I'm back problem (Double Euler's?)

We begin by dealing with the RHS of the equation given, with the aid of Euler's identity $e^{i\theta} = \cos \theta + i\sin \theta$ .

$\exp(-\ln 3+i\sin^{-1} \frac{24}{25}) = e^{-\ln 3}e^{i\sin^{-1} \frac{24}{25}} = \frac{1}{3}\left( \cos \sin^{-1} \frac{24}{25} + i\sin \sin^{-1} \frac{24}{25} \right)$

Now, employing that $cos \theta = \sqrt{1-\sin^2 \theta}$ ,

$\frac{1}{3}\left( \cos \sin^{-1} \frac{24}{25} + i\sin \sin^{-1} \frac{24}{25} \right) = \frac{1}{3}\left(\sqrt{1-(\frac{24}{25})^2} + i(\frac{24}{25}) \right) = \frac{7}{75}+\frac{8i}{25}$

Equating this with the LHS of the equation given, we have $\ln k = \frac{7}{75}+\frac{8i}{25}$ and so $k = \exp(\frac{7}{75}+\frac{8i}{25})$ . We then simplify the obtained expression for $k$ , using Euler's identity once more.

$\exp(\frac{7}{75}+\frac{8i}{25}) = e^{7/75}(e^{8i/25}) = e^{7/75}(\cos \frac{8}{25} + i\sin \frac{8}{25})$

The imaginary part of $k$ is hence simply

$\Im(k) = \boxed{e^{7/75}\sin \frac{8}{25}}$

so we have $A = 7, B = 75, C = 8$ and $D = 25$ , giving $A+B+C+D = \boxed{115}$ .

$\ln(k) = \exp\left[ -\ln(3) + i \sin^{-1} \left( \frac{24}{25} \right) \right]=\frac{\exp\left[i \sin^{-1} \left( \frac{24}{25} \right) \right]}{3}=\frac{\frac{7}{25}+i \frac{24}{25}}{3}=\frac{7}{75}+i\frac{8}{25}$

$k=\exp[\frac{7}{75}+i\frac{8}{25}]=e^{\frac{7}{75}}e^{i\frac{8}{25}}=e^{\frac{7}{75}}[\cos{\frac{8}{25}}+i\sin{\frac{8}{25}}]$

$\Im(k)=e^{\frac{7}{75}}\sin{\frac{8}{25}}$