I'll just do (2) and you can use it as a model to
do (1).
Let's draw the picture of that complex number as a
right triangle on a graph.
where ,
(a) .
Sometimes this is called the "modulus". This is the length of
the hypotenuse of that right triangle. We find it the same way
we always find the hypotenuse, the Pythagorean theorem:
We label the hypotenuse and this is the same as
(b) find when
"Argument" means "angle" (you can remember it because "argument"
and "angle" both start with "a" and their third letters are "g")
Now since we are told that the argument is between
and , we must take the argument as a negative angle
measured by rotating clockwise from the right side of the x-axis,
indicated by the blue arc below labeled , "phi".
It should be q, "theta", but I can't
get that Greek letter on the notation program for this site, so
I'll use instead.
Since the triangle is an isosceles right triangle, its interior
angles are or in radians. However since
tells us that the rotation is clockwise from the right
side of the x-axis, the angle is taken as negative. Thus we
take it as and we label it:
(c) write in polar form:
Since , therefore
and since , therefore
So
and we can factor out and get:
This is the polar form of
Therefore since and
(d) or (sqrt(2)-i*sqrt(2))^6}}}
DeMoivre's theorem says:
Therefore:
Since is coterminal with , we have
Edwin