where am I going wrong with solving this equation?

Question

solve $z^2=2e^{5{\pi}i/6}$.

Well, clearly $z={\sqrt{2}}e^{5{\pi}i/12}$ is a root so its' conjugate $z={\sqrt{2}}e^{-5{\pi}i/12}$ is the other root.

But I can also argue $z=-{\sqrt{2}}e^{5{\pi}i/12}$ is a root which when simplified is $z={\sqrt{2}}e^{-7{\pi}i/12}$

which is a distinct root from the other two. But a quadratic has only 2 roots.

Thanks

Answer 1

Consider an algebraic equation like $$az^2+bz+c=0.$$ Taking the conjugate, it becomes $$a^*(z^*)^2+b^*z^*+c^*=0.$$ If the coefficients are real, this is the same as $$a(z^*)^2+bz^*+c=0,$$ showing that if $z$ is a root, so is $z^*$.

The property does not hold for complex coefficients, as you get two different equations.

Answer 2

$$z={\sqrt{2}}e^{-5{\pi}/12}\\ z^2=2e^{-5\pi/6}\ne2e^{5\pi/6}$$

Answer 3

The conjugate of this root is not another root. Note that in general $$\begin{align} &(a+bi)^2=(a-bi)^2 \\\iff& a^2+2abi-b^2=a^2-2abi-b^2 \\\iff&2abi=-2abi \\\iff& a=0\,\,\,\,\,\text{ or }\,\,\,\,\, b=0 \end{align}$$

Hence $(z)^2=(\bar{z})^2$ if and only if $z$ is pure real or pure imaginary.