Question 460576
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3\ -\ 4i)^2]


does indeed equal


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -7\ -\ 24i]


Hence, *[tex \LARGE 3\ -\ 4i] is NOT a square root of *[tex \LARGE 7\ -\ 24i]


Use the formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{a\ +\ bi}\ =\ \sqrt{\frac{\sqrt{a^2\,+\,b^2}\ +\ a}{2}\ \pm\ i\frac{\sqrt{a^2\,+\,b^2}\ - a}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{7\ +\ 24i}\ =\ \sqrt{\frac{\sqrt{7^2\,+\,24^2}\ +\ 7}{2}\ \pm\ i\frac{\sqrt{7^2\,+\,24^2}\ - 7}{2}]


The only thing left is a little arithmetic.


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