Video Transcript
Determine the solution set of the
equation 𝑧 cubed equals four times root two minus root two 𝑖 in the set of complex
numbers, expressing the solutions in exponential form.
When we represent a number in
exponential form, we write it as 𝑧 equals 𝑟 times 𝑒 to the power of 𝑖𝜃, where
𝑟 is the modulus of the complex number and 𝜃 is its argument. And we can use De Moivre’s theorem
for roots to calculate solutions to our equation.
But before we do so, we’re going to
need to write our equation, 𝑧 cubed equals four times root two minus root two 𝑖,
in exponential form. Distributing the parentheses, and
we see we can rewrite this as four root two minus four root two 𝑖. And so we can represent this
complex number on an Argand diagram by the point whose Cartesian coordinates are
four root two, negative four root two.
The modulus of our complex number
then, let’s define that to be equal to 𝑟, is the length of the line segment that
joins this point to the origin. And so we can use the distance
formula or simply the formula for finding the modulus. And we get that 𝑟 is equal to the
square root of four root two squared plus negative four root two squared. Four root two squared becomes 16
times two, which is 32. So the modulus is the square root
of 32 plus 32, which is of course the square root of 64, which is eight. So we know the modulus of our
complex number to be equal to eight. But what about its argument?
The argument is the measure of the
angle that our line segment makes with the positive real axis, but measured in a
counterclockwise direction. Alternatively, we can define 𝜃 in
terms of the principal argument, that is, in the left-open right-closed interval
from negative 𝜋 to 𝜋, by considering the clockwise direction to be negative and
finding the measure of this angle here. And we could use the tangent ratio
to do so.
We know that the measure of the
line segment opposite 𝜃 is four root two units and the measure of the line segment
adjacent to it is also four root two units. But if we notice that the length of
these line segments is equal, we actually see we have an isosceles triangle. Specifically, it’s an isosceles
right-angled triangle. And we know that two of the angles
in an isosceles right-angled triangle are 45 degrees or 𝜋 by four radians.
Since we’re traveling in a
clockwise direction, we know that our argument must be negative 𝜋 by four
radians. And that’s great because we now can
write our complex number four root two minus root two 𝑖 as eight 𝑒 to the power of
negative 𝜋 by four 𝑖. And that, in turn, is really
helpful because we can now use De Moivre’s theorem for roots to find the solution
set of our equation.
De Moivre’s theorem for roots says
that for a complex number of the form 𝑟𝑒 to the 𝑖𝜃, its 𝑛th roots given in
exponential form are 𝑟 to the power of one over 𝑛 times 𝑒 to the power of 𝜃 plus
two 𝜋𝑘 over 𝑛𝑖 for values of 𝑘 from zero to 𝑛 minus one. Now, of course, we’re solving the
equation 𝑧 cubed equals eight 𝑒 to the power of negative 𝜋 over four 𝑖. And so we are going to let 𝑛 be
equal to three. And we’re essentially finding the
cube roots to this equation.
By applying De Moivre’s theorem, we
see that the general form of the roots are eight to the power of one-third times 𝑒
to the power of negative 𝜋 by four plus two 𝜋𝑘 over three 𝑖. And since 𝑘 takes values from zero
to 𝑛 minus one, our 𝑘 needs to take values from zero to three minus one, which is
two.
Now, in fact, eight to the power of
one-third is the cube root of eight. And so that’s two. And so this is the general form to
the roots of our equation. We’re now going to substitute 𝑘
equals zero, 𝑘 equals one, and 𝑘 equals two in. When 𝑘 is equal to zero, our root
is two 𝑒 to the power of negative 𝜋 by four plus zero over three 𝑖. But negative 𝜋 by four plus zero
over three is negative 𝜋 by four over three, which in turn is negative 𝜋 over
12. So the first root to our equation
is two 𝑒 to the power of negative 𝜋 over 12𝑖.
Next, we substitute 𝑘 equals one
in. And we get two 𝑒 to the power of
negative 𝜋 by four plus two 𝜋 over three 𝑖. The numerator of that expression
becomes seven 𝜋 over four. And seven 𝜋 over four divided by
three can be written as seven 𝜋 by 12. And so our second root is two 𝑒 to
the power of seven 𝜋 over 12𝑖.
Finally, we substitute 𝑘 equals
two in. And when we do, the two 𝜋𝑘 part
becomes two 𝜋 times two, which is four 𝜋. So we get two 𝑒 to the power of
negative 𝜋 over four plus four 𝜋 over three 𝑖. And then the numerator here
simplifies to 15𝜋 over four. 15𝜋 over four divided by three
becomes 15𝜋 over 12 or five 𝜋 over four.
Now, it’s quite general to
represent the argument in terms of the principal argument, where the principal
argument takes values from the left-open right-closed interval from negative 𝜋 to
𝜋. And to achieve this, we simply add
or subtract multiples of two 𝜋 to the given argument. We’re going to subtract two 𝜋 from
five 𝜋 over four. And that gives us negative three 𝜋
by four. And so we can say that our final
root is two 𝑒 to the power of negative three 𝜋 by four 𝑖.
Now, of course, this question asked
us to determine the solution set to the equation. So we’re going to represent our
answer slightly differently. And so the solution set to the
equation 𝑧 cubed equals four times root two minus root two 𝑖 is the set containing
the elements two 𝑒 to the power of negative 𝜋 by 12𝑖, two 𝑒 to the power of
seven 𝜋 by 12𝑖, and two 𝑒 to the power of negative three 𝜋 by four 𝑖.
