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Length of the arc of locus of a complex number
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Let z be a complex number satisfying $$argbigg(frac{z^3-1}{z^3+1}bigg) = frac{pi}{2}$$ on the Argand plane. Then, length of the arc of the locus of z for which $Re(z)>0$ and $Img(z)<0$ (where '$Re$' and '$Img$' represent real and imaginary part respectively) is?
Since it's argument is $frac{pi}{2}$, the complex number will be purely imaginary.
So, its conjugate would be equal to the negative of itself.
Solving that gives me, $|z|^6 = 1$
Since $|z|$ is a positive real number, so the only solution is $|z| = 1$
So, the answer should be $frac{pi}{2}$
But the answer given is $frac{pi}{6}$
Any help would be appreciated.
complex-numbers locus
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
asked 1 hour ago
Piano Land
377114
add a comment |
up vote
1
down vote
favorite
Let z be a complex number satisfying $$argbigg(frac{z^3-1}{z^3+1}bigg) = frac{pi}{2}$$ on the Argand plane. Then, length of the arc of the locus of z for which $Re(z)>0$ and $Img(z)<0$ (where '$Re$' and '$Img$' represent real and imaginary part respectively) is?
Since it's argument is $frac{pi}{2}$, the complex number will be purely imaginary.
So, its conjugate would be equal to the negative of itself.
Solving that gives me, $|z|^6 = 1$
Since $|z|$ is a positive real number, so the only solution is $|z| = 1$
So, the answer should be $frac{pi}{2}$
But the answer given is $frac{pi}{6}$
Any help would be appreciated.
complex-numbers locus
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
asked 1 hour ago
Piano Land
377114
2
You're taking the argument of $frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$.
– vrugtehagel
1 hour ago
@vrugtehagel I am not sure what you're asking. For your second question, I took the conjugate of the complex number which is inside the argument i.e $frac{z^3-1}{z^3+1}$ and equated it to $-bigg(frac{z^3-1}{z^3+1}bigg)$ since its purely imaginary. Solving this equation got me to $|z|^6 = 1$.
– Piano Land
5 mins ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let z be a complex number satisfying $$argbigg(frac{z^3-1}{z^3+1}bigg) = frac{pi}{2}$$ on the Argand plane. Then, length of the arc of the locus of z for which $Re(z)>0$ and $Img(z)<0$ (where '$Re$' and '$Img$' represent real and imaginary part respectively) is?
Since it's argument is $frac{pi}{2}$, the complex number will be purely imaginary.
So, its conjugate would be equal to the negative of itself.
Solving that gives me, $|z|^6 = 1$
Since $|z|$ is a positive real number, so the only solution is $|z| = 1$
So, the answer should be $frac{pi}{2}$
But the answer given is $frac{pi}{6}$
Any help would be appreciated.
complex-numbers locus
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
asked 1 hour ago
Piano Land
377114
Let z be a complex number satisfying $$argbigg(frac{z^3-1}{z^3+1}bigg) = frac{pi}{2}$$ on the Argand plane. Then, length of the arc of the locus of z for which $Re(z)>0$ and $Img(z)<0$ (where '$Re$' and '$Img$' represent real and imaginary part respectively) is?
Since it's argument is $frac{pi}{2}$, the complex number will be purely imaginary.
So, its conjugate would be equal to the negative of itself.
Solving that gives me, $|z|^6 = 1$
Since $|z|$ is a positive real number, so the only solution is $|z| = 1$
So, the answer should be $frac{pi}{2}$
But the answer given is $frac{pi}{6}$
Any help would be appreciated.
complex-numbers locus
complex-numbers locus
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
asked 1 hour ago
Piano Land
377114
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
asked 1 hour ago
Piano Land
377114
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
edited 50 mins ago
Jean-Claude Arbaut
14.3k63361
14.3k63361
asked 1 hour ago
Piano Land
377114
asked 1 hour ago
Piano Land
377114
asked 1 hour ago
Piano Land
377114
377114
2
You're taking the argument of $frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$.
– vrugtehagel
1 hour ago
@vrugtehagel I am not sure what you're asking. For your second question, I took the conjugate of the complex number which is inside the argument i.e $frac{z^3-1}{z^3+1}$ and equated it to $-bigg(frac{z^3-1}{z^3+1}bigg)$ since its purely imaginary. Solving this equation got me to $|z|^6 = 1$.
– Piano Land
5 mins ago
add a comment |
2
You're taking the argument of $frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$.
– vrugtehagel
1 hour ago
@vrugtehagel I am not sure what you're asking. For your second question, I took the conjugate of the complex number which is inside the argument i.e $frac{z^3-1}{z^3+1}$ and equated it to $-bigg(frac{z^3-1}{z^3+1}bigg)$ since its purely imaginary. Solving this equation got me to $|z|^6 = 1$.
– Piano Land
5 mins ago
2
2
You're taking the argument of $frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$.
– vrugtehagel
1 hour ago
You're taking the argument of $frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$.
– vrugtehagel
1 hour ago
@vrugtehagel I am not sure what you're asking. For your second question, I took the conjugate of the complex number which is inside the argument i.e $frac{z^3-1}{z^3+1}$ and equated it to $-bigg(frac{z^3-1}{z^3+1}bigg)$ since its purely imaginary. Solving this equation got me to $|z|^6 = 1$.
– Piano Land
5 mins ago
@vrugtehagel I am not sure what you're asking. For your second question, I took the conjugate of the complex number which is inside the argument i.e $frac{z^3-1}{z^3+1}$ and equated it to $-bigg(frac{z^3-1}{z^3+1}bigg)$ since its purely imaginary. Solving this equation got me to $|z|^6 = 1$.
– Piano Land
5 mins ago
add a comment |
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2
You're taking the argument of $frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$.
– vrugtehagel
1 hour ago
@vrugtehagel I am not sure what you're asking. For your second question, I took the conjugate of the complex number which is inside the argument i.e $frac{z^3-1}{z^3+1}$ and equated it to $-bigg(frac{z^3-1}{z^3+1}bigg)$ since its purely imaginary. Solving this equation got me to $|z|^6 = 1$.
– Piano Land
5 mins ago