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