where z and a are complex numbers
With 
it will be 
where p>0 the modulus and the argument.
So, if 
is one solution of the equation 
then:
From the last equation we conclude
and 
,
where k integer.
The equation is true for the complex numbers
of the form:
Now, we are going to show that the v values
of z that come out from the last equation for k=0,1,2,...,v-1
are all different from each other.
Indeed let's assume that there are 
with 
such
as 
then the
difference of the arguments 
, 
will
be a multiple of 
.
So, 
which
means that the number 
which cannot be true since 0<<v.
So, we found the v different roots of the
equation .
That means that the equation has exactly v
roots.
We have proven the Theorem
has exactly different roots that can be found from the following formula:  |
