Prove that the area of the triangle with vertices $z, wz, z+wz$ is $frac{sqrt{3}}{4}|z|^2$
$begingroup$
Prove that the area of the triangle with vertices $z, wz, z+wz$ is $frac{sqrt{3}}{4}|z|^2$
My Attempt
$$
A=frac{i}{4}begin{vmatrix}
1&1&1\
z&wz&z+wz\
bar{z}&bar{w}bar{z}&bar{z}+bar{w}bar{z}\
end{vmatrix}
\
=frac{i}{4}bigg[wz(bar{z}+bar{w}bar{z})-bar{w}bar{z}(z+wz)-z(bar{z}+bar{w}bar{z})+bar{z}(z+wz)-z(bar{w}bar{z})-bar{z}(wz)bigg]\
=frac{i}{4}bigg[ w|z|^2+|w|^2|z|^2-bar{w}|z|^2-|w|^2|z|^2-|z|^2-bar{w}|z|^2+|z|^2-{w}|z|^2-bar{w}|z|^2+w|z|^2 bigg]\
=frac{i|z|^2}{4}bigg[ color{red}w+color{blue}{|w|^2}-color{green}{bar{w}}-color{blue}{|w|^2}-color{brown}1-color{orange}{bar{w}}+color{brown}1-color{red}{w}-bar{w}+w bigg]
=frac{i|z|^2}{4}(w-bar{w})
$$
How can it be same as $frac{sqrt{3}}{4}|z|^2$ ?
complex-numbers triangles
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
$endgroup$
add a comment |
$begingroup$
Prove that the area of the triangle with vertices $z, wz, z+wz$ is $frac{sqrt{3}}{4}|z|^2$
My Attempt
$$
A=frac{i}{4}begin{vmatrix}
1&1&1\
z&wz&z+wz\
bar{z}&bar{w}bar{z}&bar{z}+bar{w}bar{z}\
end{vmatrix}
\
=frac{i}{4}bigg[wz(bar{z}+bar{w}bar{z})-bar{w}bar{z}(z+wz)-z(bar{z}+bar{w}bar{z})+bar{z}(z+wz)-z(bar{w}bar{z})-bar{z}(wz)bigg]\
=frac{i}{4}bigg[ w|z|^2+|w|^2|z|^2-bar{w}|z|^2-|w|^2|z|^2-|z|^2-bar{w}|z|^2+|z|^2-{w}|z|^2-bar{w}|z|^2+w|z|^2 bigg]\
=frac{i|z|^2}{4}bigg[ color{red}w+color{blue}{|w|^2}-color{green}{bar{w}}-color{blue}{|w|^2}-color{brown}1-color{orange}{bar{w}}+color{brown}1-color{red}{w}-bar{w}+w bigg]
=frac{i|z|^2}{4}(w-bar{w})
$$
How can it be same as $frac{sqrt{3}}{4}|z|^2$ ?
complex-numbers triangles
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
$endgroup$
1
$begingroup$
Something is missing, perhaps $omega$ is complex cube root of unity.
$endgroup$
– Anurag A
Jan 16 at 8:45
add a comment |
$begingroup$
Prove that the area of the triangle with vertices $z, wz, z+wz$ is $frac{sqrt{3}}{4}|z|^2$
My Attempt
$$
A=frac{i}{4}begin{vmatrix}
1&1&1\
z&wz&z+wz\
bar{z}&bar{w}bar{z}&bar{z}+bar{w}bar{z}\
end{vmatrix}
\
=frac{i}{4}bigg[wz(bar{z}+bar{w}bar{z})-bar{w}bar{z}(z+wz)-z(bar{z}+bar{w}bar{z})+bar{z}(z+wz)-z(bar{w}bar{z})-bar{z}(wz)bigg]\
=frac{i}{4}bigg[ w|z|^2+|w|^2|z|^2-bar{w}|z|^2-|w|^2|z|^2-|z|^2-bar{w}|z|^2+|z|^2-{w}|z|^2-bar{w}|z|^2+w|z|^2 bigg]\
=frac{i|z|^2}{4}bigg[ color{red}w+color{blue}{|w|^2}-color{green}{bar{w}}-color{blue}{|w|^2}-color{brown}1-color{orange}{bar{w}}+color{brown}1-color{red}{w}-bar{w}+w bigg]
=frac{i|z|^2}{4}(w-bar{w})
$$
How can it be same as $frac{sqrt{3}}{4}|z|^2$ ?
