Dayv O. answered • 12/25/20
Attentive Reliable Knowledgeable Math Tutor
To solve this fun problem, complex algebra is used; that
is, to solve the problem the constant =square root of -1,
(in notation the letter i) is used. (-1)^(1/2)=i
the answer is depending on n, there are
n-1 many z that satisfy equation.
z=(1/2)*(1+i*cot(πk/n)) k=1,2,...,(n-1)
The technique used is
(z-1)^n=z^n is same as (z-1)^n=z^n*1
so z-1=z*1^(1/n), taking n-th root both sides.
z(1-1^(1/n))=1,
z=1/(1-1^(1/n))
z is reciprocal of 1-1^(1/n)
1-[1^(1/n)]=1-[cos(2πk/n)+isin(2πk/n)] k=0,1,...(n-1)
because 1^(1/n)=[cos(2πk/n)+isin(2πk/n)] k=0,1,...(n-1)
notice the square root of -1 =i is introduced, this
formula is found in complex algebra books
next
1-[cos(2πk/n)+isin(2πk/n)]
=2*sin2(πk/n)+i*2*sin(πk/n)*cos(πk/n)
[half angle/double angle formulas from trigonometry.]
so far 1-1^(1/n)=2*sin(πk/n)*[sin(πk/n)+icos(πk/n)]
k=0.1,...,(n-1)
z= the reciprocal of this quantity
= the values z satisfying (z-1)^n=z^n
=1/(1-1^(1/n))=1/[2*sin(πk/n)*(sin(πk/n)-icos(πk/n)
=(1/2)*(1+i*cot(πk/n))
cleared denominator of i by multiplying by 1=
by (sin(πk/n)+icos(πk/n))/ (sin(πk/n)+icos(πk/n))
note that: sin^2(πk/n)+cos^2(πk/n) =1
notice that k must go from 1,2,...,(n-1)
k cannot equal 0, since z=(1/2)*(1+icot(πk/n)) doesn't exist for k=0
this is good because n cannot equal 1
there is no z such that z^1=(z-1)^1
and k was supposed to range to n-1
for n=2, k=1 only and z=1/2+0i since cot(π/2)=0
((1/2)-1)^2=(1/2)^2
for n=3, there is a quadradic to solve
3z^2-3z+1=0 the same two z are found
with z=(1/2)*(1+icot(πk/3)) k=1,2
for n=4 and above
z=(1/2)*(1+icot(πk/n)) k=1,2,...,(n-1)
Mark M.
"1^(1/n)=[cos(2πk/n)+isin(2πk/n)] k=0,1,...(n-1)" how is this established?12/26/20
Tom K.
I answered this correctly this morning. If you factor out cos to make up 1/2, which is the real part, you get sin/cos, so it would be tan. 180 degrees is also a legitimate answer for the n'th power. For n even, you can't include k = n/2. Note that, with the original polynomial and n even, the z^n's cancel.12/26/20
Mark M.
I have not seen your solution. How do you determine that 1^(1/n) is imaginary?12/26/20
Dayv O.
Tom, yes I upvoted your answer to (1-z)^n=z^n, and it is tan is part of solution. But the real problem I wanted to post was (z-1)^n=z^n, for which cot is involved. It is kind of of a cool problem.12/26/20
Dayv O.
Hi Mark, I see you are involved very much positively answering questions. 1=(cos(2*pI)+isin(2+pI)) so 1 has "i" involved but just times zero. When 1^(1/n) is evaluated, since 1/n is not an integer, there are n solutions, It is Demoivre's therom.12/26/20
Dayv O.
I meant 1=(cos(2*pI)+isin(2*pI))12/26/20
Dayv O.
when it is stated 625^(1/4)=+5 or -5, two complex number answers are omitted. 625^(1/4) also is equal to 5*(cos(pi/2)+isin(pi/2)=5i and is also equal to 5*(cos(3*pi/4)+isin(3*pi/4)=-5i. Note i^4=1 so (5i)^4=625 and (-5i)^4=62512/26/20
Dayv O.
I meant -5i=%*(cos(3*pi/2)+isin(3*pi/2)12/26/20
Dayv O.
-5i=5*(cos(3*pi/2)+isin(3*pi/2))12/26/20
Mark M.
OK, understood that 1 can have an unlimited number of complex roots (Demoivre's). Yet the proposal only states n > 1 that implies real roots.12/26/20
Dayv O.
we had to take nth root of 1 to clear the equation (z-1)^n=(z^n)*1 of the n exponentiation. That is where fraction 1/n shows up.12/26/20
Mark M.
My comment is not in regard to the ^(1/n). It is to the assumption that complex numbers are included in the domain. Nothing to that is stated in the problem.12/26/20
Dayv O.
if n=2 problem asks something simple. when is (z-1)^2=z^2? As you might say, just expand left side of equation, subtract right side of equation, and solve. If n=3, same thing except the resulting quadradic would indeed have it's two solutions in the complex domain. When n=4, just expanding the left side and subtracting right side leaves a more challenging problem, but roots obviously might be in complex domain. All three (1/2+i*cot(pi*k/4)), k=1,2,3) are in complex domain.12/27/20
Dayv O.
And isn't that if the exponent is fractional rational, n=p/q, the solutions of z^n=1 are n numbered? But if n is irrational( like n=π) or if n itself is of form a+i*b, then the solutions are infinite in number? It is a very unwell known fact that e^(i*π) not only equals -1 , but has an infinite amount of solutions in the complex domain.12/27/20
Dayv O.
sorry, e was z in w=z^(i*π), so -1 and the other infinite outcomes are in the range of e^(i*π). Again, the outcome of e^i*π)=-1 and an infinite other outcomes (I think -e^(2*π* π) is one and -e(4*π*π) is another in the sequence of outcomes)Here 0+i*π is the w in z^w. Let's look at z^n-1=w. When w=0 we have roots. So z=1^(1/n) is in the domain, and is of n values if n is integer (1/n is fractional rational).12/27/20
Dayv O.
no, not e^(i*π)=-e^(2*π*π).........It is e^(i*π)=-e^(π*π).... Notice (-e^(π*π))^(1/(i*π))=e12/27/20
Mark M.
1^(1/n) = 1 and -1 for n even, and 1^(1/n) = 1 for n odd.12/26/20