Quickest way to calculate argument of a complex number

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Discussion Overview

The discussion revolves around calculating the argument of the complex number $\displaystyle \pi e^{-\frac{3i\pi}{2}}$. It explores different approaches to determine the argument, including the principal value and general forms of the argument.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks for the quickest method to calculate the argument of $\displaystyle \pi e^{-\frac{3i\pi}{2}}$.
  • Another participant confirms the modulus of the complex number as $|\displaystyle \pi e^{-\frac{3i\pi}{2}}| = \pi$.
  • A different participant explains that the argument can take values of the form $-\dfrac{3\pi}{2} + 2k\pi$, where $k$ is an integer, and discusses how to find the principal value based on the chosen range.
  • The same participant suggests that for the principal value, selecting $k = 1$ yields $\dfrac{\pi}{2}$ as the argument.
  • A later reply expresses gratitude for the information provided.

Areas of Agreement / Disagreement

Participants present different aspects of calculating the argument, but there is no explicit consensus on the method or the preferred range for the principal value.

Contextual Notes

The discussion includes considerations of different definitions for the principal range of the argument, which may affect the choice of $k$.

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What's the quickest way to calculate the argument of $\displaystyle \pi e^{-\frac{3i\pi}{2}}$?
 
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To be sure, I know that $|\displaystyle \pi e^{-\frac{3i\pi}{2}}| = \pi.$
 
Guest said:
What's the quickest way to calculate the argument of $\displaystyle \pi e^{-\frac{3i\pi}{2}}$?
The argument of $\displaystyle \pi e^{-\frac{3i\pi}{2}}$ can take any value of the form $-\dfrac{3\pi}2 + 2k\pi$, where $k$ is an integer. If you want the principal value of the argument then you need to choose $k$ so as to get a value in the range $(-\pi,\pi]$ (or maybe $[0,2\pi)$, depending on which definition you are using for the principal range). In this example, you would want $k = 1$, giving the principal value of the argument as $\dfrac\pi2.$
 
Opalg said:
...
Thank you very much! :D
 

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