**Graphically Understanding Complex Roots**

If you have studied complex numbers then you’ll be familiar with the idea that many polynomials have complex roots. For example x^{2 } + 1 = 0 has the solution x = i and -i. We know that the solution to x^{2 } – 1 = 0 ( x = 1 and -1) gives the two x values at which the graph crosses the x axis, but what does a solution of x = i or -i represent graphically?

There’s a great post on Maths Fun Facts which looks at basic idea behind this and I’ll expand on this in a little more detail. This particular graphical method only works with quadratics:

**Step 1**

You have a quadratic graph with complex roots, say y = (x – 1)^{2} + 4. Written in this form we can see the minimum point of the graph is at (1,4) so it doesn’t cross the x axis.

Step 2

Reflect this graph downwards at the point of its vertex. We do this by transforming y = (x – 1)^{2} + 4 into y = -(x – 1)^{2} + 4

**Step 3**

We find the roots of this new equation using the quadratic formula or by rearranging – leaving the plus or minus sign in.

-(x – 1)^{2} + 4 = 0

(x-1) = ± 2

x = 1 ± 2

Plot a circle with centre (1,0) and radius of 2. This will touch both roots.

**Step 4**

We can now represent the complex roots of the initial equation by rotating the 2 real roots we’ve just found 90 degrees anti-clockwise, with the centre of rotation the centre of the circle.

The points B and C on the diagram are a representation of the complex roots (if we view the graph as representing the complex plane). The complex roots of the initial equation are therefore given by x = 1 ± 2i.

**General case**

It’s relatively straightforward to show algebraically what is happening:

If we take the 2 general equations:

1) y = (x-a)^{2 } + b

2) y = -(x-a)^{2 } + b (this is the reflection at the vertex of equation 1 )

(b > 0 ). Then the first equation will always have complex roots. The roots of both equations will be given by:

1) a ±i√b

2) a ±√b

So we can think of (2) as representing a circle of radius √b, centred at (a,0). Therefore multiplying √b by i has the effect of rotating the point (√b, 0 ) 90 degrees anti-clockwise around the point (a,0). Therefore the complex roots will be graphically represented by those points at the top and bottom of this circle. (a, √b) and (a, -√b)

**Graphically finding complex roots of a cubic**

There is also a way of graphically calculating the complex roots of a cubic with 1 real and 2 complex roots. This method is outlined with an algebraic explanation here

**Step 1**

We plot a cubic with 1 real and 2 complex roots, in this case y = x^{3} – 9x^{2} + 25x – 17.

**Step 2**

We find the line which goes through the real root (1,0) and which is also a tangent to the function.

**Step 3
**

If the x co-ordinate of the tangent intersection with the cubic is a and the gradient of the tangent is m, then the complex roots are a ± (√m)i. In this case the tangent x intersection is at 4 and the gradient of the tangent is 1, therefore the complex roots are 4 ± 1i.

[2nd last sentence edited – thanks for the comments below!]

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## 6 comments

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September 30, 2015 at 6:41 am

SavelyThanks for simple explanation!

May 10, 2016 at 8:32 pm

Doug KuhlmannI believe that for the complex roots of a cubic the slope of the tangent line is the square of of the imaginary part. So if the line were 3x+4, the complex roots would be 3+2i and 3-2i.

February 3, 2017 at 9:05 pm

MHIf p is the real root and the complex roots are a + bi and a – bi, the equation of the tangent line is y = b^2 (x – p). The tangent line intersects the graph of the function at x = a.

February 28, 2018 at 3:01 am

JohnThe the tangent intercept is represented by (h,k) and the Tangent line is represented by Slope Intercept form (y=mX+b) than the actual zero is ; h±(the square root of m)

March 2, 2018 at 2:40 am

dkuhlmannblogA more complete answer: All of the discussion so far (including my response of almost two years ago) assumes the leading coefficient is 1 which will guarantee the slope of the tangent line in question is positive. However if the cubic has leading coefficient of A (not equal to zero) the real part of the complex solutions remains the first coordinate of the intersection point but the imaginary parts are +/- the square root of m/A where m is the slope of the tangent line.

March 2, 2018 at 6:55 am

Ibmathsresources.comthanks for the replies – I have (belatedly!) updated.