Video Transcript
What is the general form for the
seventh roots of unity in polar form?
In this question, we are being
asked to find the complex numbers given in the polar form 𝑧 equals 𝑟 cos 𝜃 plus
𝑖 sin 𝜃 that satisfy the equation 𝑧 to the power of seven equals one. Before doing any calculations, we
should note that there is a formula for which the 𝑛th roots of unity are defined
for any 𝑛. While it is possible to use this
formula to answer the question directly, we will instead take a more general
approach and answer this question using de Moivre’s theorem.
We recall that de Moivre’s theorem
for roots states that for a complex number given in polar form, the 𝑛th roots are
given by 𝑟 to the power of one over 𝑛 times cos of 𝜃 plus two 𝜋𝑘 over 𝑛 plus
𝑖 sin of 𝜃 plus two 𝜋𝑘 over 𝑛. And this is for values of 𝑘 that
go from zero up to 𝑛 minus one. In other words, de Moivre’s theorem
gives us the explicit formula for the roots of a complex number given in polar
form. And in the case of 𝑛 equals seven,
we have seven roots.
Since we are looking for complex
numbers that satisfy 𝑧 to the power of seven equals one, we can see that this is
the same thing as 𝑧 equals one to the power of one-seventh. Therefore, if we treat one as a
complex number, we can find its seventh roots using this theorem. This means we’ll need to write one
in polar form.
To do this, we consider its
position on an Argand diagram. The real part of one is one, and
the imaginary part is zero, so its position on an Argand diagram is represented by
the point one, zero. Its modulus, 𝑟, is the distance of
this point from the origin, which we can see must be one. Meanwhile, its argument, 𝜃, is the
measure of this angle that the line segment joining this point to the origin makes
with the positive real axis when measured counterclockwise. We can see that this is zero. That is to say we can write one in
polar form as one times cos of zero plus 𝑖 sin of zero.
We now want to apply de Moivre’s
theorem by considering the seventh roots of one. We do this by substituting 𝑟
equals one, 𝜃 equals zero, and 𝑛 equals seven into the right-hand side of the
underlined equation. If we do this, we get one to the
power of one-seventh times cos of zero plus two 𝜋𝑘 over seven plus 𝑖 sin of zero
plus two 𝜋𝑘 over seven. This can be simplified, since one
to the power of one-seventh is just one and multiplying by one leaves the expression
unchanged, while the zero inside the cos and sin terms will vanish, leaving us with
just two 𝜋𝑘 over seven.
Therefore, our final answer will be
cos of two 𝜋𝑘 over seven plus 𝑖 sin of two 𝜋𝑘 over seven, for values of 𝑘
between zero and six.
