If I have z^n = (z + 1)^n = 1 from which I derived |z| = |z+1| = 1, how can I find all the possible values of z in exponential form? So far I have |z|^n = 1 is e^{(2\pi i)/n} for n^{th} roots of unity. Is that on the right track? I don't even know how to approach the exponential for for (z+1)^n=1

Saphia1123 Mar 16, 2023

#1**0 **

You're on the right track! From the equation z^n = 1, we know that z must be an nth root of unity, so we can write:

z = e^(2πik/n)

where k is an integer such that 0 ≤ k < n. This gives us n distinct solutions for z, one for each value of k.

To find the solutions for (z+1)^n = 1, we can first notice that if we define w = z+1, then the equation becomes w^n = 1. So we know that:

w = e^(2πik/n)

for some integer k between 0 and n-1. Substituting z+1 for w, we get:

z + 1 = e^(2πik/n)

which we can rearrange to get:

z = e^(2πik/n) - 1

Now we have n distinct solutions for z, one for each value of k between 0 and n-1.

To write these solutions in exponential form, we can use Euler's formula:

e^(iθ) = cos(θ) + i sin(θ)

So we can write each solution as:

z = e^(2πik/n) - 1 = cos(2πk/n) + i sin(2πk/n) - 1

= (cos(2πk/n) - 1) + i sin(2πk/n)

= -2 sin²(πk/n) + i 2 sin(πk/n) cos(πk/n)

where we've used the identity cos(θ) - 1 = -2 sin²(θ/2) and the double-angle formula sin(2θ) = 2 sin(θ) cos(θ).

So the n solutions for z in exponential form are:

z = -2 sin²(πk/n) + i 2 sin(πk/n) cos(πk/n)

for k = 0, 1, ..., n-1.

Guest Mar 16, 2023