question :
$\sinh^{-1}[\ln(-1$)] is equal to (find all values):
(a) $\ln \left[\pi+\sqrt{\pi^2-1}\right]+\dfrac{\pi i}{2}+2ki\pi,\,\,\forall k=0,\pm1,\pm2,.....$
(b)$\ln \left[\pi-\sqrt{\pi^2-1}\right]+\dfrac{\pi i}{2}+2ki\pi,\,\,\forall k=0,\pm1,\pm2,.........$
(c)$\ln \left[(2k+1)\pi+\sqrt{(2k+1)^2\pi^2-1}\right]+\dfrac{\pi i}{2}+2mi\pi\,\forall\,k,m=0,\pm1,\pm2,\cdot\cdot\cdot\cdot\cdot$
(d)$\ln \left[-(2k+1)\pi+\sqrt{(2k+1)^2\pi^2-1}\right]+\dfrac{\pi i}{2}+2mi\pi\,\forall\,k,m=0,\pm1,\pm2,\cdot\cdot\cdot\cdot\cdot$
my attempt:
$\ln(-1)=\ln(e^{i\pi}e^{2k\pi i})=(2k+1)i\pi$
we know,
$\sinh^{-1}z=\ln(z+\sqrt{z^2+1})+2mi\pi $ where $m=0$ corresponds to principal branch
$\sinh^{-1}[\ln(-1)]=(\ln\left[(2k+1)i\pi+\sqrt{1-(2k+1)^2\pi^2}\right])+2mi\pi$
$=\left(\ln\left[(2k+1)i\pi+i\sqrt{(2k+1)^2\pi^2-1}\right]\right)+2mi\pi$
$=\left(\ln\left[(2k+1)\pi+\sqrt{(2k+1)^2\pi^2-1}\right]\right)+\ln i+2mi\pi$
$=\left(\ln\left[(2k+1)\pi+\sqrt{(2k+1)^2\pi^2-1}\right]\right)+\ln e^{\frac{i\pi}{2}}+2mi\pi$
$=\left(\ln\left[(2k+1)\pi+\sqrt{(2k+1)^2\pi^2-1}\right]\right)+{\dfrac{i\pi}{2}}+2mi\pi,\,\,\forall k,m =0,\pm1,\pm2........$
so,i marked option (c) but in exam's answer key option (c,d) both are correct
i don't know why (d) is correct
help me in finding my mistake and give me procedure to find right answer
too.....thanks
Your mistake (more of an omission than a mistake) is that the logarithm of a complex number involves absolute values, i.e. $$\text {Log} (z) = \ln |z| + i \arg z,$$ where $|z| = \pm \sqrt {x + yi}$ and $\arg z = \arctan (y/x)$.
In your case, $$\ln (-1) = |(2k + 1)|i \pi,$$ or $$\ln (-1) = \pm (2k + 1)i \pi.$$ You can check in the above definition of $\sinh^{-1} (z)$ to see that $-(2k + 1)i \pi$ also works, and thus (c) and (d) are correct.