If $3 - 5i$ is a square root of $z$, find the other root. 
If $3 - 5i$ is a square root of $z$, find the other root.

Well, I was under the impression that only the sign in front of the imaginary part would change so the other root would be $3 + 5i$.
However, when I solve for the complex root using de Moivre's theorem, then I get $-3 + 5i$.
Both the real and imaginary parts have changed signs - this seems to go against what I thought was the complex conjugate root theorem?
 A: If we apply DeMoivre's Theorem, the second root will be a $180$ degree rotation around the pole. So the other root is in fact, $-3 + 5i$ (you're right!). You are confusing the usage of the complex conjugate root theorem, which only applies to polynomials with real coefficients, not square roots of complex numbers in the complex plane.
A: We have the general property that: $t^2 =(-t)^2$. This holds in $\Bbb C$ as much as it does in $\Bbb  R$. Hence if $z=(3-5i)^2$ then $z=(-(3-5i))^2=(-3+5i)^2$ holds also. On the other hand, we have:
$$(\bar{z})^2=\overline {z^2},$$
which is obvious if you square out $a+bi$ and $a-bi$ or represent them and their squares on an Argand Diagram.
The complex conjugate root theorem is based on
$$(x-z)(x-\bar z)=x^2-(z+\bar z)x+ z\bar z ,$$
for which all coefficients are real, thusly we can see complex solutions to real polynomials. However, the polynomial we might generate from this problem;
$$z^2+16+30i=0$$ does not have real coefficients, so that method can't be used.
A: Polar coordinates:
$z_1= re^{i\alpha}$, given.
$z_1^2= r^2e^{i(2\alpha)}=: z$.
The  roots of $z$ are:
$z_1= re^{i(2\alpha)/2}$,and
$z_2= re^{i[(2\alpha)/2+ (2π)/2]}$;
$z_2= z_1\cdot(e^{iπ})= -z_1$.
Hence $z_2=-3+i5$.
