find the fifth roots of -32i
Notice that -32i = 2%5E5%2Acis%28270%5Eo%29.
Therefore, due to DeMoivre formulas, the 5-th degree roots of -32i are
(1) 2cis(54°); ( notice that 54° = 270%5Eo%2F5 );
(2) 2cis%2854%5Eo+%2B+360%5Eo%2F5%29 = = 2cis(54° + 72°) = 2cis(126°); ( Notice that 72° = )
(3) 2cis%2854%5Eo+%2B+2%2A%28360%5Eo%2F5%29%29 = 2cis(54° + 2*72°) = 2cis(198°);
(4) 2cis%2854%5Eo+%2B+3%2A%28360%5Eo%2F5%29%29 = 2cis(54° + 3*72°) = 2cis(270°);
(5) 2cis%2854%5Eo+%2B+4%2A%28360%5Eo%2F5%29%29 = 2cis(54° + 4*72°) = 2cis(342°).
To find the fifth roots of -32i, we can start by expressing -32i in polar form.
Step 1: Convert -32i to polar form.
We can express -32i as -32 * (cos(π/2) + i * sin(π/2)) in polar form.
Step 2: Simplify the expression.
-32 * (cos(π/2) + i * sin(π/2)) = -32 * cis(π/2).
Step 3: Find the principal fifth root of -32i.
The principal fifth root, denoted as z, of a complex number in polar form, r * cis(θ), is given by z = (r)^(1/5) * cis(θ/5).
In this case, r = 32 and θ = π/2.
z = (32)^(1/5) * cis((π/2)/5).
Step 4: Calculate the principal fifth root.
(32)^(1/5) = 2^(5/5) = 2.
cis((π/2)/5) can be determined using the formula cis(θ) = cos(θ) + i * sin(θ).
cos((π/2)/5) = cos(π/10) ≈ 0.9877
sin((π/2)/5) = sin(π/10) ≈ 0.1564
Therefore, cis((π/2)/5) ≈ 0.9877 + i * 0.1564.
Step 5: Calculate the fifth roots.
The fifth roots of -32i are given by:
z1 = 2 * cis((π/2)/5) ≈ 2 * (0.9877 + i * 0.1564) ≈ 1.9754 + i * 0.3128
z2 = 2 * cis((π + 2π)/5) ≈ 2 * (0.5403 + i * 0.8415) ≈ 1.0806 + i * 1.683
z3 = 2 * cis((π + 4π)/5) ≈ 2 * (-0.5403 + i * 0.8415) ≈ -1.0806 + i * 1.683
z4 = 2 * cis((π + 6π)/5) ≈ 2 * (-0.9877 + i * 0.1564) ≈ -1.9754 + i * 0.3128
z5 = 2 * cis((π + 8π)/5) ≈ 2 * (-0.9877 - i * 0.1564) ≈ -1.9754 - i * 0.3128
Hence, the fifth roots of -32i are:
z1 ≈ 1.9754 + i * 0.3128
z2 ≈ 1.0806 + i * 1.683
z3 ≈ -1.0806 + i * 1.683
z4 ≈ -1.9754 + i * 0.3128
z5 ≈ -1.9754 - i * 0.3128
To find the fifth roots of -32i, we can start by writing -32i in polar form.
Step 1: Convert -32i into polar form
To find the magnitude (r) and argument (θ) of -32i, we can use the following formulas:
r = √(real part^2 + imaginary part^2)
= √(0^2 + (-32)^2)
= √(0 + 1024)
= √1024
= 32
θ = arctan(imaginary part / real part)
= arctan(-32 / 0)
(Note that arctan(∞) is undefined, but we can still determine the value of θ)
Since -32i lies in the third quadrant of the complex plane, θ = arctan(-∞) = -π/2.
Therefore, -32i in polar form is: 32(cos(-π/2) + isin(-π/2))
Step 2: Finding the fifth roots
To find the fifth roots of -32i, we need to find values of z that satisfy the equation z^5 = -32i.
Since we have -32i in polar form, we can express the fifth roots using De Moivre's formula:
z_k = (√r)(cos((θ + 2kπ)/n) + isin((θ + 2kπ)/n))
where k is an integer from 0 to n-1, and n is the root index (in this case, n = 5).
Plugging in the values, we get:
z_0 = (√32)(cos((-π/2 + 2(0)π)/5) + isin((-π/2 + 2(0)π)/5))
= 4(cos(-π/10) + isin(-π/10))
z_1 = (√32)(cos((-π/2 + 2(1)π)/5) + isin((-π/2 + 2(1)π)/5))
= 4(cos(7π/10) + isin(7π/10))
z_2 = (√32)(cos((-π/2 + 2(2)π)/5) + isin((-π/2 + 2(2)π)/5))
= 4(cos(13π/10) + isin(13π/10))
z_3 = (√32)(cos((-π/2 + 2(3)π)/5) + isin((-π/2 + 2(3)π)/5))
= 4(cos(19π/10) + isin(19π/10))
z_4 = (√32)(cos((-π/2 + 2(4)π)/5) + isin((-π/2 + 2(4)π)/5))
= 4(cos(25π/10) + isin(25π/10))
Therefore, the fifth roots of -32i are:
z_0 = 4(cos(-π/10) + isin(-π/10))
z_1 = 4(cos(7π/10) + isin(7π/10))
z_2 = 4(cos(13π/10) + isin(13π/10))
z_3 = 4(cos(19π/10) + isin(19π/10))
z_4 = 4(cos(25π/10) + isin(25π/10))