Consider the complex number z=-2sqrt3+2i. What is z^(3) ? Hint: z has a modulus of 4 and an argument of 150^(@). Choose 1 answer: (A) -6-10.4 i (B) 64 i (c) -10.4+6i (D) -32-55.4 i

Question

Consider the complex number  z=-2sqrt3+2i. What is  z^(3) ? Hint:  z has a modulus of 4 and an argument of  150^(@). Choose 1 answer: (A)  -6-10.4 i (B)  64 i (c)  -10.4+6i (D)  -32-55.4 i

Bytelearn · Accepted Answer

Given complex number: We are given the complex number z = -2√3 + 2i. We are also given that z has a modulus of 4 and an argument of 150 degrees. To find z^3, we can use De Moivre's Theorem, which states that (r(cos(θ) + i sin(θ)))^n = r^n (cos(nθ) + i sin(nθ)), where r is the modulus and θ is the argument of the complex number.Convert to radians: First, we convert the argument of 150 degrees to radians because trigonometric functions in the complex plane are typically expressed in radians. 150 degrees is equivalent to 150 * (π/180) = 5π/6 radians.Apply De Moivre's Theorem: Now we can apply De Moivre's Theorem. We have r = 4 and θ = 5π/6. We want to find z^3, so we raise the modulus to the power of 3 and multiply the argument by 3. r^3 = 4^3 = 64 θ * 3 = 5π/6 * 3 = 5π/2Simplify argument: Using De Moivre's Theorem, we get: z^3 = 64 * (cos(5π/2) + i sin(5π/2)) Since the sine and cosine functions are periodic with a period of 2π, we can simplify the argument 5π/2 to π/2 by subtracting 2π (which is one full period) from 5π/2. 5π/2 - 2π = π/2Find cosine and sine: Now we can find the cosine and sine of π/2: cos(π/2) = 0 sin(π/2) = 1 So, z^3 = 64 * (0 + i * 1) = 64i

Consider the complex number $z=-2 \sqrt{3}+2 i$ . $\newline$ What is $z^{3}$ ? $\newline$ Hint: $z$ has a modulus of $4$ and an argument of $150^{\circ}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-6-10.4 i$ $\newline$ (B) $64 i$ $\newline$ (C) $-10.4+6 i$ $\newline$ (D) $-32-55.4 i$

Full solution

More problems from Domain and range

Related topics