Find all complex numbers z satisfying
z3 = 27i
and give your answer as a comma-separated list of values in the form a+bi.
z^3 = 27i
z = 3?i
i = 1 cis(?/2)
?i = cis((1/3)*(?/2 + 2k?)) for k=0,1,2
= cis ?/6, 5?/6, 9?/6
See
http://www.wolframalpha.com/input/?i=z%5E3+%3D+27i
To find the complex numbers, z, satisfying the equation z^3 = 27i, we can use the cube root property.
Step 1: Express 27i in polar form.
To express 27i in polar form, we can represent it as r * e^(iθ), where r is the modulus (magnitude) and θ is the argument.
The modulus r is found by taking the square root of the sum of the squares of the real and imaginary parts:
r = √(0^2 + 27^2) = √(0 + 729) = √729 = 27
The argument θ is found using the formula:
θ = arctan(y/x) = arctan(27/0) = π/2
Therefore, 27i can be expressed in polar form as 27 * e^(iπ/2).
Step 2: Apply the cube root property.
The cube root property states that if w^3 = z, then w = ∛z, where ∛ denotes the principal cube root.
In our case, we have z^3 = 27i. Taking the principal cube root, we can write:
z = ∛(27i)
Step 3: Calculate the principal cube root.
To calculate the principal cube root of 27i, we can use De Moivre's theorem.
De Moivre's theorem states that if z = r * e^(iθ), then ∛z = ∛r * e^(iθ/3).
Applying this to our case, where z = 27 * e^(iπ/2), we can calculate the principal cube root as follows:
∛(27i) = ∛27 * e^((iπ/2)/3)
The principal cube root of 27 is 3, so we have:
∛(27i) = 3 * e^((iπ/2)/3)
Step 4: Convert back to rectangular form.
To convert the polar form to rectangular form, we can use Euler's formula, which states that: e^(ix) = cos(x) + i*sin(x).
Applying this formula to our case, we have:
∛(27i) = 3 * (cos((π/2)/3) + i*sin((π/2)/3))
Now, we simplify the expression:
∛(27i) = 3 * (cos(π/6) + i*sin(π/6))
= 3 * (√3/2 + i*(1/2))
= 3√3/2 + 3i/2
Therefore, the complex numbers z satisfying the equation z^3 = 27i are: 3√3/2 + 3i/2.
To find all complex numbers satisfying the equation z^3 = 27i, we can start by expressing 27i in its polar form.
Step 1: Write 27i in polar form
To do this, we note that the magnitude of 27i is the absolute value of 27 times the absolute value of i, which is 27*1 = 27.
The argument, or angle, of 27i is π/2 since i lies on the positive imaginary axis in the complex plane.
So, in polar form, 27i can be written as 27(cos(π/2) + isin(π/2)) or 27cis(π/2).
Step 2: Find the cube root of 27cis(π/2)
To find the cube root, we need to find all angles that result in the value of π/2 when cubed. The cube roots will have arguments of π/6, π/2, and 5π/6.
So the cube roots of 27cis(π/2) are:
z₁ = 3cis(π/6) = 3(cos(π/6) + isin(π/6)) = 3(√3/2 + 1/2i) = (3√3/2) + (3/2)i
z₂ = 3cis(π/2) = 3(cos(π/2) + isin(π/2)) = 3(0 + 1i) = 3i
z₃ = 3cis(5π/6) = 3(cos(5π/6) + isin(5π/6)) = 3(-√3/2 + 1/2i) = (-3√3/2) + (3/2)i
Therefore, the complex numbers that satisfy z^3 = 27i are (3√3/2) + (3/2)i, 3i, and (-3√3/2) + (3/2)i.
In comma-separated form, the complex numbers are:
(3√3/2) + (3/2)i, 3i, (-3√3/2) + (3/2)i