(\sqrt{i})^{\sqrt{i}} = e^{\frac{-\pi}{4\sqrt{2}}}\left(\cos\frac{\pi}{4\sqrt{2}} + i\sin\frac{\pi}{4\sqrt{2}}\right)

To verify the identity \((\sqrt{i})^{\sqrt{i}} = e^{\frac{-\pi}{4\sqrt{2}}}\left(\cos\frac{\pi}{4\sqrt{2}} + i\sin\frac{\pi}{4\sqrt{2}}\right)\), let's break down the complex exponentiation and trigonometric forms of complex numbers step by step.

### Step 1: Express \(\sqrt{i}\) in polar form

Start with \(i\), which in polar form is written as:

\[
i = e^{i\frac{\pi}{2}}
\]

This is because \(i\) corresponds to a complex number on the unit circle at an angle of \(\frac{\pi}{2}\) radians. Now, we take the square root of \(i\), which means finding \(\sqrt{i}\). The square roots of a complex number \(r e^{i\theta}\) are given by:

\[
\sqrt{r e^{i\theta}} = \sqrt{r} e^{i\frac{\theta}{2}}
\]

For \(i = e^{i\frac{\pi}{2}}\), we get:

\[
\sqrt{i} = e^{i\frac{\pi}{4}}
\]

### Step 2: Raise \(\sqrt{i}\) to \(\sqrt{i}\)

Now, we need to compute:

\[
\left( \sqrt{i} \right)^{\sqrt{i}} = \left( e^{i\frac{\pi}{4}} \right)^{e^{i\frac{\pi}{4}}}
\]

Using the property of exponentiation for complex numbers, we can rewrite this as:

\[
\left( e^{i\frac{\pi}{4}} \right)^{e^{i\frac{\pi}{4}}} = e^{i\frac{\pi}{4} \cdot e^{i\frac{\pi}{4}}}
\]

Next, compute the product \(i\frac{\pi}{4} \cdot e^{i\frac{\pi}{4}}\). Expanding this using Euler’s identity:

\[
e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}
\]

Now multiply \(i\frac{\pi}{4}\) by this expression:

\[
i\frac{\pi}{4} \cdot \left( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right)
\]

Using distributive multiplication:

\[
= \frac{i\pi}{4\sqrt{2}} + \frac{-\pi}{4\sqrt{2}}
\]

This simplifies to:

\[
-\frac{\pi}{4\sqrt{2}} + i\frac{\pi}{4\sqrt{2}}
\]

### Step 3: Final exponentiation

Thus, we now have:

\[
\left( e^{i\frac{\pi}{4}} \right)^{e^{i\frac{\pi}{4}}} = e^{-\frac{\pi}{4\sqrt{2}} + i\frac{\pi}{4\sqrt{2}}}
\]

Using Euler’s formula \(e^{ix} = \cos x + i\sin x\), this becomes:

\[
e^{-\frac{\pi}{4\sqrt{2}}} \left( \cos\frac{\pi}{4\sqrt{2}} + i\sin\frac{\pi}{4\sqrt{2}} \right)
\]

### Conclusion

This matches the given identity:

\[
(\sqrt{i})^{\sqrt{i}} = e^{\frac{-\pi}{4\sqrt{2}}}\left(\cos\frac{\pi}{4\sqrt{2}} + i\sin\frac{\pi}{4\sqrt{2}}\right)
\]

Thus, the equation is correct.

---

Would you like more details on any step, or have further questions?

### Relative questions:
1. How do you convert complex numbers into polar form?
2. What are the properties of complex exponentiation?
3. How do you compute powers of complex numbers?
4. What is Euler's identity and how is it applied in complex exponentiation?
5. How do you simplify trigonometric expressions in complex form?

### Tip:
When dealing with powers and roots of complex numbers, using polar form simplifies both the calculations and understanding of their geometric interpretation.

The query submitted by the user: (\sqrt{i})^{\sqrt{i}} = e^{\frac{-\pi}{4\sqrt{2}}}\left(\cos\frac{\pi}{4\sqrt{2}} + i\sin\frac{\pi}{4\sqrt{2}}\right)

Explore the step-by-step solution to verifying the identity (\sqrt{i})^{\sqrt{i}} = e^{\frac{-\pi}{4\sqrt{2}}}(\cos\frac{\pi}{4\sqrt{2}} + i\sin\frac{\pi}{4\sqrt{2}}). This problem covers key concepts in complex numbers, including polar form and Euler's formula. Suitable for advanced high school students and undergraduates.

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