Video Transcript
Find the coordinates of the vertices of a regular pentagon centred at the origin with
one vertex at three, three.
Since we’re dealing with a pentagon, let’s see how we can link this with the fifth
root of a complex number. We know that, on an Argand diagram, the fifth roots of unity form a regular
pentagon. That pentagon is inscribed within a unit circle whose centre is the origin. And one of the vertices lies at the point whose cartesian coordinates are one,
zero. So let’s consider the cartesian plane to be an Argand diagram containing our regular
pentagon.
We can transform this pentagon to a regular pentagon centred at the origin with a
vertex at 𝑧 one by multiplying each of the fifth roots of unity by 𝑧 one. And that’s the equivalent to finding the fifth root of 𝑧 one to the power of
five.
Now we could use De Moivre’s theorem or simply recall that the fifth roots of unity
are one, 𝜔, 𝜔 squared, 𝜔 cubed, and 𝜔 to the power of four, where 𝜔 is 𝑒 to
the two 𝜋 by five 𝑖. Since one of the vertices of our pentagon lies at the point three, three, which
represents the complex number three plus three 𝑖, we can say that 𝑧 one is equal
to three plus three 𝑖.
Now we know that if 𝑧 one is a root to the equation 𝑧 𝑛 minus 𝑤 equals zero and
one 𝜔 all the way through to 𝜔 to the power of 𝑛 minus one are the 𝑛th roots of
unity, then the roots of 𝑧 to the power of 𝑛 minus 𝑤 equals zero are 𝑧 one, 𝑧
one times 𝜔, 𝑧 one times 𝜔 squared, all the way through to 𝑧 one times 𝜔 to the
power of 𝑛 minus one. So we can find the coordinates of all the vertices of our regular pentagon by
multiplying our 𝑧 one by the fifth roots of unity.
Before we do that though, we’ll need to write it in exponential form. The modulus of 𝑧 one is the square root of the sum of the squares of the real and
imaginary parts. That’s the square root of three squared plus three squared, which is three root
two. And since both its real and imaginary parts are positive, we know it lies in the
first quadrant. So its argument is the arctan of three divided by three, which is 𝜋 by four. And we can say that 𝑧 one is equal to three root two 𝑒 to the 𝜋 by four 𝑖.
The rest of the roots and therefore the other vertices of our pentagon will be given
by 𝑧 one times 𝜔, 𝑧 one times 𝜔 squared, and all the way through to 𝑧 one times
𝜔 to the power of four. To find 𝑧 one times 𝜔, it’s three root two 𝑒 to the 𝜋 by four 𝑖 times 𝑒 to the
two 𝜋 by five 𝑖. And remember, to multiply complex numbers in exponential form, we multiply their
moduli and then we add their arguments. This means our second root is three root two 𝑒 to the 13𝜋 by 20𝑖.
Now since we’re trying to find the coordinates, we’ll need to represent this in
algebraic form. And to convert from exponential to algebraic form, we first convert it to polar
form. That’s three root two times cos 13𝜋 by 20 plus 𝑖 sin of 13𝜋 by 20. Distributing the parentheses, we see this is the same as three root two cos of 13𝜋
by 20 plus three root two 𝑖 sin of 13𝜋 by 20. And we can see that the second vertex of our pentagon will lie at the point with
cartesian coordinates three root two cos of 13𝜋 by 20, three root two sin of 13𝜋
by 20.
We repeat this process with the third vertex. And we subtract two 𝜋 from the argument so that we can express the argument within
the range for the principal argument. And we see that the third solution is three root two 𝑒 to the negative 19 over
20𝜋𝑖. Once again, representing this in polar form and distributing the parentheses, we find
the coordinates here are three root two cos of negative 19𝜋 by 20, three root two
sin of negative 19𝜋 by 20.
We can repeat this process for 𝑧 one times 𝜔 cubed and 𝑧 one times 𝜔 to the power
of four. And we find the vertices of the pentagon to lie at the point whose cartesian
coordinates are three, three; three root two cos 13𝜋 by 20, three root two sin 13
𝜋 by 20, three root two cos of negative 19 𝜋 by 20, three root two sin of negative
19𝜋 by 20. We have three root two cos of negative 11𝜋 by 20, three root two sin of negative
11𝜋 by 20, and three root two cos of negative three 𝜋 by 20, three root two sin of
negative three 𝜋 by 20.
