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Let z be a complex number satisfying $$\arg\bigg(\frac{z^3-1}{z^3+1}\bigg) = \frac{\pi}{2}$$ 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.

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    $\begingroup$ You're taking the argument of $\frac{z^3-1}{z^3+1}$, not $z$ itself. Not sure how you got to $|z|^6=1$. $\endgroup$ – vrugtehagel Nov 19 '18 at 9:11
  • $\begingroup$ @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 $|z|^6 = 1$. $\endgroup$ – Piano Land Nov 19 '18 at 10:39
  • $\begingroup$ @vrugtehagel $$\arg\bigg(\frac{z^3-1}{z^3+1}\bigg) = \frac{\pi}{2}$$ means that $\frac{z^3-1}{z^3+1} = \lambda i$ for some $\lambda > 0$ and you get $z^3 = \frac{1+\lambda i}{1-\lambda i}$. Hence $\lvert z \rvert^6 = z^3 \cdot \overline{z^3} = 1$. $\endgroup$ – Paul Frost Nov 19 '18 at 13:05
  • $\begingroup$ I see. In fact, just calculating $|z^3|$ with $\frac{1+\lambda i}{1-\lambda i}$ will yield $|z|^3=1$. I'm unfamiliar with the terminology "length of the arc of the locus" however so I'm afraid I can't help any further. $\endgroup$ – vrugtehagel Nov 19 '18 at 13:09
  • $\begingroup$ @PianoLand $\lvert z \rvert = 1$ is necessary, but not sufficient for $(*) \phantom{x} \arg\bigg(\frac{z^3-1}{z^3+1}\bigg) = \frac{\pi}{2}$. Take $z=1$. Moreover, as vrugtehagel said, it is not really clear what you mean by "length of the arc of the locus of $z$ for which ...". I guess you mean the set of solutions $z$ of $(*)$? $\endgroup$ – Paul Frost Nov 19 '18 at 13:23
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We have \begin{align}\operatorname{arg}\left(\frac{z^3-1}{z^3+1}\right)=\frac{\pi}{2}&\iff \exists \lambda >0,\frac{z^3-1}{z^3+1}=\lambda i \\&\iff \exists \lambda>0, z^3=\frac{1+\lambda i}{1-\lambda i} \\&\iff \exists \lambda>0, z^3=\frac{(1+\lambda i)^2}{1+\lambda^2} \\&\iff \exists \lambda>0, z^3=\frac{(1-\lambda^2)+2\lambda i}{1+\lambda^2} \\&\iff |z|=1 \wedge \Im z^3>0 \\&\iff |z|=1 \wedge \operatorname{arg} z^3\in (0,\pi) \\&\iff \exists \theta \in(0,\pi/3)\cup (2\pi/3,\pi)\cup (4\pi/3,5\pi/3),z=e^{i\theta}.\end{align} If in addition, $\Re z>0$ and $\Im z<0$, then $$\operatorname{arg}\left(\frac{z^3-1}{z^3+1}\right)=\frac{\pi}{2}\iff \exists \theta\in (3\pi/2,5\pi/3), z=e^{i\theta}.$$ So, your book is right, the answer is $5\pi/3-3\pi/2=\pi/6$.

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  • $\begingroup$ There is minor mistake: $\exists \lambda>0, z^3=\frac{(1-\lambda)^2+2\lambda i}{1+\lambda^2} \iff |z|=1 \wedge \Re z^3>0 \wedge \Im z^3 > 0\iff |z|=1 \wedge \operatorname{arg} z^3\in (0,\pi/2)$. $\endgroup$ – Paul Frost Nov 19 '18 at 14:53
  • $\begingroup$ Moreover, $\operatorname{arg} z^3\in (0,\pi/2) \iff \operatorname{arg} z \in (0,\pi/6) \cup (2\pi/3, 2\pi/3 + \pi/6) \cup (4\pi/3, 4\pi/3 + \pi/6)$. $\endgroup$ – Paul Frost Nov 19 '18 at 15:04
  • $\begingroup$ @PaulFrost I think there is a typo: $(1-\lambda)^2$ should be $1-\lambda^2$. So, no, the real part of $z^3$ is not necessarily positive. That is why Snookie got $\operatorname{arg}z^3\in \left(0,\frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2},2\pi\right)$. But the imaginary part has to be positive. $\endgroup$ – user593746 Nov 19 '18 at 15:45
  • $\begingroup$ @Snookie Please fix this. $\endgroup$ – user593746 Nov 19 '18 at 15:46
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    $\begingroup$ That is because the cubing map $z\mapsto z^3$ is a $3$-fold map. The equation $z^3=w$ has three solutions. So, $z^3=e^{3i\theta}$ must have an argument in $(0,\pi)$, and it can mean $3\theta\in (0,\pi)$, or $3\theta\in (2\pi,3\pi)$, or $3\theta \in (4\pi,5\pi)$. $\endgroup$ – user614671 Nov 20 '18 at 12:08

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