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Let $a,b,c\in\mathbb C\setminus\{0\}$ be complex numbers such that $$b^2-4ac \neq 0.$$ We consider the equation $$ax^2+bx+c=0.$$ I am interested in general statements about the roots of this equation from the geometric point of view and how they are located in the complex plane. In particular, I seak sufficient conditions on $a,b,c$ to obtain roots whose imaginary parts have opposite signs. For instance, if $a=1,\mathrm{Im}b>0,c=1,$ then this is true. Indeed,if $x_1,x_2$ are the two solutions of $$x^2+bx+1,$$ then $$x_1x_2=1$$ and thus $$\mathrm{Im}x_1=\mathrm{Im}\frac{1}{x_2}.$$

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The conjecture about imaginary parts is certainly false, for the equation $$ P(x) = (x - i) (x - (1+i)) $$ has roots $i$ and $1 + i$, whose imaginary parts have the same sign.

In general, there are two roots, and just as in the real case, the sum of the roots is $-\frac{b}{2a}$, while the product of the roots is $c/a$. Other than that, there's not a lot to say.

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A general quadratic polynomial over $\mathbb C$ has the form $a(z-w_1)(z-w_2)$, where the roots $w_1, w_2 \in \mathbb C$ are arbitrary. So, there isn't much more to say about their nature and location.

The restriction $b^2-4ac \neq 0$ says that $w_1 \ne w_2$.

The restriction $b \neq 0$ says that $w_1 \ne -w_2$.

The restriction $c \neq 0$ says that $w_1 \ne 0$ and $w_2 \ne 0$.

The roots have imaginary parts of opposite signs iff $b/a$ is real.

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It's not true that the imaginary parts of the two roots will be conjugates of each other. In fact, for real numbers $a$ and $b$, $$(z - (a+bi))(z-(a-bi)) = z^2-2az+(a^2+b^2)$$ So a quadratic polynomial has this property if and only if it has real coefficients. In fact, you can make the roots of a complex quadratic polynomial anything you want; just pick two complex numbers $z_1$ and $z_2$. Then the polynomial $$(z-z_1)(z-z_2)$$ will have $z_1$ and $z_2$ as roots.

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