Precalculus proof

Question

If z is a complex number and n is a positive integer:

z^n=(z+1)^n=1

How can you prove that |z|=|z+1|=1?

How can I find the the values of z in exponential form?

How can I prove that n must be divisible by 6?

Answer 1

okay, if z=e^iθ since 1=e^i2πk ,,,,if z=1^1/n then θ=2πk/n,,,k=0,1,...,n-1

yay

say w=e^iθ+1 and (e^iθ+1)=1^1/n

then e^iθ=e^i2πk/n-1

have e^iθ=(cos(2πk/n)-1+isin(2πk/n),,,k=0,1,...n-1

=2sin(πk/n)[-sin(πk/n-1)+icos(πk/n)]

=-2isin(πk/n)[e^iπk/n]

=2sin(πk/n)[e^{i(3π/2+πk/n)}}

note that the magnitude of e^iθ is 1

therefore πk/n=π/6 since sin(π/6)=1/2,,, n/k=6

in turn

make the magnitude of right hand side =1

by the way there is a similar problem

find z such that zⁿ=(z+1)ⁿ

say n=1, there is no number z where z=z+1

n=2, have z²=z²+2z+1

z=(-)1/2.

n=3 have z³=z³+3z²+3z+1,,,,, using quadradic formula, can solve for complex numbers z_1,2

z_1,2=(-)1/2+/-(i√3)/6

there is a method to generalize the answer for all n>1.

that method by the way first does this:,,,

find z for n>1

zⁿ=(z+1)ⁿ

1=(z+1)ⁿ/zⁿ,,,,,here is "1" on left hand side

1^(1/n)=1+1/z

and continues from there.