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If AAA and BBB are real numbers, and the following is true:
Aeiπ/3+Beiπ/6=2e−iπ/4\large{A e^{i \pi/3} + B e^{i \pi/6} = \sqrt{2} e^{-i \pi/4}}Aeiπ/3+Beiπ/6=2e−iπ/4
What is AB\large{\frac{A}{B}}BA?
Note: i=−1i = \sqrt{-1}i=−1
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