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Mathematics Teaching Self-evaluation Tools

You are viewing a limited version of the NCETMâ€™s self-evaluation tools. Any answers you save during this session will be removed after seven days. Log in or sign up to view, and use, the full version of the tools.

Here you can see a summary of the areas in which you are confident and those in which you are less confident; there are some ideas and suggestions which may help you in your professional learning.

Showing all next steps for the selected topic.Click on a question to show more information.

you understand the language of complex numbers?(show/hide all)

What this might look like in the classroom

Simplify:

a. b. c.

Solutions:

a.

b.

c.

Taking this mathematics further

Thinking back over the sorts of mathematics one can carry out with “regular” numbers – and wondering how these operations could be carried out with complex numbers.

Making connections

The key here is the completeness that complex numbers allows the mathematician. It is often helpful to trace through a history of number in order to strengthen the need for imaginary numbers.

Counting starts with integers and all addition is possible, multiplication too is possible with just integers. But the two inverses – subtraction and division give a need for some different sorts of numbers. Subtraction draws us to negative numbers and division draws us to fractions.

In the same way there is something quite ugly about not being able to square root a negative number – and thus there is a sensible argument for creating a new form of number to enable these sorts of square roots.

Later study shows that complex numbers are complete – that the four operations, and powers and roots performed on complex numbers yield a complex answer and do not lead to any further “can’t” situations.

you understand how to solve any quadratic equation?(show/hide all)

What this might look like in the classroom

Solve the following equations:

a.

b.

c.

Solutions:

a.

b.

c.

It is now possible to solve any random quadratic – there is no longer any need to avoid equations with no real roots.

So dice, or other random number generators, could be used to generate the coefficients in a quadratic equation to enable practice.

Taking this mathematics further

It is worth considering, as an introduction to later work, that if a quadratic can be solved then the product of two quadratics – a quartic equation – can also be solved.

It would be necessary to find a way of splitting the quartic equation into two quadratics.

Making connections

It would help students to look back at their previous study – when such equations were labelled as having “no real roots” and considering how such equations can now be solved.

Students should be able to appreciate why “no real roots” now makes good sense.

you can recognise that a complex number is zero if both real and imaginary parts are zero?(show/hide all)

What this might look like in the classroom

This is a very simple statement and should be easily understood.

More complicated equations can be set up – in the form of .

Looking at just the real parts gives and then the imaginary parts gives (Note that when writing these imaginary parts it is usual to omit the from the equations).

Taking this mathematics further

Making connections

This is much like comparing coefficients – students will have used this concept in earlier mathematics to help them solve equations.

The idea of being able to look at part of an equation and compare with another part of the same equation is a valuable one.

you can recognise that solutions to polynomial equations occur in conjugate pairs?(show/hide all)

What this might look like in the classroom

Read the next section to give an example of using this idea in practice.

Taking this mathematics further

This concept is a short-cut method to finding a second root – given one complex root then its conjugate must also be a root. Combining this knowledge with the factor theorem allows a method of solving a cubic equation given a complex root to begin with.

Making connections

Related information and resources

from the NCETMfrom other sites

Read here about the fundamental theorem of algebra.

This link explains how to solve quadratic equations with complex roots.

you can solve equations with real coefficients, of cubic or higher?(show/hide all)

What this might look like in the classroom

Two of the roots of a quartic equation are and .

Hence find the remaining roots and construct the original equation.

Solution:
If is a root then we know that its conjugate, will also be a root.

This gives two factors of and , similarly gives and hence also and .

Multiplying the first pair gives

and the second pair gives

.

So the quartic is found by multiplying these two quadratic expressions:

.

So the equation was .

Taking this mathematics further

For a quartic equation the above process can be repeated, assuming there are two real roots to be found. When all four roots are complex then a method of solution is rather harder.

So far the student has had a general method of solving a quadratic equation – and complex numbers have allowed the solution of all possible quadratic equations.

Now we are able to find a method of solving any polynomial equation.

Let us consider a cubic equation. We know that it has at least one real root.

The Factor theorem can often be used to find a real factor, which is then turned into a root.

The cubic can then be divided by this root to give a quadratic, which is now always solvable.

Example:

Solve

Let this be written as and then we are intending to solve .

Since then we know that is a root and thus is a factor.

We can divide by to produce and thus we can solve giving .

Finally we can write that the solutions to are and the conjugate pair .

you can represent a complex number in modulus-argument form?(show/hide all)

What this might look like in the classroom

Revision of right angled trigonometry, and Pythagoras should help the connection between Cartesian and modulus-argument form become clear.

Taking this mathematics further

With some practice it should be possible to quickly move from the (Cartesian) form of a complex number to the modulus-argument form – without needing to consider Pythagoras or trigonometry.

Making connections

When considering the modulus-argument form it is worth considering what loci would be formed if the modulus was fixed, but the argument allowed to vary.

Alternatively, a fixed argument and a varying modulus. Although this is A2 work, this discussion would help cement the concepts of modulus and argument.

Related information and resources

from the NCETMfrom other sites

These links explain how to use the modulus-argument form:

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