Suppose for $z, a, b, c \in \mathbb{C}$ that we have $az^2+bz+c=0,$ how do you define solutions in the complex plane? Can you even use the quadratic formula at all for the non-real case?
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4Does this answer your question? How to find the solution of a quadratic equation with complex coefficients? – Ryan Shesler Apr 30 '20 at 03:39
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1Why not? Many quadratic Equations have complex roots – Anand Apr 30 '20 at 03:40
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It doesn't quite because that problem seems to assume a real variable, whereas I have a complex variable with any coefficient. Basically I want to know when a solution is valid, I don't want to be bogged down having to analyze 50 different branch technicalities. – AskingRandomQuestions Apr 30 '20 at 03:43
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https://math.stackexchange.com/questions/928516/quadratic-formula-for-complex-variable-with-real-coefficients/928553#928553 – Ryan Shesler Apr 30 '20 at 03:48
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Both of those are half-right. One of them says real variable complex coefficients, the other says complex variable real coefficients. I'm looking for complex variable with complex coefficients. – AskingRandomQuestions Apr 30 '20 at 05:21
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Yes, it works the same way it does in the "real case". If the discriminant, $b^2-4ac\ne0$, then there are two complex roots. If the discriminant is zero, then there is one complex root.
At any rate, the formula provides the two solutions guaranteed by the fundamental theorem of algebra.
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If one starts with a quadratic having 2 real roots (and positive discriminant), then multiplies its coefficients all by $i,$ the discriminant gets its sign changed. But still the roots are real. – coffeemath Apr 30 '20 at 04:01
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So the fact that the square root function is a completely different function in the complex plane than in the real plane is somehow irrelevant? What's the point of complex analysis then? – AskingRandomQuestions Apr 30 '20 at 04:04
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Sorry again. My statement about the discriminant was incorrect. Of course we can't compare $b^2-4ac$ and $0$ in general. – Apr 30 '20 at 04:41