### Video Transcript

Determine the solution set of the
equation π§ squared equals two plus two root three π in the set of complex
numbers.

To solve this equation for π§,
weβre going to need to find the square root of both sides. Now, on the right-hand side of our
equation, we see we have a complex number. It has a real part of two and an
imaginary part of two root three. To find roots of complex numbers,
we know we can use De Moivreβs theorem. De Moivreβs theorem for roots
states that we can find the πth root of a complex number of the form π cos π plus
π sin π by calculating π to the power of one over π times cos of π plus two π
π over π plus π sin of π plus two ππ over π, where π takes integer values
from zero to π minus one. Now, in the case of our equation,
we said weβre going to solve by finding the square root of both sides. So weβre going to let π be equal
to two. This means π will be equal to
zero, and it will be equal to one.

But then we notice that the complex
number in our equation is written in rectangular form, rather than polar form. And so letβs sketch the complex
number two plus two root three π on an Argand diagram to help us convert it. Since the complex number has a real
part of two and an imaginary part of two root three, we represent it by a point
whose Cartesian coordinates are two, two root three. We know that π is the modulus, and
thatβs the length of the line segment that joins this point to the origin. We can use the distance formula to
find the modulus. That is, we find the square root of
the sum of the squares of the real and imaginary parts. So thatβs the square root of two
squared plus two root three squared. Two root three squared becomes four
times three, which is 12. So we get the square root of four
plus 12. Thatβs root 16, which is of course
equal to four. The length of the line segment is
four units, and so π, the modulus of our complex number, is four.

Next, we want to find the
argument. Now we know that the argument is
the angle this line segment makes with the positive real axis measured in a
counterclockwise direction. And so, for complex numbers plotted
in the first quadrant, we can use the tangent ratio. On this occasion, the side opposite
the included angle of π is two root three units, and the side adjacent to it is two
units. We divide both the numerator and
denominator of this fraction by two. And we get that tan π is equal to
the square root of three. Now, in fact, we know this
value. We know that tan of π by three
radians is equal to the square root of three. And so the argument π is π by
three. And so we have our complex number
written in polar form. Itβs four cos π by three plus π
sin of π by three.

Weβre now ready to apply De
Moivreβs theorem with π equals two. The modulus is four to the power of
one-half. But of course, four to the power of
one-half is the square root of four, so itβs just two. The general form of the argument to
our solution is π by three plus two ππ over two. To find the exact solutions, weβre
going to substitute π equals zero and π equals one into this expression. When π is equal to zero, we get
two cos of π by three plus zero over two plus π sin of π by three plus zero over
two. But of course, π by three plus
zero is just π by three. So our numerator is π by three,
and we now need to divide that by two.

When we do, we find the first
solution to our equation to be two times cos of π by six plus π sin of π by
six. But cos of π by six is root three
over two and sin of π by six is one-half, and so we can distribute these
parentheses. Two times the square root of three
over two is the square root of three. And two times one-half π is just π, and so
we have the first solution to our equation. To find the second, weβre going to
substitute π equals one into the general form. This time, we get an argument of π
by three plus two π over two. π by three plus two π becomes
seven π over three. And then when we divide that by
two, we get seven π by six. And so we have two times cos of
seven π by six plus π sin of seven π by six.

We evaluate cos of seven π by six,
and we get negative root three over two. Similarly, sin of seven π by six
is negative one-half. Finally, we distribute those
parentheses, and that gives us the second solution to our equation as negative root
three minus π. And so weβve used De Moivreβs
theorem to solve the equation. We write this in set notation as
shown. And so we see that the solution set
to our equation in the set of complex numbers is the set including the elements root
three plus π and negative root three minus π.