1. **State the problem:** Find the real and imaginary parts of $$\left(\sqrt{i}\right)^{\sqrt{i}}$$. 2. **Recall the formula and rules:** - Complex number $i$ can be written in polar form as $$i = e^{i\frac{\pi}{2}}$$. - The square root of a complex number $z = re^{i\theta}$ is $$\sqrt{z} = \sqrt{r} e^{i\frac{\theta}{2}}$$. - For any complex number $a$ and $b$, $$a^b = e^{b \ln a}$$ where $\ln a$ is the complex logarithm. 3. **Find $\sqrt{i}$:** - Since $i = e^{i\frac{\pi}{2}}$, then $$\sqrt{i} = e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$$. 4. **Express the power:** $$\left(\sqrt{i}\right)^{\sqrt{i}} = e^{\sqrt{i} \ln(\sqrt{i})}$$. 5. **Calculate $\ln(\sqrt{i})$:** - Since $$\sqrt{i} = e^{i\frac{\pi}{4}}$$, $$\ln(\sqrt{i}) = i \frac{\pi}{4}$$. 6. **Calculate the exponent:** - $$\sqrt{i} = e^{i\frac{\pi}{4}}$$, so $$\sqrt{i} \ln(\sqrt{i}) = e^{i\frac{\pi}{4}} \times i \frac{\pi}{4}$$. - Write $e^{i\frac{\pi}{4}}$ as $\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}$: $$\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) \times i \frac{\pi}{4} = i \frac{\pi}{4} \times \frac{\sqrt{2}}{2} + i \frac{\pi}{4} \times i \frac{\sqrt{2}}{2}$$ - Simplify: $$= i \frac{\pi \sqrt{2}}{8} + i^2 \frac{\pi \sqrt{2}}{8} = i \frac{\pi \sqrt{2}}{8} - \frac{\pi \sqrt{2}}{8}$$ 7. **Rewrite the exponent:** $$\sqrt{i} \ln(\sqrt{i}) = - \frac{\pi \sqrt{2}}{8} + i \frac{\pi \sqrt{2}}{8}$$ 8. **Calculate the value:** $$\left(\sqrt{i}\right)^{\sqrt{i}} = e^{- \frac{\pi \sqrt{2}}{8} + i \frac{\pi \sqrt{2}}{8}} = e^{- \frac{\pi \sqrt{2}}{8}} \times e^{i \frac{\pi \sqrt{2}}{8}}$$ 9. **Use Euler's formula:** $$e^{i \theta} = \cos \theta + i \sin \theta$$, so $$\left(\sqrt{i}\right)^{\sqrt{i}} = e^{- \frac{\pi \sqrt{2}}{8}} \left( \cos \frac{\pi \sqrt{2}}{8} + i \sin \frac{\pi \sqrt{2}}{8} \right)$$ 10. **Final answer:** - Real part: $$\text{Re} = e^{- \frac{\pi \sqrt{2}}{8}} \cos \frac{\pi \sqrt{2}}{8}$$ - Imaginary part: $$\text{Im} = e^{- \frac{\pi \sqrt{2}}{8}} \sin \frac{\pi \sqrt{2}}{8}$$