Let $\cos ^{-1} \dfrac {3i}{4} =x+iy$

$ \therefore \dfrac {3i}{4}=\cos (x+iy)=\cos x \cosh y - i \sin x \sinh y$

Comparing real and imaginary parts,

$\cos x \cosh y=0 \ \therefore \cos x =0 \ \therefore x = \dfrac {\pi}{2}$

$-\sin x \sinh y = \dfrac {3i}{4} \ \text{ but }\sin x = \sin \left ( \dfrac {\pi}{2} \right )=1$

$\therefore \sinh y=\dfrac {-3}{4}$

$\therefore y = \sin h^{-1} \dfrac {-3}{4} = \log \left ( \dfrac {-3}{4} + \sqrt{1+\dfrac {9}{16}} \right )=\log \left ( \dfrac {1}{2} \right )= -\log 2$

Hence,

real part $=\dfrac {\pi}{2} \ and$

imaginary parts = -log 2