, we get $a=0$, and since we also have $z_1^2+z_2^2+z_3^2=0$, the identity
$$(z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)$$
yields $b=0$.
Thus, $f=t^3-c$, hence, since
$$f(z_1)=f(z_2)=f(z_3)=0$$
we get
$$z_1^3=z_2^3=z_3^3=c$$
so $|z_1|^3=|z_2|^3=|z_3|^3$, which yields $|z_1|=|z_2|=|z_3|$.
answered 12 mins ago
quasi
34.5k22561
answered 12 mins ago
quasi
34.5k22561
answered 12 mins ago
quasi
34.5k22561
answered 12 mins ago
quasi
34.5k22561
34.5k22561
add a comment |Â
add a comment |Â
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