Evaluating z^{60} for z = -1 + i√3

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Homework Help Overview

The discussion revolves around evaluating \( z^{60} \) for \( z = -1 + i\sqrt{3} \), focusing on the application of polar coordinates and complex exponentiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express \( z \) in polar form and calculates \( z^{60} \) based on that representation. Some participants question the accuracy of the result obtained from Maple, while others consider the potential for rounding errors in the computation.

Discussion Status

The discussion is ongoing, with participants exploring the discrepancies between the manual calculation and the output from Maple. There is a recognition of the challenges posed by large numbers and the possibility of errors in numerical methods.

Contextual Notes

Participants note the use of a limited number of significant digits in Maple's calculations, which may affect the results. There is also an implicit assumption regarding the methods used by Maple for evaluating complex powers.

nicksauce
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Homework Statement


Evaluate z^{60}, where z = -1 + i\sqrt{3}

Homework Equations


The Attempt at a Solution


z = 2(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})
z^{60} = 2^{60}(\cos(40\pi) + i\sin(40\pi))
z^{60} = 2^{60}

Is this right? Maple doesn't seem to agree.
 
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Unless I'm misremembering the various formulas involved, that seems correct. What is Maple giving as the answer?
 
z := -1 + I sqrt(3)
a := z^60
evalf(a, 5)
1.1544 10^18 - 8.5175 10^14 I

Perhaps the imaginary part is just due to some kind of rounding error?
 
Probably so. After all, you're dealing with some pretty big numbers and using, I'm guessing, 5 digits. Bound to introduce a pretty big error.
 
I'm quite surprised maple wouldn't use exact methods, such as I used here, to calculate such numbers!
 

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