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I can graph the parabola $y = x^2 + 1$. I see that it does not intersect the $x$-axis, and therefore it must have complex roots, namely $+i$ and $-i$. I can plot these roots on an Argand diagram at $(0, +1)$ and $(0, -1)$.

My questions is this: is there any intuition of how the graph of a function relates to the graph of its roots on the Argand diagram?

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up vote 7 down vote accepted

Yes, you can find the complex roots of a quadratic equation from its graph. I'll quote from a very nice book you might want to read, An Imaginary Tale: the Story of $\sqrt{-1}$ by Paul J. Nahin.

Say we have a quadratic polynomial $f(x)=ax^2+bx+c$. Suppose the discriminant $b^2-4ac$ is negative, so that there are two conjugate complex roots $p \pm i q$. Now, since that is the case, we can also write the quadratic polynomial as

$$f(x) = a(x-p-iq)(x-p+iq) = a[(x-p)^2+q^2] \; .$$

From this, we see that if $a>0$ the polynomial has a minimum in $p$ and if $a<0$ the polynomial has a maximum in $p$. The value of that extremum is in both cases $aq^2$. Now, from that extremum, we draw a vertical line up to $2 aq^2$ in the y-coordinate. From there, we draw a horizontal to the right until we cross the parabola, the point we arrive at has x-coordinate $p+q$ because

$$\begin{eqnarray}2aq^2 & = & a[(x-p)^2+q^2] \\ aq^2 & = & a(x-p)^2 \\ q & = & \pm(x-p)\end{eqnarray}$$

and since we already found $p$, we found the complex roots.

The book also describes how to do this for a cubic. It's a very nice book, definitely worth having on your bookshelf.

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I second the recommendation for Nahin; it's a nice one. –  J. M. Apr 30 '11 at 8:26
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