Math Problem Statement

(\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)

Solution

To verify the identity (i)i=eπ42(cosπ42+isinπ42)(\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 i\sqrt{i} in polar form

Start with ii, which in polar form is written as:

i=eiπ2i = e^{i\frac{\pi}{2}}

This is because ii corresponds to a complex number on the unit circle at an angle of π2\frac{\pi}{2} radians. Now, we take the square root of ii, which means finding i\sqrt{i}. The square roots of a complex number reiθr e^{i\theta} are given by:

reiθ=reiθ2\sqrt{r e^{i\theta}} = \sqrt{r} e^{i\frac{\theta}{2}}

For i=eiπ2i = e^{i\frac{\pi}{2}}, we get:

i=eiπ4\sqrt{i} = e^{i\frac{\pi}{4}}

Step 2: Raise i\sqrt{i} to i\sqrt{i}

Now, we need to compute:

(i)i=(eiπ4)eiπ4\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:

(eiπ4)eiπ4=eiπ4eiπ4\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 iπ4eiπ4i\frac{\pi}{4} \cdot e^{i\frac{\pi}{4}}. Expanding this using Euler’s identity:

eiπ4=cosπ4+isinπ4=12+i12e^{i\frac{\pi}{4}} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}

Now multiply iπ4i\frac{\pi}{4} by this expression:

iπ4(12+i12)i\frac{\pi}{4} \cdot \left( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right)

Using distributive multiplication:

=iπ42+π42= \frac{i\pi}{4\sqrt{2}} + \frac{-\pi}{4\sqrt{2}}

This simplifies to:

π42+iπ42-\frac{\pi}{4\sqrt{2}} + i\frac{\pi}{4\sqrt{2}}

Step 3: Final exponentiation

Thus, we now have:

(eiπ4)eiπ4=eπ42+iπ42\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 eix=cosx+isinxe^{ix} = \cos x + i\sin x, this becomes:

eπ42(cosπ42+isinπ42)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:

(i)i=eπ42(cosπ42+isinπ42)(\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.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Form
Complex Exponentiation
Euler's Formula

Formulas

i = e^{i\frac{\pi}{2}}
\sqrt{r e^{i\theta}} = \sqrt{r} e^{i\frac{\theta}{2}}
e^{ix} = \cos x + i\sin x

Theorems

Euler's Identity

Suitable Grade Level

Undergraduate/Advanced High School