Question 624561
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"I just subtracted 1 from both sides..." is your error.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (1\ +\ i)^5\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(1\ +\ i)^5\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ln(1\ +\ i)\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(1\ +\ i)\ =\ \frac{\ln(2)}{5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\frac{\ln(2)}{5}}\ =\ 1\ +\ i]


NOW you add -1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i\ =\ e^{\frac{\ln(2)}{5}}\ -\ 1]


Run the RHS through your calculator, convert to percent, and round to two places.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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