Find Two Distinct Square Roots Of I

Finding Two Distinct Square Roots of i: A Comprehensive Guide

Finding the square roots of a complex number like i might seem daunting, but with a bit of understanding of complex numbers and some algebraic manipulation, it becomes surprisingly straightforward. This article will guide you through the process, explaining the concepts involved and demonstrating how to arrive at the two distinct square roots. This process involves leveraging polar form and DeMoivre's Theorem.

Let's dive in and discover the two distinct square roots of the imaginary unit, i.

Understanding Complex Numbers in Polar Form

Before tackling the square root of i, it's crucial to understand how complex numbers are represented in polar form. A complex number z can be written as z = a + bi, where a is the real part and b is the imaginary part. The polar form represents this number using its magnitude (or modulus) r and its argument (or angle) θ:

z = r(cos θ + i sin θ)

where r = √(a² + b²) and θ = arctan(b/a). The argument θ is often expressed in radians.

Applying DeMoivre's Theorem

DeMoivre's Theorem provides a powerful tool for finding roots of complex numbers. It states that for any complex number z = r(cos θ + i sin θ) and any integer n, the nth roots of z are given by:

z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]

where k = 0, 1, 2, ..., n - 1. This formula generates n distinct roots.

Finding the Square Roots of i

Now let's apply this to find the square roots of i. First, we express i in polar form:

i = 1(cos(π/2) + i sin(π/2))

Here, r = 1 and θ = π/2. Since we're looking for square roots (n = 2), DeMoivre's Theorem gives us:

i^(1/2) = 1^(1/2) [cos((π/2 + 2kπ)/2) + i sin((π/2 + 2kπ)/2)]

We need to calculate this for k = 0 and k = 1 to obtain the two distinct square roots:

• For k = 0:

i^(1/2) = cos(π/4) + i sin(π/4) = √2/2 + i√2/2

• For k = 1:

i^(1/2) = cos(5π/4) + i sin(5π/4) = -√2/2 - i√2/2

Therefore, the two distinct square roots of i are: √2/2 + i√2/2 and -√2/2 - i√2/2. These can also be expressed as (1 + i)/√2 and (-1 - i)/√2.

Conclusion

By understanding complex numbers in polar form and applying DeMoivre's Theorem, we successfully found the two distinct square roots of i. This method can be extended to find the nth roots of any complex number, demonstrating the power and elegance of complex number analysis. Remember to always consider the multiple solutions provided by DeMoivre's theorem when dealing with roots of complex numbers. This comprehensive approach ensures you’ll always arrive at the correct answer, even when dealing with more intricate complex number calculations.