%5E12.jpeg "(1-i)^12")
Demystifying (1-i)^12
This article explores the process of simplifying the complex expression (1 - i)^12.
Understanding Complex Numbers
Before diving into the calculation, let's understand the basics of complex numbers. A complex number is a number 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 (i² = -1).
Utilizing De Moivre's Theorem
De Moivre's theorem offers a powerful tool for simplifying expressions involving complex numbers raised to powers. It states:
(cos θ + i sin θ)^n = cos(nθ) + i sin(nθ)
To apply this theorem, we need to convert our complex number (1 - i) into polar form.
Converting to Polar Form
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Magnitude: The magnitude or modulus of a complex number (a + bi) is calculated as: √(a² + b²). For (1 - i), the magnitude is √(1² + (-1)²) = √2.
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Angle: The angle (or argument) is determined using the arctangent function: θ = arctan(b/a). In our case, θ = arctan(-1/1) = -π/4. Since (1 - i) lies in the fourth quadrant, we need to add 2π to get the angle in the range of 0 to 2π: θ = 7π/4.
Therefore, the polar form of (1 - i) is √2(cos(7π/4) + i sin(7π/4)).
Applying De Moivre's Theorem
Now we can apply De Moivre's theorem:
(1 - i)^12 = [√2(cos(7π/4) + i sin(7π/4))]^12
= (√2)^12 [cos(12 * 7π/4) + i sin(12 * 7π/4)]
= 2^6 [cos(21π) + i sin(21π)]
= 64 [cos(π) + i sin(π)]
= 64 (-1 + 0i)
= -64
Final Result
Therefore, (1 - i)^12 simplifies to -64.