1. **State the problem:** Solve the equation $$Z^{i} - 1 = 0$$ where $Z$ is a complex number and $i$ is the imaginary unit. 2. **Rewrite the equation:** We want to find $Z$ such that $$Z^{i} = 1$$. 3. **Recall the exponential form of complex numbers:** Any complex number $Z$ can be written as $$Z = re^{i\theta}$$ where $r$ is the modulus and $\theta$ is the argument. 4. **Apply the power:** $$Z^{i} = (re^{i\theta})^{i} = r^{i} e^{i \cdot i \theta} = r^{i} e^{-\theta}$$ because $i \cdot i = -1$. 5. **Express $r^{i}$ using exponentials:** $$r^{i} = e^{i \ln r}$$ so $$Z^{i} = e^{i \ln r} e^{-\theta} = e^{-\theta + i \ln r}$$. 6. **Set $Z^{i} = 1$:** Since $1 = e^{2k\pi i}$ for any integer $k$, we have $$e^{-\theta + i \ln r} = e^{2k\pi i}$$. 7. **Equate real and imaginary parts:** - Real part: $$-\theta = 0 \implies \theta = 0$$ - Imaginary part: $$\ln r = 2k\pi$$ 8. **Solve for $r$:** $$r = e^{2k\pi}$$ for any integer $k$. 9. **Write the general solution:** $$Z = r e^{i\theta} = e^{2k\pi} e^{i \cdot 0} = e^{2k\pi}$$ 10. **Interpretation:** The solutions are all positive real numbers of the form $$e^{2k\pi}$$ for integer $k$. 11. **Argand diagram representation:** All solutions lie on the positive real axis at points $$e^{0} = 1, e^{2\pi}, e^{4\pi}, e^{6\pi}, \ldots$$ which are spaced exponentially along the positive real axis. **Final answer:** $$Z = e^{2k\pi} \quad \text{for all integers } k$$.