Complex Log

$\displaystyle \sin(\ln(i^i)) = ? \\ \displaystyle e^{i\pi} = -1 \\ \displaystyle e^{\dfrac{i\pi}{2}} = i\\ \\ \displaystyle \left(e^{\dfrac{i\pi}{2}}\right)^i = i^i \\ \displaystyle \sin(\ln(i^i)) = \sin\left(\ln\left(e^{\dfrac{i\pi}{2}}\right)^i\right) \\ \displaystyle = \sin\left(i\ln\left(e^{\dfrac{i\pi}{2}}\right)\right) \\ \displaystyle = \sin\left(i\dfrac{i\pi}{2}\ln\left(e\right)\right)\\ \displaystyle = \sin\left(-1 \times \dfrac{\pi}{2}\right) \\ \displaystyle = - \sin\left(\dfrac{\pi}{2}\right) \\ \displaystyle = -1 \\ \displaystyle \boxed{\therefore -1 = a + ib \implies a = -1, b = 0 \implies a + b = -1}$

$sin$ $log i^i$ = $sin$ $log (e^{i \frac{\pi}{2}})^i$ = $sin$ $log (e^{- \frac{\pi}{2}})$ = $sin(-\frac{\pi}{2})$ = $-1$