$\sin(x+iy)$

$\begin{aligned} \sin z & = \frac{3}{4} + \frac{i}{4} \\ \frac{i}{2}(e^{-zi}-e^{zi}) & = \frac{3}{4} + \frac{i}{4} \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }-4i} \\ 2e^{-zi}-2e^{zi} & = -3i + 1 \quad \quad \small \color{#3D99F6}{\text{Multiply both sides by }e^{zi} \text{ and rearrange}} \\ \implies 2e^{2zi} + (1-3i)e^{zi} - 2 & = 0 \\ \implies e^{zi} & = \frac{-1+3i \pm \sqrt{(1-3i)^2+16}}{4} \\ & = \frac{-1+3i \pm \sqrt{8-6i}}{4} \\ & = \frac{-1+3i \pm (3 - i)}{4} \quad \quad \small \color{#3D99F6}{\text{Let }z = x+yi} \\ \implies e^{-y+xi}& = \begin{cases} \dfrac{1+i}{2} = \dfrac{e^{\frac{\pi i}{4}}}{\sqrt{2}} \\ - 1 + i = \dfrac{\sqrt{2}} {e^{\frac{\pi i}{4}}} \end{cases} \\ \implies y & = \frac{ \ln 2}{2} \quad \quad \small \color{#3D99F6}{(\text{One of the solution})} \\ x & = \frac{\pi}{4} \\ \implies z & = \boxed{\dfrac{\pi}{4} + \dfrac{i \ln 2}{2}} \end{aligned}$