Question 766224
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z^2\ -\ (3\ +\ 2i)z\ +\ (1\ +\ 3i)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{1,2}\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


where *[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -(3\,+\,2i)\ \ ], and *[tex \LARGE c\ =\ (1\,+\,3i)]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{1,2}\ =\ \frac{3\,+\,2i\ \pm\ \sqrt{(3\,+\,2i)^2\ -\ 4(1)(1\,+\,3i)}}{2(1)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{1,2}\ =\ \frac{3\,+\,2i\ \pm\ \sqrt{9\,+\,12i\,-\,4\ -\ 4\,-\,12i}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_{1,2}\ =\ \frac{3\,+\,2i\ \pm\ \sqrt{1}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_1\ =\ 1\ +\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_2\ =\ 2\ +\ i]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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