complex-numbers triangles
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
$endgroup$
Prove that the area of the triangle with vertices $z, wz, z+wz$ is $frac{sqrt{3}}{4}|z|^2$
My Attempt
$$
A=frac{i}{4}begin{vmatrix}
1&1&1\
z&wz&z+wz\
bar{z}&bar{w}bar{z}&bar{z}+bar{w}bar{z}\
end{vmatrix}
\
=frac{i}{4}bigg[wz(bar{z}+bar{w}bar{z})-bar{w}bar{z}(z+wz)-z(bar{z}+bar{w}bar{z})+bar{z}(z+wz)-z(bar{w}bar{z})-bar{z}(wz)bigg]\
=frac{i}{4}bigg[ w|z|^2+|w|^2|z|^2-bar{w}|z|^2-|w|^2|z|^2-|z|^2-bar{w}|z|^2+|z|^2-{w}|z|^2-bar{w}|z|^2+w|z|^2 bigg]\
=frac{i|z|^2}{4}bigg[ color{red}w+color{blue}{|w|^2}-color{green}{bar{w}}-color{blue}{|w|^2}-color{brown}1-color{orange}{bar{w}}+color{brown}1-color{red}{w}-bar{w}+w bigg]
=frac{i|z|^2}{4}(w-bar{w})
$$
How can it be same as $frac{sqrt{3}}{4}|z|^2$ ?
complex-numbers triangles
complex-numbers triangles
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
edited Jan 16 at 9:22
ss1729
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
asked Jan 16 at 8:28
ss1729ss1729
2,04711124
2,04711124
1
$begingroup$
Something is missing, perhaps $omega$ is complex cube root of unity.
$endgroup$
– Anurag A
Jan 16 at 8:45
add a comment |
1
$begingroup$
Something is missing, perhaps $omega$ is complex cube root of unity.
$endgroup$
– Anurag A
Jan 16 at 8:45
1
1
$begingroup$
Something is missing, perhaps $omega$ is complex cube root of unity.
$endgroup$
– Anurag A
Jan 16 at 8:45
$begingroup$
Something is missing, perhaps $omega$ is complex cube root of unity.
$endgroup$
– Anurag A
Jan 16 at 8:45
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
HINT...It is easier just to draw a picture. Assuming that you mean $omega$ is a cube root of unity, the three points form an isosceles triangle whose two equal sides have length $|z|$, since $|omega z|=|z|$, and the angle between these sides is $120^o$
The answer you obtained is equivalent to the answer given because $omega=cos120^o+isin120^o$
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
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1 Answer
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$begingroup$
HINT...It is easier just to draw a picture. Assuming that you mean $omega$ is a cube root of unity, the three points form an isosceles triangle whose two equal sides have length $|z|$, since $|omega z|=|z|$, and the angle between these sides is $120^o$
The answer you obtained is equivalent to the answer given because $omega=cos120^o+isin120^o$
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
HINT...It is easier just to draw a picture. Assuming that you mean $omega$ is a cube root of unity, the three points form an isosceles triangle whose two equal sides have length $|z|$, since $|omega z|=|z|$, and the angle between these sides is $120^o$
The answer you obtained is equivalent to the answer given because $omega=cos120^o+isin120^o$
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
HINT...It is easier just to draw a picture. Assuming that you mean $omega$ is a cube root of unity, the three points form an isosceles triangle whose two equal sides have length $|z|$, since $|omega z|=|z|$, and the angle between these sides is $120^o$
The answer you obtained is equivalent to the answer given because $omega=cos120^o+isin120^o$
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
HINT...It is easier just to draw a picture. Assuming that you mean $omega$ is a cube root of unity, the three points form an isosceles triangle whose two equal sides have length $|z|$, since $|omega z|=|z|$, and the angle between these sides is $120^o$
The answer you obtained is equivalent to the answer given because $omega=cos120^o+isin120^o$
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
edited Jan 16 at 11:58
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
answered Jan 16 at 10:07
David QuinnDavid Quinn
24.1k21141
24.1k21141
add a comment |
add a comment |
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$begingroup$
Something is missing, perhaps $omega$ is complex cube root of unity.
$endgroup$
– Anurag A
Jan 16 at 8:45