1. **State the problem:** We need to find all values of $z$ such that $$e^{iz} = i.$$ 2. **Recall Euler's formula:** $$e^{i\theta} = \cos \theta + i \sin \theta.$$ 3. **Express $i$ in exponential form:** We know that $$i = e^{i\frac{\pi}{2} + 2n\pi i}$$ for any integer $n$, because $$\cos \frac{\pi}{2} = 0$$ and $$\sin \frac{\pi}{2} = 1.$$ 4. **Set the exponents equal:** Since $$e^{iz} = e^{i\left(\frac{\pi}{2} + 2n\pi\right)},$$ we have $$iz = i\left(\frac{\pi}{2} + 2n\pi\right).$$ 5. **Solve for $z$:** Divide both sides by $i$ (noting $i \neq 0$), $$z = \frac{\pi}{2} + 2n\pi,$$ where $n$ is any integer. 6. **Interpret the solution:** The general solution is $$z = \frac{\pi}{2} + 2n\pi.$$ 7. **Check the options:** - Option a: $$\frac{\pi}{2} + n\pi$$ (includes extra $n\pi$ term) - Option b: $$(2n + 1) \frac{\pi}{2} + n \pi$$ (more complicated) - Option c: $$n \pi$$ (does not match) Our derived solution matches option a if $n$ is replaced by $2n$, but the exact form is $$z = \frac{\pi}{2} + 2n\pi.$$ **Final answer:** $$\boxed{z = \frac{\pi}{2} + 2n\pi}.$$