Question

Применяя формулу муавра найти Z^n z=3-3i n=8 помогите пожалуйста

Answer 1

Формула Муавра позволяет заменить z^n  на r^n*(cos na + i*sin na),
где r - модуль комплексного числа, а - угол
r=(3^2+(-3)^2)^1/2=3*(2)^1/2
tg(a)=-3/3=-1    a=(пи)/4
Подставляем в формулу:
(3*(2)^1/2)^8*(cos 2(пи)+i*sin 2(пи))=3^8*2^4=104976 

Answer 2

Applying the Formula of De Moivre

To find the value of Z^n, where Z = 3 - 3i and n = 8, we can use the formula of De Moivre. This formula allows us to raise a complex number to a power.

The formula of De Moivre states that for any complex number Z = a + bi and any positive integer n, we have:

Z^n = (r^n) * (cos(nθ) + i * sin(nθ))

where r is the modulus of Z given by r = sqrt(a^2 + b^2), and θ is the argument of Z given by θ = arctan(b/a).

In this case, Z = 3 - 3i, so a = 3 and b = -3. Let's calculate the modulus and argument of Z:

r = sqrt(3^2 + (-3)^2) = sqrt(18) = 3√2

θ = arctan((-3)/3) = arctan(-1) = -π/4

Now, we can substitute these values into the formula of De Moivre to find Z^n:

Z^n = (3√2)^8 * (cos(8 * (-π/4)) + i * sin(8 * (-π/4)))

Simplifying further:

Z^n = 144 * (cos(-2π) + i * sin(-2π))

Using the trigonometric identity cos(-θ) = cos(θ) and sin(-θ) = -sin(θ), we can rewrite the expression as:

Z^n = 144 * (cos(2π) - i * sin(2π))

Since cos(2π) = 1 and sin(2π) = 0, the expression simplifies to:

Z^n = 144 * (1 - 0i) = 144

Therefore, Z^n = 144.

Conclusion

Using the formula of De Moivre, we found that Z^n = 144 when Z = 3 - 3i and n = 8.