Question

[cos(2П/3)+isin(2П/3)]*5[cos(-П/2)+isin(-П/2)]=

Answer 1

0+-(1)*0+-(1)=-1

Вот

мы тоже сейчас это проходим

Answer 2

To simplify the expression, let's start by performing the multiplication of the two complex numbers:

Let z1 = cos(2π/3) + isin(2π/3) and z2 = cos(-π/2) + isin(-π/2).

Now, the product of these two complex numbers is:

z1 * z2 = (cos(2π/3) + isin(2π/3)) * (cos(-π/2) + isin(-π/2))

To multiply complex numbers, we use the property that (a + bi)(c + di) = (ac - bd) + (ad + bc)i.

Applying this property:

z1 * z2 = (cos(2π/3) * cos(-π/2) - sin(2π/3) * sin(-π/2)) + (cos(2π/3) * sin(-π/2) + sin(2π/3) * cos(-π/2))i

Now, recall the trigonometric identities:

cos(θ) * cos(φ) = cos(θ + φ) sin(θ) * cos(φ) = sin(θ + φ)

Using these identities:

cos(2π/3) * cos(-π/2) - sin(2π/3) * sin(-π/2) = cos(2π/3 - π/2) = cos(π/6) cos(2π/3) * sin(-π/2) + sin(2π/3) * cos(-π/2) = sin(2π/3 - π/2) = sin(π/6)

Therefore, the simplified expression is:

z1 * z2 = cos(π/6) + isin(π/6)

Since cos(π/6) = √3/2 and sin(π/6) = 1/2, the final result is:

z1 * z2 = (√3/2) + i(1/2)