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

Thread starter btbam91
Start date Sep 16, 2010
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The discussion centers on the mathematical expansion of 2^(1-i) into the form 2cos(ln2) - 2i(sin(ln2). Participants suggest starting with the exponential form, using the property exp(ln[a^b]) = a^b. The key steps involve applying Euler's formula, which connects complex exponentials to trigonometric functions. Understanding the natural logarithm of 2 and its implications in the expansion is crucial. The conversation emphasizes the importance of these mathematical principles in solving the equation.

Sep 16, 2010

btbam91

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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Sep 16, 2010

logarithmic

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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