Download Practice with Polar Form and Roots

Math 4000: Modern Algebra and Geometry I
Spring 2014, Dr. Klipper
Extra Examples for Week 6
More Practice with Polar Form and Roots in C
In this handout, we give more examples of how to work with polar form and roots. The
textbook has some useful examples, but they provide fairly few details about some tricky
steps of the work. This handout will go more slowly through some similar examples. It
may help to review some trigonometry, especially about how to find angle measurements in
quadrants other than Quadrant I.
At the end of the handout, we mention some other facts about the roots of unity.
Polar Form Practice
Writing a complex number z 6= 0 in polar form is very useful for computing products and
roots. We start with z in a+bi form (rectangular form) and convert it to z = r(cos θ +i sin θ)
where r = |z| is the length of z and θ is the angle made with the positive x-axis. Note that
θ is only unique if you restrict it to [0, 2π) (we say θ is unique modulo 2π).
If a or b is zero, then z lies on an axis, and you can spot the angle fairly quickly from a
picture; the angle must be a multiple of π/2. When a, b 6= 0, I recommend the reference
√ angle
method. I’ll show the steps of the method demonstrated on the example of z = − 3 + i.
1. Make a picture with your number z =
a+bi plotted in the plane, and draw a right
triangle whose base is the x-axis. The
legs of this triangle are |a| and |b|, the hypotenuse is the length |z|. The acute angle
inside that triangle is the reference angle
which we call θR .
Example: On√ the right, we see the drawing of z =
q−√ 3 + i. Here,√the hypotenuse
is |z| = ( 3)2 + 12 = 4 = 2. Notice
how both θ and θR are pictured.
2. Use any of the inverse trigonometric functions to get θR . You can do arccos(|a|/|z|),
arcsin(|b|/|z|), or arctan(|b|/|a|). Sometimes the arctan choice is easier because it
doesn’t require computation of |z|, but you may find special values of trig functions
easier to recognize for arccos or arcsin.
◦
Example: Here, I have sin(θR ) = opp/hyp
also
√ = 1/2, so θR = π/6 (i.e. 30 ). I could’ve √
done θR = arccos(adj/hyp) = arccos( 3/2) or θR = arctan(opp/adj) = arctan(1/ 3)
instead to get the same answer.
Caution: Except for a few special values, we generally do not have simplified values of
θR memorized. If θR doesn’t equal one of the “special angles” like π/6, π/4, or π/3,
you’ll have to leave θR unsimplified as an inverse trig value.
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3. Get θ from θR by accounting for the quadrant.
Example: Here, our point is in Quadrant II, so θ and θR combine to form a half-circle
of π radians. Thus, we have θ = π − θR = π − π/6 = 5π/6.
√
For our example, the polar form is − 3 + i = 2(cos(5π/6) + i sin(5π/6)) .
Finding Roots
In class, we showed the following process to find all the nth roots of a√complex number
w. We’ll demonstrate it with a procedure to find the fifth roots of w = − 3 + i.
1. Get w in polar form r(cos(θ) + i sin(θ)). Based on DeMoivre’s Theorem, we can get
one of the roots z0 in polar form: its length is |z0 | = |w|1/n , and its polar angle is θ/n.
√
Example: In our previous work in this document, we found that − 3 + i has length 2
and angle 5π/6. Therefore, one fifth root is z0 = 21/5 (cos(π/6) + i sin(π/6)).
2. Compute the simplest nth root of unity, which we called ω:
2π
2π
+ i sin
ω = cos
n
n
(Basically, ω has length 1 and has angle 2π/n, so it represents a rotation of 1/nth of
the unit circle.)
Example: For fifth roots, ω uses the angle 2π/5. (Since that’s not a special angle, we
won’t be able to simplify the values of cos(2π/5) and sin(2π/5).)
3. To get the other nth roots, multiply z0 by the powers of ω from ω 0 up to ω n−1 . In
other words, keep adding 2π/n to the polar angle of z0 until you have n roots in total.
Example: Since z0 has angle π/6, the five angles for the fifth roots are
2π
π
2π
π
2π
π π 2π π
, +
, +2
, +3
, +4
6 6
5 6
5
6
5
6
5
(In simplified form, they are π/6, 17π/30, 29π/30, 41π/30, and 53π/30.)
