Question 1123447
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  z^2\ +\ (2i\ -\ 3)z\ +\ 5\ -\ i\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ -3\ +\ 2i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 5\ -\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{-(-3\,+\,2i)\ \pm\ \sqrt{(-3\,+\,2i)^2\ -\ (4)(1)(5\,-\,i)}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{3\,-\,2i\ \pm\ \sqrt{5\,-\,12i\,-\,(20\,-\,4i)}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{3\,-\,2i\ \pm\ \sqrt{-15\,-\,8i}}{2}]


As an interim step, we need *[tex \Large \sqrt{-15\,-\,8i}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{-15\,-\8i}\ =\ a\ +\ bi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -15\ -\ 8i\ =\ a^2\ +\ 2abi\ -\ b^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^2\ -\ b^2\ =\ -15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  b\ =\ \frac{-8}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^2\ -\ \frac{64}{4a^2}\ =\ -15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^4\ +\ 15a^2\ -\ 16\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (a^2\ +\ 16)(a^2\ -\ 1)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a\ =\ \pm\ 1\ \Right\ b\ =\ \mp\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a\ =\ \pm\ 4i\ \Right\ b\ =\ \mp\ 1i]


Either way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sqrt{-15\,-\,8}\ =\ \pm(1\ -\ 4i)]


Using *[tex \Large 1\ -\ 4i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{3\,-\,2i\ \pm\ 1\,-\,4i}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_1\ =\ 2\ -\ 3i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z_2\ =\ 1\ +\ i]


You get the same result if you use *[tex \Large -1\ +\ 4i].  Verification 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
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

</font>