Let $a$, $b$, and $c$ be complex numbers with $a\neq0$. Show that the solutions of $az^2+bz+c=0$ are $z_1,z_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, just as they are in the case when $a$, $b$, and $c$ are real numbers.
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See page 101 of https://books.google.com/books?id=Z9z7iliyFD0C&printsec=frontcover&dq=gelfand+algebra&hl=en&sa=X&ved=0ahUKEwjKgNaKlejOAhVL2GMKHfo5A8kQ6AEIHjAA#v=onepage&q=quadratic&f=false – avs Aug 30 '16 at 03:12
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3@Analysis15 Do you know how to complete the square? That's the way how to prove the quadratic formula, and that is also the way how to solve quadratic equations with complex coefficients... – imranfat Aug 30 '16 at 03:25
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If $b^2-4ac=x+yi \in \mathbb{C}$, then
$$\sqrt{x+yi}=\pm \left( \sqrt{\frac{\sqrt{x^2+y^2}+x}{2}} +\frac{iy}{|y|}\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} \right)$$
Ng Chung Tak
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