All of these roots have the same length
21/5 . As the drawing shows, these roots all
lie on a circle of radius 21/5 about the origin,
and they are equally spaced apart from one
another.
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More Difficult Ways to get Exact Values
The algorithm for finding nth roots produces the answers in polar form. Sometimes, the
answers involve non-special angles, so we don’t usually have the exact values of these trig
functions memorized. I won’t require you to get exact values for these functions, but it can
sometimes be useful to get more exact values. (See Example 9 in the textbook, for instance.)
Half-angle solving:
Suppose that you want a value cos(θ) where θ is not a special angle, but 2θ is. In other
words, you have a convenient value for cos(2θ), and you want to use it to get a value for
cos(θ). The key is to exploit the double-angle identity
cos(2θ) = 2 cos2 (θ) − 1
Plug in the value of cos(2θ) that you know, and algebraically solve for cos2 (θ). You get two
possibilities (of opposite signs), but you should be able to tell which sign is appropriate by
looking at θ’s quadrant.
You can do something similar to get sin(θ) from sin(2θ)’s double-angle identity, but that’s
more complicated because
p that identity has both sin(θ) and cos(θ) in it. A better option is
to note that sin(θ) = ± 1 − cos2 (θ).
Example:
Let’s say θ = π/12; we’ll use
√ this method to find cos(π/12), given that we know the
value
of
cos(2θ)
=
cos(π/6)
=
3/2. When we substitute this
√
√ value in the identity, we get
2
2
3/2 = 2 cos (θ) − 1, which we solve to get cos (θ) = (2 + 3)/4. Taking square roots, we
obtain
s
p
√
√
2+ 3
2+ 3
=±
cos(π/12) = ±
4
2
Because θ is an acute angle (it’s less than π/2), its cosine should be positive, so we take the
+ sign.
p
Now, we use that information with sin(θ) = 1 − cos2 (θ) (only the positive square root
makes sense for acute angles) to get
v
p
u
√ ! s
√
√
u
2
+
3
2
−
3
2− 3
t
sin(π/12) = 1 −
=
=
4
4
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Variant of the above technique:
The above technique can sometimes be adapted to higher multiples of θ. For instance,
there is a formula for cos(4θ) in terms of only cos(θ) (see the Challenge Problems of HW 5).
If you know a convenient value for cos(4θ), then you can use it to solve for a value of cos(θ).
We will not go into more detail here.
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Powers of Roots of Unity
In our root-finding approach, we start with a “basic” root of unity ω and keep taking
powers of it repeatedly until we obtain all n roots of unity. This raises a good question: does
the same behavior happen if you take powers of any root of unity? In other words, if c ∈ C
is any of the nth roots of unity, does it follow that c0 , c1 , up to cn−1 give us the total set of
n roots of unity?
If c is a root of unity where the first n powers c0 through cn−1 are all different, and then
cn = 1 again, we call c a primitive root of unity. Thus, ω is a primitive root of unity, but
there may be other possibilities for primitive roots. Let’s try some concrete examples first.
• Let’s consider n = 5 and use c = ω 2 (so it has polar angle 4π/5 instead of 2π/5). Then
c0 is 1 (with polar angle 0), c1 has angle 4π/5, c2 has angle 8π/5, c3 has angle 12π/5
(which reduces back to 2π/5 when you treat this mod 2π), and c4 has angle 16π/5
(which reduces to 6π/5). Finally, c5 has angle 20π/5, which reduces to 0 mod 2π, so
c5 = 1 again.
Thus, ω 2 is a primitive root of unity here!
The picture on the right shows a bit of
what’s going on. There are dots for each
of the five primitive roots, and you see that
every dot is some power of c.
• Consider n = 6 and c = ω 4 . Hence, c0 = 1, c1 = ω 4 , and then c2 = ω 8 = (ω 6 )(ω 2 ) = ω 2 .
(Note how we use the fact that ω is a 6th root of unity here!) Continuing on this path,
you get the picture on the right.
Note, however, that you only get 3 powers
of c before you finally return to 1. Thus, c
is not a primitive root of unity: its powers
only account for 3 roots out of 6!
Your HW has you simplify powers of primitive roots, and if you look at some of the
Challenge Problems, you’ll get to figure out exactly which roots are primitive! (There’s
actually a lot of similarity between the situation here and units in Zn ...)
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