$z^{200} = 1+i$ If $z_1$ and $z_2$ roots of equation then also $z_1*z_2$ and $z_1+z_2$ roots of equation

Question

$z^{200} = 1+i$ If $z_1$ and $z_2$ roots of equation then also $z_1*z_2$ and $z_1+z_2$ roots of equation

for $z_1*z_2$ I think its false because let $z_1 = rcis x$ and $z_2 = rcis y$ so $z_1*z_2=(r^2)cis(x+y)$ this cann't be root because it has different $|z|$

Is this enough to prove it and how to prove $z_1+z_2$ ?

Answer 1

For $z_1z_2$ your argument re the modulus of roots is correct.

For any $z_1$ let $z_2=-z_1$ then $z_1+z_2=0$ which is clearly not a root.

Answer 2

$$(z_1*z_2)^{200}= z_1^{200}*z_2^{200} = (1+i)(1+i) = 2i$$

So it is not true for $z_1*z_2$

The second part is obviously wrong because if $z_1$ is a solution then $z_2=-z_1$ is also a solution and the sum is not a solution.