### Video Transcript

1) Find the roots of π§ to the power of eight plus 16 equals zero. 2) The complex numbers representing the roots of π§ to the power of eight plus 16
equals zero are each squared to form the vertices of a new shape. What is the area of the shape?

Letβs begin with part one. To solve this equation, weβll subtract 16 from both sides to get π§ to the power of
eight equals negative 16. And then weβll find the eighth roots of both sides. But to apply De Moivreβs theorem for roots, negative 16 is going to need to be
expressed in exponential or polar form.

Letβs write it in exponential form. Its modulus is 16. And since itβs a purely real number, which would lie on the negative real axis of an
Argand diagram, its argument is π radians. And therefore, the solutions to our equation are the eighth roots of 16π to the
ππ. Applying De Moivreβs theorem and we see that the roots are given by 16 to the power
of one-eighth times π to the π plus two ππ over eight π, where π takes values
from zero through to seven.

So our roots whose arguments are expressed in the range for the principal argument
are root two π to the π by eight π, root two π to the three π by eight π, all
the way through to root two π to the negative π by eight π. And since these are the eighth roots of a complex number, it follows that they will
form the vertices of a regular octagon inscribed in a circle whose radius is root
two and centre is the origin.

And what about part two? Well, to square a complex number in exponential form, we square its modulus and we
double its argument. Notice how our eight roots have diminished to just four. These roots represent the vertices of a square this time, inscribed in a circle
radius two units. Its area can be found by using the Pythagorean theorem. And that will generate a side length of two root two units. So its area is two root two squared. Itβs eight square units.

Now did you predict what might happen when we squared the roots? It actually makes a lot of sense that the number of roots would halve. Essentially, itβs a little like simply finding the fourth roots of our original
equation, which we know would form the vertices of a square.