Precalc
(1+i)^3
and
the cube root of 8i
(1+i) = (√2,pi/4)
so, (1+i)^3 = (2√2,3pi/4) = (-2+2i)
8i = (8,pi/2)
(8i)^(1/3) = (2,pi/6) = (√3+i)
To solve these precalculus problems, we will need to use basic algebra and complex number properties.
1) Calculating (1+i)^3:
To find the value of (1+i)^3, we'll use the binomial expansion formula, which states that (a + b)^n can be expanded using Pascal's triangle coefficients.
For (1+i)^3, we have a = 1, b = i, and n = 3. According to the binomial expansion formula, we expand as follows:
(1+i)^3 = C(3,0)(1)^3(i)^0 + C(3,1)(1)^2(i)^1 + C(3,2)(1)^1(i)^2 + C(3,3)(1)^0(i)^3
= (1)(1) + (3)(1)(i) + (3)(-1)(1) + (1)(-i)
= 1 + 3i - 3 + (-i)
= -2 + 2i
Therefore, (1+i)^3 simplifies to -2 + 2i.
2) Finding the cube root of 8i:
To find the cube root of 8i, we'll use the polar form of complex numbers.
First, let's convert 8i to polar form:
The magnitude (r) of 8i is 8, and the argument (θ) can be found using the equation θ = atan(Im/Re) where Im is the imaginary part and Re is the real part.
For 8i, the argument θ = atan(8/0) = 90 degrees or π/2 radians.
Now, we can express 8i in polar form:
8i = 8(cos(π/2) + i*sin(π/2))
To find the cube root of 8i, we need to take the cube root of the magnitude and divide the argument by 3:
(cbrt(8))(cos(π/6) + i*sin(π/6))
Simplifying further, the cube root of 8i is:
2(cos(π/6) + i*sin(π/6))
The final answer is 2(cos(π/6) + i*sin(π/6)).