Writing 2^{1-i} using Euler's Formula

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SUMMARY

The discussion focuses on converting the expression 2^{1-i} into the form a + ib using Euler's formula. The solution involves recognizing that 2 can be expressed as e^{log(2)}, leading to the transformation 2^{1-i} = (e^{log(2)})^{1-i} = e^{log(2) * (1-i)}. This method effectively utilizes the properties of logarithms and exponentials to simplify the expression into the desired format.

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Homework Statement



Use Euler's formula to write [itex]2^{1-i}[/itex] in the form a+ib.

The Attempt at a Solution



I know this has to be so simple because I could do this easy if the question was to write [itex]e^{1-i}[/itex] in the form a+ib using Euler's. So what I do not understand is how does 1-i being raised on 2 rather than e change what I need to do to solve it. Thank you.
 
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2^(1-i)=(e^log(2))^(1-i)=e^(log(2)*(1-i)).
 

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