**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!]

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

## 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.