Question

Вычислить: (1 - i)^42.

Answer 1

$(1-i)^2 = 1 - 2i + i^2 = -2i\\(1-i)^{42} = \underbrace{(1-i)(1-i)\cdots(1-i)}_{42} = \underbrace{(-2i)(-2i)\ldots(-2i)}_{21} = -2^{21}\cdot i^{21} =\\ -2^{21}\cdot (i^2)^{10}\cdot i = -2^{21}\cdot i$

Answer 2

To calculate the value of (1 - i)^42, you can use the De Moivre's theorem. The formula is as follows:

z^n = r^n * (cos(nθ) + i * sin(nθ))

where z is a complex number in the form a + bi, r is the magnitude of the complex number (sqrt(a^2 + b^2)), and θ is the argument of the complex number (tan^(-1)(b/a)).

In this case, z = 1 - i, so a = 1, and b = -1.

Step 1: Calculate r (magnitude of z) r = sqrt(a^2 + b^2) = sqrt(1^2 + (-1)^2) = sqrt(2)

Step 2: Calculate θ (argument of z) θ = tan^(-1)(b/a) = tan^(-1)(-1/1) = -π/4

Step 3: Apply De Moivre's theorem (1 - i)^42 = r^42 * (cos(42θ) + i * sin(42θ)) = (sqrt(2))^42 * (cos(42 * (-π/4)) + i * sin(42 * (-π/4)))

Step 4: Simplify Since cos(-π/4) = cos(7π/4) = sqrt(2)/2 and sin(-π/4) = sin(7π/4) = -sqrt(2)/2, we get:

(1 - i)^42 = (sqrt(2))^42 * (sqrt(2)/2 + i * (-sqrt(2)/2)) = 2^21 * (sqrt(2)/2 - sqrt(2)/2) = 2^21 * 0 = 0

Therefore, (1 - i)^42 equals 0.