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

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SUMMARY

The discussion focuses on solving the equation z3 = i using De Moivre's Theorem in polar form. The correct polar representation of the complex number i is R = 1 and θ = π/2, leading to the solutions z = CIS(π/6), CIS(5π/6), and CIS(-π/6). Participants emphasized the importance of accurately converting complex numbers to polar form before applying De Moivre's Theorem to find the cube roots. The modulus and argument were clarified as essential components in this process.

PREREQUISITES
  • Understanding of complex numbers and their polar form
  • Familiarity with De Moivre's Theorem
  • Knowledge of trigonometric functions and their relation to complex numbers
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the conversion of complex numbers to polar form
  • Learn how to apply De Moivre's Theorem for finding roots of complex numbers
  • Explore the geometric interpretation of complex numbers on the Argand plane
  • Investigate the properties of cis notation in complex analysis
USEFUL FOR

Students studying complex analysis, mathematicians interested in polar coordinates, and anyone looking to deepen their understanding of De Moivre's Theorem and its applications in solving complex equations.

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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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