Question 517383
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First perform *[tex \Large (3\ -\ 2i)^2] using FOIL:  *[tex \Large 9\ -\ 6i\ -\ 6i\ +\ (-1)4\ =\ 5\ -\ 12i] remembering that *[tex \Large i^2\ =\ -1]


Then do *[tex \Large (5\ -\ 12i)(3\ +\ 2i)\ =\ 15\ +\ 10i\ -\ 36i\ -\ (-1)24\ =\ 39\ -\ 26i]


Now you have:  *[tex \Large \frac{39\ -\ 26i}{5\ -\ i}]


To rationalize the denominator, multiply the whole fraction by 1 in the form of the conjugate of the denominator divided by itself.  Remember that if you have a number *[tex \Large a\ +\ bi], then its conjugate is  *[tex \Large a\ -\ bi] and that the product of two conjugates is the difference of two squares, so  *[tex \Large (a\ +\ bi)(a\ -\ bi)\ =\ a^2\ -\ (-1)b^2\ =\ a^2\ +\ b^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{39\ -\ 26i}{5\ -\ i}\right)\left(\frac{5\ +\ i}{5\ +\ i}\right)]


With careful arithmetic you will get an integer denominator that can be divided into each of the resulting coefficients in the numerator.  Remember that 8.5 is the same as 17/2.


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