Powers and Roots

To find the roots of a complex number we use the above rule but we need to consider the periodicity of the trig

functions. The statement

but the trig functions are periodic with period . So

and therefore

is another root. And so on.

So when do we stop?

The above statements show that we can continue to get roots for every

where

is a positive integer. However, once reaches then we are at

which is the same as our first root. So, to summarize, the

th complex roots of are found at the angles

(*)

where . Notice that this gives use roots.

Example: Find the three cube roots of -8.

Solution: We write -8 in polar form as

By letting

in (*) the first root we get is

Then by letting in (*) we get a root of

Then by letting we get a root of

then the others are spread out with arguments

apart on a circle of radius 2.

It's easy to see that, in general, the

roots of any

complex number will all have the same magnitude

and will be situated on a circle centered at 0. They

will be equally spaced around the circle so that they

form the vertices of an

-side polygon.

which satisfies and is a line segment.

Find its length.

A solution to this problem is posted below.

Green Question: (Needham VCA) (modified) Given two starting numbers , let us build up an infinite sequence

with this rule: each new number is twice the difference of the previous two. For example, if and

we obtain: Our aim is to find a formula for the th number .

(i) Our generating rule can be written succinctly as

and show that if and are arbitrary complex numbers,

to get and then plug both

into (*) to get

.

(iii) If we want only real solutions of the recurrence relation, show that

by using the Binomial Theorem. Then

deduce from this result that

we must have . Do

this by recognizing that

is a constant and therefore you may use two of the terms in the series to find the real and

imaginary parts of

. Then, rewriting this in polar form, deduce that

(v) Check that this formula predicts and use a calculator to verify this.

[Note that this method can be applied to any recurrence relation of the form .]

STEP 2000, Math II, #4: Prove that

and that, for every positive integer

,

By considering or otherwise, prove that

Prove also that

is another notation for . ]

Roots of unity

The

then

Therefore the entire list of roots is

where

We can write the full factorization of the equation

as

Using Synthetic division to factor out

we get

Then plugging in

we get

(***)

So that the sum of the

th roots of unity is always 0.

2010, Math III, #3: For any given positive integer

th root of unity if and there is no integer such that and . Write down

the two primitive 4th roots of unity.

is one and the equation has no repeated roots. Show that .

, and such that .

STEP 2013, Math III, #4: Show that .

Write down the th roots of -1 in the form , where , and deduce that

show that, when is even,

(ii) Show that, when

is odd,

You may use without proof the fact that

.

In the previous section on cubics we were able to use clever algebraic tricks to find the real solution to any cubic.

Now we can find all three solutions for any cubic.

Our solution of the cubic will proceed in three steps:

!. Show that all cubics can be transformed into new cubics with no squared term.

2. Find a single solution to the transformed cubic.

3. Generate all three solutions using roots of unity.

Step 1.

A general cubic can be written as

Now use the substitution

to get

For our purposes we only need notice that the squared term is gone. So without loss of generality we will solve a

cubic without a squared term. The form we will use is

(*)

Step 2.

We will make the "inspired guess" that

and are complex numbers.

Substitute into (*) to get

and regroup to get

and so that

and

Combining these we get

This equation is quadratic in so

Since

and

we get

so our general solution becomes

Here is one of them. We multiply out using the Binomial Theorem and make use of the fact that .

and we're done.

is a root of (*).

You can similarly show that

as defined above is a root of (*). The three roots of

are

and are real, then

think about the three cases

What kind of roots does the cubic have in each case? Think graphically as well about the three possibilities.