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**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!]

Essential resources for IB students:

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Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.