Video Transcript
Write a general form for the roots
on 𝑧 to the 𝑛th power equals one, giving your answer in polar form.
We begin by recalling that if 𝑧 is
an 𝑛th root of unity, then it satisfies the relation 𝑧 to the 𝑛th power equals
one. We use de Moivre’s theorem to help
us solve this equation for 𝑧, and that gives us a general form for the 𝑛th roots
of unity. Remember, de Moivre’s theorem for
roots states that for a complex number of the form 𝑟 multiplied by cos 𝜃 plus 𝑖
sin 𝜃, then the 𝑛th roots are given by 𝑟 to the power of one over 𝑛 multiplied
by cos of 𝜃 plus two 𝜋𝑘 over 𝑛 plus 𝑖 sin of 𝜃 plus two 𝜋𝑘 over 𝑛 for
values of 𝑘 equal to zero, one, two, and so on up to 𝑛 minus one.
Now, our equation is 𝑧 to the 𝑛th
power equals one, and so we’re going to begin by writing the number one in polar
form. The real part of one is one and the
imaginary part is zero. And so we can represent the number
one on an Argand diagram by the point whose Cartesian coordinates are one, zero. The modulus 𝑟 of this number is
the length of the line segment that joins this point to the origin, so 𝑟 is equal
to one. The argument 𝜃 is the measure of
the angle that this line segment makes with the positive real axis measured in a
counterclockwise direction. And we can see quite clearly from
the diagram that 𝜃 must be equal to zero. The number one can therefore be
written in polar or trigonometric form as one multiplied by cos of zero plus 𝑖 sin
of zero. And as a result, the original
equation in our question is as shown.
To solve this equation and find the
roots, we are going to raise both sides to the power of one over 𝑛. 𝑧 to the 𝑛th power to the power
of one over 𝑛 is simply 𝑧. And then, we are going to apply de
Moivre’s theorem for roots to the right-hand side. The modulus becomes one to the
power of one over 𝑛. And then 𝜃 plus two 𝜋𝑘 over 𝑛
becomes zero plus two 𝜋𝑘 over 𝑛, and our expression for 𝑧 is as shown. We can then simplify this, since
one to the power of one over 𝑛 is simply one and zero plus two 𝜋𝑘 over 𝑛 is just
two 𝜋𝑘 over 𝑛. And so we’ve solved the equation,
finding the roots of the possible values of 𝑧.
The general form of these roots is
cos two 𝜋𝑘 over 𝑛 plus 𝑖 sin two 𝜋𝑘 over 𝑛, where 𝑘 will take integer values
from zero through to 𝑛 minus one.
