Question 210457
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-2+4i}{\ \,6+3i}]


Multiply the fraction by 1 in the form of the conjugate of the denominator divided by itself:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-2+4i}{\ \,6+3i}\right)\left(\frac{6-3i}{6-3i}\right)]


Apply FOIL to the numerator binomials and use the fact that the product of two conjugates is the difference of two squares (and remember that *[tex \Large i^2 = -1]), then reduce to lowest terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-12+6i+24i+12}{\ \,36-(-9)}\ =\ \frac{30i}{45}\ =\ \frac{2i}{3}]


But standard form is *[tex \Large a + bi].  Since there is no real part, *[tex \Large a = 0].  Therefore the properly constituted answer is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0 + \left(\frac{2}{3}\right)i]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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