Question

Вычислить (1/32)*(-1+sqrt(3)*i)^5

Answer 1

z=x+iy

|z|=√(x²+y²)

cosφ=x/|z|

sinφ=y/|z|

z=-1+√3i

x=-1; y=√3

|z|=√((-1)²+(√3)²)=2

cosφ=x/|z|=-1/2

sinφ=y/|z|=√3/2

φ=2π/3

Тригонометрическая форма:

z=2(cos(2π/3)+isin(2π/3))

z⁵=2⁵(cos5·(2π/3)+isin5·(2π/3))=32(cos(10π/3)+isin(10π/3))=

=32(cos(4π-(2π/3))+isin(4π-(2π/3))=

=32(cos(-(2π/3))+isin(-(2π/3))=

=32(cos(2π/3)-isin(2π/3)=32(1-√3i)

О т в е т. 1-√3i

Answer 2

To compute the expression (1/32)*(-1+sqrt(3)*i)^5, follow these steps:

Step 1: Find the value of (-1 + √3i)^5 Step 2: Divide the result by 32

Let's begin:

Step 1: Finding (-1 + √3i)^5

(-1 + √3i)^5 can be computed using the binomial theorem, which states that (a + b)^n = Σ(choose(n, k) * a^(n-k) * b^k), where Σ is the sum from k = 0 to n, and choose(n, k) is the binomial coefficient, given by choose(n, k) = n! / (k! * (n-k)!).

In this case, a = -1 and b = √3i. Also, n = 5.

So, (-1 + √3i)^5 = choose(5, 0) * (-1)^(5-0) * (√3i)^0 + choose(5, 1) * (-1)^(5-1) * (√3i)^1 + choose(5, 2) * (-1)^(5-2) * (√3i)^2 + choose(5, 3) * (-1)^(5-3) * (√3i)^3 + choose(5, 4) * (-1)^(5-4) * (√3i)^4 + choose(5, 5) * (-1)^(5-5) * (√3i)^5

Now, let's calculate each term:

choose(5, 0) = 1, and (-1)^5 = -1, (√3i)^0 = 1 choose(5, 1) = 5, and (-1)^4 = 1, (√3i)^1 = √3i choose(5, 2) = 10, and (-1)^3 = -1, (√3i)^2 = 3i choose(5, 3) = 10, and (-1)^2 = 1, (√3i)^3 = 3√3i choose(5, 4) = 5, and (-1)^1 = -1, (√3i)^4 = 9i^2 = -9 choose(5, 5) = 1, and (-1)^0 = 1, (√3i)^5 = 27i^2 = -27

Putting it all together:

(-1 + √3i)^5 = 1 * (-1) * 1 + 5 * 1 * √3i + 10 * (-1) * 3i + 10 * 1 * 3√3i + 5 * (-1) * (-9) + 1 * 1 * (-27) = -1 + 5√3i - 30i + 30√3i + 45 - 27 = 18 + 35√3i - 30i

Step 2: Divide the result by 32

Now, divide the value obtained above by 32:

(1/32) * (-1 + √3i)^5 = (1/32) * (18 + 35√3i - 30i) = (18/32) + (35/32)√3i - (30/32)i

Finally, simplify the expression:

(1/32) * (-1 + √3i)^5 = 9/16 + (35/32)√3i - (15/16)i

So, the final result is 9/16 + (35/32)√3i - (15/16)i.