1. **State the problem:** Solve the equation $$z^{i} - 1 = 0$$ and represent the solution on an Argand diagram. 2. **Rewrite the equation:** We want to find all complex numbers $$z$$ such that $$z^{i} = 1$$. 3. **Recall the formula for complex exponentiation:** For a complex number $$z = re^{i\theta}$$ (in polar form), and any complex exponent $$w$$, we have $$z^{w} = e^{w \log z}$$ where $$\log z = \ln r + i\theta$$. 4. **Apply this to our problem:** $$z^{i} = e^{i \log z} = 1$$. 5. **Since $$1 = e^{2\pi i k}$$ for any integer $$k$$, we have:** $$e^{i \log z} = e^{2\pi i k}$$ 6. **Equate exponents:** $$i \log z = 2\pi i k$$ 7. **Divide both sides by $$i$$:** $$\cancel{i} \log z = 2\pi \cancel{i} k \implies \log z = 2\pi k$$ 8. **Recall $$\log z = \ln r + i\theta$$, so:** $$\ln r + i\theta = 2\pi k$$ 9. **Separate real and imaginary parts:** - Real part: $$\ln r = 2\pi k$$ - Imaginary part: $$\theta = 0$$ 10. **Solve for $$r$$:** $$r = e^{2\pi k}$$ 11. **Since $$\theta = 0$$, the complex number $$z$$ lies on the positive real axis:** $$z = r e^{i0} = r$$ 12. **Therefore, the solutions are:** $$z = e^{2\pi k}$$ for all integers $$k$$. 13. **Interpretation:** The solutions are all positive real numbers of the form $$e^{2\pi k}$$, which form a discrete set on the positive real axis. 14. **Argand diagram representation:** Plot points on the positive real axis at positions $$..., e^{-4\pi}, e^{-2\pi}, 1, e^{2\pi}, e^{4\pi}, ...$$. **Final answer:** $$\boxed{z = e^{2\pi k}, \quad k \in \mathbb{Z}}$$