What Are Complex Numbers and How Can Beginners Start Learning About Them?

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Complex numbers can be approached by starting with the equation involving cosine and Euler's formula, specifically \(\cos z = \frac{1}{2}(e^{iz} + e^{-iz})\). To solve equations like \(\cos(z) = 2\), it is crucial to consider the periodic nature of cosine, which leads to multiple solutions. The complex logarithm is introduced to find solutions, with the principal value being \(\mathrm{Cos}^{-1}(2) = \imath \mathrm{Cosh}^{-1}(2) = \imath \mathrm{Ln}(2+\sqrt{3})\). The discussion emphasizes that all solutions can be expressed by adding integer multiples of \(2\pi\) to the principal value, reflecting the periodicity of the cosine function. Understanding these concepts is essential for beginners learning about complex numbers.
  • #31
MissP.25_5 said:
Since the instruction says to find all solutions, doesn't that mean ln have to be multivalued? Multivalued means k>0, right? k=0 would be the principle value, which is single valued, isn't it?
Can you check the attachment?
That attachment is fine.
It is a minor technicality. To find all solutions we can use a multivalued inverse, or we can use a single value inverse to generate all solutions. k=0 can be the principal value if it is set up that way

your last post is confusing
where did i^0 come from?
third line from the bottom should have 2k pi i
then it should in the next line become 2k pi when you multiply both sides by -i
 
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  • #32
lurflurf said:
That attachment is fine.
It is a minor technicality. To find all solutions we can use a multivalued inverse, or we can use a single value inverse to generate all solutions. k=0 can be the principal value if it is set up that way

your last post is confusing
where did i^0 come from?
third line from the bottom should have 2k pi i
then it should in the next line become 2k pi when you multiply both sides by -i

Sorry, it should be i*0, this is due to iargZ. And argZ here is 0.
 
  • #33
lurflurf said:
That attachment is fine.
It is a minor technicality. To find all solutions we can use a multivalued inverse, or we can use a single value inverse to generate all solutions. k=0 can be the principal value if it is set up that way

your last post is confusing
where did i^0 come from?
third line from the bottom should have 2k pi i
then it should in the next line become 2k pi when you multiply both sides by -i

So, is this okay now?
 

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  • #34
^Yes that looks good.
 
  • #35
lurflurf said:
^Yes that looks good.

Between the 2 terms, (regarding the final answer) the symbol is just + or is it +/- ? I mean, how to simplify it?
 
Last edited:
  • #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.
 
  • #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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