Question

даны комплексные числа z1=3+3i ; z2=-2-4i; найти : z1+z2;z1-z2;3z1-4z2;z1*z2;z1/z2;

Answer 1

1) z1-z2

(3+3i)+(-2-4i) = 3+3i-2-4i = 1-i

2) z1-z2

(3+3i)-(-2-4i) = 3+3i+2+4i = 5+7i

3) 3z1-4z2

3(3+3i)-4(-2-4i) = 9+9i+8+16i = 17+25i

4) z1*z2

(3+3i)(-2-4i) = -6-12i-6i-12i² = -6-18i+12 = 6-18i

Зная, что i² = -1

5)

$\frac{3 + 3i}{- 2 - 4i} = \frac{(3 + 3i)( - 2 + 4i)}{(- 2 - 4i)( - 2 + 4i)} = \frac{ - 6 + 12i - 6i + 12{i}^{2} }{4 - 8i + 8i - 16 {i}^{2} } = { \frac{ - 18 + 6i}{20} } = - 0.9 + 0.3i$