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Consider the quadratic $y=2x^2+3x+23$, this has no real solutions, so doesn't intercept the $x$-axis, but still has a complex conjugate pair as solutions.

My question is: Do these complex solutions have any meaning graphically on an $(x,y)$ coordinate axis (not an argand diagram)?

I have heard that a Reimann surface may have something to do with this, but I am not too sure what that is, thanks.

Ted Shifrin
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Jamminermit
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  • In casual settings, I like to say that it intersects the $x$-axis "behind and in front of" the paper. – Arthur Sep 13 '19 at 07:04
  • I mean, do the solutions to a polynomial with no real solutions, for example the one above, have any significance to the sketch of the polynomial. So do the complex solutions correspond to a specific point on the polynomial when sketched on the (x,y) coordinate system? – Jamminermit Sep 13 '19 at 07:05
  • @Jamminermit, in this case, no. Since complex plane is two dimensional, if the root doesn't lie on the real line, then there's no "special" real point on the plot of the polynomial that has any significance in relation to the root. – Yuriy S Sep 13 '19 at 07:08
  • Take a look at this question I asked some time ago. – Jean Marie Sep 13 '19 at 07:10

2 Answers2

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If $z_0=u+iv$ and $ \overline{z_0}=u-iv$ are the solutions of the equation $ 2x^2+3x+23=0$, then the graphical representations on the real Cartesian plane are

$$(u,v)$$

and

$$(u,-v).$$

Fred
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  • I think this is complex cartesian plane consisting of a complex axis and real axis. Isn't it? – Aditya Jain Sep 13 '19 at 07:10
  • Ok, so when you plot (u,v) and (u, -v), will they lie on the graph? – Jamminermit Sep 13 '19 at 07:11
  • @Jamminermit, clearly not. Since your polynomial has no real roots, it lies wholly in the upper or the lower halfplane. So, both of those points can't even in principle lie on the curve – Yuriy S Sep 13 '19 at 07:24
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Since you say "not on an Argand diagram" I suspect you are asking about whether or not the complex solutions can be seen on a Cartesian graph of the function $y= 2x^2+ 3x+ 23$. Yes, but it is subtle!

Complete the square: $2x^2+ 3x+ 23= 2(x^2+ (3/2)x)+ 23= 2(x^2+ (3/2)x+ 9/16- 9/16)+ 23= 2(x+ 3/4)^2- 9/8+ 23= 2(x+ 3/4)^2+ 193/8$. Solving for x we get $x= -3/4\pm\frac{\sqrt{193}}{4}$.

And from $y= 2(x+ 3/4)+ 193/8$ we can see that the graph is a parabola with vertex at (-3/4, 193/8). The x coordinate of the vertex is the real part of the solution. The imaginary part is a little subtler. It is the square root of the of the y coordinate divided by the leading coefficient.

user247327
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  • Ok, I just saw that if a quadratic has no solutions, then it was not intersect the x-axis, but when you take the mirror image of the parabola in the horizontal line passing through the vertex, then your new parabola will intersect the x-axis at the complex roots. – Jamminermit Sep 16 '19 at 15:39