Question 1120650
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ +\ bi)^2\ =\ i]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ +\ bi)(a\ +\ bi)\ =\ 0\ +\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ +\ 2abi\ -\ b^2\ =\ 0\ +\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a^2\ -\ b^2)\ +\ 2abi\ =\ 0\ +\ i]


Since *[tex \Large z_1\ =\ z_2] if and only if *[tex \Large \text{Re}\{z_1\}\ =\ \text{Re}\{z_2\}] and *[tex \Large \text{Im}\{z_1\}\ =\ \text{Im}\{z_2}\]


It must be true that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2\ =\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2ab\ =\ 1]


So *[tex \Large a\ =\ b] and then *[tex \Large a^2\ =\ \frac{1}{2}]


Hence *[tex \Large a\ =\ b\ =\ \pm\frac{\sqrt{2}}{2}]


The positive square root of *[tex \Large i] is then *[tex \Large \frac{\sqrt{2}}{2}\ +\ \frac{\sqrt{2}}{2}i]


Check the answer:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(\frac{\sqrt{2}}{2}\ +\ \frac{\sqrt{2}}{2}i\)^2\ =^?]


The arithmetic is left as an exercise for the student.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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