Question 374909
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First remember the properties of exponents:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^{m\,\cdot\,n}\ =\ a^m^n]


So, rewrite your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \left(\left(\sqrt{2}\,-\,i\right)^2\right)^2]


Next, remember the pattern for squaring a binomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ -\ b)^2\ =\ a^2\ -\ 2ab\ +\ b^2]


And don't forget that *[tex \Large i^2\ =\ -1]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt{2}\,-\,i\right)^2\ =\ 2\ -\ 2i\sqrt{2}\ -\ 1\ =\ 1\ -\ 2i\sqrt{2}]


And so far we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \left(1\ -\ 2i\sqrt{2}\right)^2]


Now repeat the process:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ 1\ -\ 4i\sqrt{2}\ -\ 8\ =\ -7\ -\ 4i\sqrt{2}]


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