Solving z^3 = i using De Moivre's Theorem

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To solve the equation z^3 = i using De Moivre's Theorem, the first step is to express the complex number i in polar form, which is R cis θ. The modulus of i is 1, leading to r^3 = 1, so r = 1. The angle θ for i is π/2, resulting in the polar form of i as 1 cis (π/2). By applying De Moivre's Theorem, the cube roots of i are found to be at angles π/6, 5π/6, and -π/6.
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Homework Statement


solve below using de Moivre's theorem, in polar form
z^3 = i


Homework Equations


r CIS theta

Answer:CIS pi/6, CIS 5pi/6, CIS (-pi/6)

The Attempt at a Solution


r^3 = sqrt(0^2+1^2)
r^3 = sqrt (1)
r = sqrt (1)

no idea how to get the angle
 
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Mathysics said:

Homework Statement


solve below using de Moivre's theorem, in polar form
z^3 = i


Homework Equations


r CIS theta

Answer:CIS pi/6, CIS 5pi/6, CIS (-pi/6)

The Attempt at a Solution


r^3 = sqrt(0^2+1^2)
r^3 = sqrt (1)
r = sqrt (1)

no idea how to get the angle

It is not true that i = 1, which is essentially what you have done in solving the equation. 1 is simply the modulus of the complex number i.

Try writing the complex number i in polar form i.e. in the form R cis \theta, and then taking the cube root of both sides of the equation.
 

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