A Problem With Powers of 1+i | Problem 360

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Опубликовано:

24 сен 2024

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Комментарии : 10

@Goofyahhgameenjoyer 9 дней назад

Eulers identity has got to be the greatest identity in all of mathematics. A true genius

@aplusbi 7 дней назад

Absolutely!!! 😍

@kevinmadden1645 10 дней назад

Too long and drawn out. Once you have 1+I expressed in Polar Form use De Moivre's Theorem with theta equal to 7pi/4. You end up with 2 raised to the power of n/2 equal to 2 to the power of 7/2. Therefore n equals 7.

@crazysin96 9 дней назад

You are using Euler's Theorem and the polar coordinate anyways, why not just write LHS as 2^(n/2)*e^(n*pi*i/4) and RHS as 2^(7/2)*e^(7*pi*i/4). From this, n is clearly 7, comparing either the phase or the magnitude of LHS and RHS. In general, I always like thinking of powers of complex numbers as a rotation+scaling in the argand plane.

@vikrantbhadouriya 9 дней назад

arg(8-8i) = 315 degrees arg(1+i) = 45 degrees n * arg(1+i) = arg(8-8i) n = 7

@key_board_x 9 дней назад

a = 1 + i ← this is a complex number The modulus of a is: m = √[(1)² + (1)²] = √2 The argument of a is β such as: tan(β) = 1/1 = 1 → β = π/4 b = 8 - 8i ← this is a new complex number The modulus of b is: M = √[(8)² + (- 8)²] = √(64 + 64) = √128 = 8√2 The argument of b is ω such as: tan(ω) = (- 8)/8 = - 1 → ω = 7π/4 You want to get: a^(n) = b For the modulus: m^(n) = M (√2)^(n) = 8√2 (√2)^(n) = [2^(3)] * 2^(1/2) (√2)^(n) = 2^[3 + (1/2)] (√2)^(n) = 2^(7/2) (√2)^(n) = [(√2)^(2)]^(7/2) (√2)^(n) = (√2)^(7) → n= 7 For the argument: n * β = ω n * (π/4) = 7π/4 n * (π/4) = 7 * (π/4) → n = 7 You can conclude that: n = 7 → let's check it (1 + i)² = 1 + 2i + i² (1 + i)² = 1 + 2i - 1 (1 + i)² = 2i (1 + i)⁴ = [(1 + i)²]² (1 + i)⁴ = [2i]² (1 + i)⁴ = 4i² (1 + i)⁴ = - 4 (1 + i)⁷ = (1 + i)⁴.(1 + i)².(1 + i) (1 + i)⁷ = (- 4).(2i).(1 + i) (1 + i)⁷ = - 8i.(1 + i) (1 + i)⁷ = - 8i - 8i² (1 + i)⁷ = - 8i + 8 (1 + i)⁷ = 8 - 8i