Question 399134
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{-32}\ =\ \sqrt{-1}\sqrt{32}\ =\ i\sqrt{32}\ =\ i\sqrt{16}\sqrt{2}\ =\ 4i\sqrt{2}]


Check your work:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4i\sqrt{2})^2\ =\ 4^2\,\cdot\,i^2\,\cdot\,\sqrt{2}^2\ =\ 16\,\cdot\,(-1)\,\cdot\,2\ =\ -32]



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