Can I find the conjugate of the complex number: $\sqrt{a+ib}$?
Actually my maths school teacher says and argues with each and every student that we can't conjugate $\sqrt{a+ib}$ to $\sqrt{a-ib}$ because according to him $\sqrt{a+ib}$ isn't a complex number.
Please give some proofs, or some good explanations along with replies.
On
The notation $w=\sqrt z$ is used to denote one of the two solutions to the equation: $$ w^2 = z. $$ If $w=\sqrt z$ is one such solution, the other one is $-w=-\sqrt z$.
Now if $w=\sqrt{a+ib}$ you have $w^2 = a+ib$ and $(\bar w)^2 = \overline{w^2} = \overline{a+ib} = a - ib$. So it is true that the conjugate of $w$ is one of the two square roots of $a-ib$: $$ \overline{\sqrt{a+ib}} = \pm \sqrt{a-ib}. $$
You first have to define what is meant by "the" square root of $\sqrt{a+ib}$, because there are two candidates. We can see this by writing $a+ib$ in polar form $re^{i\theta}$. Then $r^{1/2} e^{i(\theta / 2)}$ and $-r^{1/2}e^{i(\theta / 2)} = r^{1/2}e^{i(\theta/2 + \pi)}$ both square to the correct value.
One can resolve this ambiguity by choosing, for example, whichever one has the smaller angle as measured from the positive real axis.
In any case, both numbers can be conjugated (as can any complex number), respectively, to $r^{1/2} e^{-i(\theta / 2)}$ and $r^{1/2}e^{-i(\theta/2 + \pi)}$.