Solving Euler's Formula: 2^(1-i) Explained

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The discussion focuses on the mathematical expansion of the expression 2^(1-i) using Euler's formula. It is established that 2^(1-i) can be rewritten as exp(ln[2^(1-i)]), which leads to the conclusion that it expands to 2cos(ln2) - 2i(sin(ln2). This transformation utilizes properties of logarithms and complex exponentials, specifically applying Euler's formula to derive the cosine and sine components.

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Hey guys, I'm having trouble on understanding how:

2^(1-i) expands to 2cos(ln2)-2i(sin(ln2))

Thanks in advance!
 
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btbam91 said:
Hey guys, I'm having trouble on understanding how:

2^(1-i) expands to 2cos(ln2)-2i(sin(ln2))

Thanks in advance!

2^(1-i) = exp(ln[2^(1-i)]), go from there.
 

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