Arithmetic Progression solution

[honest jewel]

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[buoyant obsidian]

Suppose k != 0 (for instance, let's consider k=1)
Then $a_n=2\pi$ for some $n$, which means $a_0+nd=2\pi\to a_0=2\pi-nd$

[keen geyser]

Omegabet_

[buoyant obsidian]

but you can show 2pi-nd is never a natural number unless it is exactly 0

r is also not defined

what's b_n?

But yeah, $1=zz^{-1}=e^{ia_n}e^{ia_m}=e^{i(a_n+a_m)}$, so $a_n+a_m=2k\pi$

to which $a_0+nr+(a_0+mr)=2k\pi\to 2a_0+nr+mr=2k\pi$

[keen geyser]

Omegabet_

Omegabet_

[buoyant obsidian]

but you showed already $a_0=0$

[keen geyser]

Omegabet_

[buoyant obsidian]

so $nr+mr=2k\pi$

[keen geyser]

Omegabet_

[buoyant obsidian]

no

z^-1 is in M

so there exists some a_m such that 1/z = cis(a_m)

dont define stuff that isnt needed

$M\leq\mathbb{C}^\times$ means if $z\in M$, then $\frac{1}{z}\in M$, which by definition of $M$ means $\frac{1}{z}=\cos(a_m)+i\sin(a_m)$ for some index $m$.

[keen geyser]

Omegabet_