Video Transcript
Let π§ one equal π to the π
two π by three and π§ two equal π to the negative π two π by three be the
complex cubic roots of unity. 1) Evaluate π§ one squared. How does this compare with π§
two? 2) Evaluate π§ two squared. How does this compare with π§
one?
Notice how π§ one and π§ two
are the complex solutions to π§ cubed equals one, written in exponential
form. This means we can use De
Moivreβs theorem to evaluate π§ one squared. This says that, for a complex
number of the form π§ is equal to ππ to the ππ, π§ to the power of π is
equal to π to the power of π times π to the πππ. Remember, π is the modulus and
π is the argument. We can see that the modulus of
π§ one is simply one. And the argument of π§ one is
two π by three. So π§ one squared is one
squared times π to the π two π by three times two.
One squared is one. And two π by three multiplied
by two is four π by three. So we can see that π§ one
squared is equal to π to the π four π by three. The argument of π§ one squared
is outside of the range for the principal argument. So weβll subtract two π. And we find that the principal
argument of π§ one squared is negative two π by three. And we can now see that π§ one
squared is equal to π§ two.
Letβs repeat this process for
question two. Letβs begin by making a
prediction. We saw that π§ one squared is
equal to π§ two. So it might seem to follow that
π§ two squared will be equal to π§ one. But letβs check. Once again, the modulus of π§
two is one. But this time, its argument is
negative two π by three. Negative two π by three
multiplied by two is negative four π by three. Once again, the argument of π§
two squared is outside of the range for the principal argument.
This time, weβll add two
π. Remember, weβre allowed to add
or subtract any multiple of two π to achieve an argument thatβs within the
range for the principal argument. This time, the argument of π§
two squared is two π by three. And we now see that π§ two
squared is equal to π§ one as we predicted. So π§ one squared is equal to
π§ two. And π§ two squared is equal to
π§ one, where π§ one and π§ two are the complex cubic roots of unity.