Simplifying Complex Number Exponents: Solving sqrt[i]{-1}

In this article, we will guide you through the process of simplifying the complex exponentiation problem: sqrt[i]{-1}. We will explore how to apply Euler's formula and the properties of the complex logarithm to find the solution.

Introduction to the Problem

We start by expressing -1 in its exponential form:

-1  e^{ipi} 

Next, we use the property of roots of complex numbers: sqrt[w]{z} z^{1/w}. Applying this to our problem, we have:

sqrt[i]{-1}  sqrt[i]{e^{ipi}}  (e^{ipi})^{1/i}  e^{ipi/i}  e^{pi} 

This is the primary result. However, we can further explore the role of the complex logarithm and phase factors to gain a more comprehensive understanding.

Using the Complex Logarithm

To fully address the problem, we must consider the complex logarithm. The natural logarithm of a negative number is not defined in the real number system, so we use the complex logarithm. The complex logarithm of a complex number z is given by:

Ln(z)  ln(|z|)   iArg(z)   2ipi n 

For -1, we have:

Arg(-1)  pi 

Thus, the complex logarithm of -1 is:

Ln(-1)  ln(1)   i(pi)   2ipi n  i(pi)   2ipi n 

Now, we can express the square root in terms of the complex logarithm:

sqrt[i]{-1}  (-1)^{1/i}  e^{{1/i} Ln(-1)}  e^{{1/i} (i(pi)   2ipi n)} 

Simplifying the exponent:

sqrt[i]{-1}  e^{pi   2pi n} 

This expression shows that the result is not unique, as it depends on the chosen branch of the logarithm, represented by the integer n.

Final Results and Considerations

When n 0, the result simplifies to:

sqrt[i]{-1}  e^{pi} 

This is the primary solution. However, by varying n, we can obtain other possible values:

sqrt[i]{-1}  e^{pi   2pi n} 

Here are a few examples:

n 0: e^{pi} n 1: e^{pi 2pi} e^{pi(1 2)} e^{3pi} n -1: e^{pi - 2pi} e^{pi(-1 2)} e^{pi}

Each choice of n provides a different value for the square root, reflecting the non-uniqueness of the complex logarithm.