Exploring the World of Complex Numbers: Beginners Guide

In summary: If t is any solutions and k is an integer$$2k\, \pi\pm t$$is all the solutions because$$\cos(2k\, \pi\pm t)=\cos(t)$$
  • #36
^Which 2? We need +/- either in front of log or between 2 and √3. We do not need it with 2k π unless we require k to not be negative.
 
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  • #37
lurflurf said:
^Which 2? We need +/- either in front of log or between 2 and √3. We do not need it with 2k π unless we require k to not be negative.

Okay, I got it now! Thank you so much for being patient with me.
 

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