Cfrgtkky

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.

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)+2lambda i}{1+lambda^2}
\&iff |z|=1 wedge Im z^3>0
\&iff |z|=1 wedge operatorname{arg} z^3in (0,pi)
\&iff exists theta in(0,pi/3)cup (2pi/3,pi)cup (4pi/3,5pi/3),z=e^{itheta}.end{align}
If in addition, $Re z>0$ and $Im z<0$, then $$operatorname{arg}left(frac{z^3-1}{z^3+1}right)=frac{pi}{2}iff exists thetain (3pi/2,5pi/3), z=e^{itheta}.$$
So, your book is right, the answer is $5pi/3-3pi/2=pi/6$.