Задача 75858 Даны комплексные числа z1 и z2 (табл....
Условие
записать в тригонометрической и алгебраической формах. Отметить полученные
числа на комплексной плоскости.
Решение
r = sqrt((-sqrt(12))^2 + 6^2) = sqrt(12 + 36) = sqrt(48)
a = -sqrt(12)/sqrt(48) = -sqrt(12/48) = -sqrt(1/4) = -1/2
b = 6/sqrt(48) = sqrt(36/48) = sqrt(3/4) = sqrt(3)/2
z1 = -sqrt(12) + 6i = sqrt(48)*(-1/2 + sqrt(3)/2*i )
[m]z2= \frac{-6}{1-i} = \frac{-6(1 + i)}{(1-i)(1+i)} = \frac{-6(1+i)}{1+1} = -3(1-i) = -3+3i[/m]
r = sqrt((-3)^2 + 3^2) = sqrt(9 + 9) = sqrt(18)
a = -3/(3sqrt(2)) = -1/sqrt(2)
b = 3/(3sqrt(2)) = 1/sqrt(2)
z2 = -3 + 3i = sqrt(18)*(-1/sqrt(2) + 1/sqrt(2)*i )
1. а) В тригонометрической форме.
z1 = sqrt(48)*(-1/2 + sqrt(3)/2*i ) = sqrt(48)*(cos (5π/3) + i*sin (5π/3))
z2 = sqrt(18)*(-1/sqrt(2) + 1/sqrt(2)*i ) = sqrt(18)*(cos (7π/4) + i*sin (7π/4))
б) z1*z2 = (-sqrt(12) + 6i)(-3+3i) = 3sqrt(12) - 18i - 3sqrt(12)*i + 18i^2 =
= 6sqrt(3) - 18 + i*(6sqrt(3) + 18)
[m]\frac{z1}{z2} = \frac{-\sqrt{12} + 6i}{-3+3i} = \frac{(-\sqrt{12} + 6i)(-3-3i)}{(-3+3i)(-3-3i)} = \frac{3\sqrt{12} - 18i + 3\sqrt{12}*i - 18i^2}{9-9i^2} = \frac{6\sqrt{3} + 18 + i*(6\sqrt{3} - 18)}{18} = \frac{\sqrt{3} + 3}{3} + i \cdot \frac{\sqrt{3} - 3}{3}[/m]