De Moivre's Theorem (-12-5i)^-3

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SUMMARY

Using De Moivre's Theorem, the expression (-12-5i)^-3 can be solved by first converting the complex number into polar form. The modulus r is calculated as 13, and the angle θ is determined using arctan(y/x), adjusted for the third quadrant. The final result is expressed as 1/2197 cis(8.241), confirming the application of the theorem in calculating powers of complex numbers.

PREREQUISITES
  • Understanding of complex numbers and their polar representation
  • Familiarity with De Moivre's Theorem
  • Knowledge of trigonometric functions, specifically cosine and sine
  • Ability to perform calculations involving the Argand diagram
NEXT STEPS
  • Study the derivation and applications of De Moivre's Theorem
  • Learn how to convert complex numbers from rectangular to polar form
  • Explore the properties of the Argand diagram in complex analysis
  • Practice solving complex number equations using polar coordinates
USEFUL FOR

Students studying complex analysis, mathematicians working with complex numbers, and educators teaching advanced algebra concepts.

Wardlaw
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Homework Statement



Using De Moivres Theorem, solve (-12-5i)^-3

Homework Equations





The Attempt at a Solution



The solution i get for this problem is different from the one given in the exercise text. This is 1/2197cis(8.241)

Note: cis is equivalent to cos(\Theta)+isin(\Theta)
 
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well you haven't actually written an equation down so there's nothing to solve.
i'm assuming you meant to rewrite it in polar form and then use de moivre

write -12-5i in polar form, that is z=r cis(theta)
where r will be 13 if my mental pythagoras is correct and theta is arctan(y/x)

then use de moivre (-12-5i)^(-3)=z^(-3)=r^(-3) cis(-3 theta)
 
latentcorpse said:
well you haven't actually written an equation down so there's nothing to solve.
i'm assuming you meant to rewrite it in polar form and then use de moivre

write -12-5i in polar form, that is z=r cis(theta)
where r will be 13 if my mental pythagoras is correct and theta is arctan(y/x)

then use de moivre (-12-5i)^(-3)=z^(-3)=r^(-3) cis(-3 theta)

Remember that in the Argand diagram,-12-5i lies in the 3rd quadrant. Thus

\theta = tan^{-1}(y/x) - \pi
 

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