(1-i)^50

Jasonbradley

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Jun 16, 2024

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Simplifying (1 - i)^50

This article will explore the process of simplifying the complex number expression (1 - i)^50.

Understanding Complex Numbers and De Moivre's Theorem

Complex numbers are numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit, defined as the square root of -1. To simplify expressions involving complex numbers raised to a power, we can utilize De Moivre's Theorem.

De Moivre's Theorem states:

(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)

This theorem allows us to express a complex number in polar form (r(cos θ + i sin θ)) and raise it to a power by simply multiplying the angle by the exponent.

Simplifying (1 - i)^50

1. Convert to Polar Form:

   First, we need to convert (1 - i) into polar form. We can do this by finding the magnitude and angle:

   • Magnitude (r): r = √(1² + (-1)²) = √2
   • Angle (θ): θ = arctan(-1/1) = -π/4 (since the point lies in the fourth quadrant)

   Therefore, (1 - i) = √2(cos(-π/4) + i sin(-π/4))

2. Apply De Moivre's Theorem:

   Now, we can apply De Moivre's Theorem to raise the complex number to the power of 50:

   (√2(cos(-π/4) + i sin(-π/4)))^50 = √2^50 (cos(-50π/4) + i sin(-50π/4))

3. Simplify:

   • √2^50 = 2^25
   • cos(-50π/4) = cos(-12.5π) = cos(-2.5π) = 0
   • sin(-50π/4) = sin(-12.5π) = sin(-2.5π) = 1

   Therefore, (1 - i)^50 = 2^25 (0 + i * 1) = 2^25 i

Conclusion

By converting (1 - i) into polar form and applying De Moivre's Theorem, we have simplified (1 - i)^50 to 2^25 i. This demonstrates how De Moivre's Theorem provides a powerful tool for working with complex numbers raised to powers.