Comparing log(i^2) and 2*log(i)

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SUMMARY

The discussion confirms that log(i^2) and 2*log(i) yield different sets of values. Specifically, log(i^2) simplifies to i*pi*(1 + 2*n), while 2*log(i) simplifies to i*pi*(1 + 4*n). The calculations utilize the logarithmic identity log z = ln|z| + i*arg z, demonstrating the distinct periodicity in the imaginary components of the two expressions. This distinction is crucial for understanding complex logarithms.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with logarithmic identities, specifically log z = ln|z| + i*arg z
  • Knowledge of the argument function for complex numbers
  • Basic proficiency in manipulating complex logarithms
NEXT STEPS
  • Study the properties of complex logarithms in depth
  • Explore the implications of periodicity in complex functions
  • Learn about the argument function and its significance in complex analysis
  • Investigate the differences between principal and multi-valued logarithms
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Mathematics students, educators, and anyone studying complex analysis or logarithmic functions will benefit from this discussion.

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Homework Statement



Show log(i^2) and 2*log(i) have different sets of values.

Homework Equations



log z=ln|z|+i*arg z

The Attempt at a Solution



log(i^2) = ln|i^2| + i*arg(i^2)
= ln|-1| + i*(pi + 2*n*pi)
= 0 + i*pi*(1 + 2*n)
= i*pi*(1+2*n)

2*log(i) = 2*(ln|i| + i*arg(i))
= 2*(ln|1| + i*(pi/2 + 2*n*pi))
= 2*(0 + i*(pi/2 + 2*n*pi ))
= i*pi*(1+4*n)

If this isn't right, what did I miss ?
 
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It's correct.
 
Thank you very much.
 

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