%5E10.jpeg "(1-i)^10")
Demystifying (1-i)^10
This article explores the process of calculating (1-i)<sup>10</sup>, where i is the imaginary unit (√-1). We will utilize De Moivre's Theorem and Euler's Formula to simplify this complex exponentiation.
Understanding De Moivre's Theorem
De Moivre's Theorem states that for any complex number in polar form, z = r(cos θ + i sin θ) and any integer n:
z<sup>n</sup> = r<sup>n</sup>(cos nθ + i sin nθ)
Expressing (1-i) in Polar Form
-
Magnitude (r):
- Find the magnitude using the Pythagorean theorem: |1-i| = √(1<sup>2</sup> + (-1)<sup>2</sup>) = √2
-
Angle (θ):
- Find the angle using the arctangent function: θ = arctan(-1/1) = -π/4. Since (1-i) lies in the fourth quadrant, we add 2π to obtain the principal argument: θ = 7π/4.
Therefore, (1-i) in polar form is √2(cos(7π/4) + i sin(7π/4)).
Applying De Moivre's Theorem
Now, we can use De Moivre's Theorem to calculate (1-i)<sup>10</sup>:
(1-i)<sup>10</sup> = (√2)<sup>10</sup>(cos(10 * 7π/4) + i sin(10 * 7π/4))
Simplify:
(1-i)<sup>10</sup> = 32(cos(17.5π) + i sin(17.5π))
Using Euler's Formula
Euler's Formula provides a relationship between exponential and trigonometric functions:
e<sup>iθ</sup> = cos θ + i sin θ
Therefore, we can rewrite the result:
(1-i)<sup>10</sup> = 32(e<sup>i17.5π</sup>)
Final Simplification
Since the period of the sine and cosine functions is 2π, we can simplify the exponent:
17.5π = 8π + 1.5π = 4(2π) + 1.5π
Therefore,
(1-i)<sup>10</sup> = 32(e<sup>i1.5π</sup>) = 32(cos(1.5π) + i sin(1.5π)) = 32(-1 + 0i) = -32
Conclusion
By using De Moivre's Theorem and Euler's Formula, we have successfully calculated (1-i)<sup>10</sup> to be -32. This demonstrates how these powerful tools can simplify complex exponentiation problems.