Let cis Θ = cos Θ + i sin Θ. Let z1 = 3 cis 25 and let z2 = 4 cis 115

Find z2^6

To find z2^6, we need to raise the complex number z2 to the power of 6.

Given that z2 = 4 cis 115, we can rewrite it using Euler's formula:

z2 = 4(cos 115° + i sin 115°)

Now, we can use De Moivre's theorem to raise z2 to the power of 6:

z2^6 = [4(cos 115° + i sin 115°)]^6

To simplify this expression, we can apply the binomial expansion. The binomial expansion states that for any real number a and b, and any positive integer n, (a + b)^n can be expanded as follows:

(a + b)^n = a^n + (nC₁)a^(n-1)b + (nC₂)a^(n-2)b^2 + ... + (nC_{n-1})ab^(n-1) + b^n

Where nCk represents the binomial coefficient, given by the formula:

nCk = n! / (k!(n-k)!)

Using the binomial expansion on z2^6, we get:

z2^6 = 4^6(cos(115°)^6 + 6C₁(4^5)(cos(115°)^5)(i sin 115°) + 15C₂(4^4)(cos(115°)^4)(i sin 115°)^2 + ... + 6C₅(4^1)(cos(115°)(i sin 115°)^5 + (i sin 115°)^6)

Simplifying further, we know that cos(115°)^2 + sin(115°)^2 = 1, and that (i)^2 = -1:

z2^6 = 4^6(cos(115°)^6 + (6C₁)(4^5)(cos(115°)^4)(sin(115°))(i) + (15C₂)(4^4)(cos(115°)^2)(sin(115°)^2) + ... + (6C₅)(4^1)(cos(115°))(sin(115°)^5) - (sin(115°)^6))

Finally, we can calculate each term in the expansion using known values:

z2^6 = 4^6(cos(690°) + (6C₁)(4^5)(cos(460°))(sin(115°))(i) + (15C₂)(4^4)(1)(1) + ... + (6C₅)(4^1)(cos(115°))(sin(575°)) - (sin(690°)))

Simplifying this expression will yield the final value of z2^6.