Can I conjugate a 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.
PS : Sorry for double post, my previous question wasn't understood properly by the reply-ers because of absence of $\sqrt{}$ symbol :(
$\endgroup$ 154 Answers
$\begingroup$Perhaps what your teacher means is that the notation $\sqrt{a + ib}$ is ambiguous, because there are two square roots of $a+ ib$, and unlike with real numbers there is no good way of distinguishing which square root you are referring to when you write $\sqrt{a + ib}$. However, you are right when you write $$\overline{\sqrt{a + ib}} = \sqrt{a - ib}$$ if you take it to mean that if $z^2 = a + ib$ then $\bar{z}^2 = a - ib$. This follows from the fact that $$\overline{zw} = \bar{z} \bar{w}$$ for all $z,w \in \mathbb{C}$.
$\endgroup$ 5 $\begingroup$The fundamental theorem of algebra allows us to state that there are two different square roots for a given complex number; but what's a positive/negative complex number? This is an explanation of Taylor's statement ("there is no good way of distinguishing which square root you are referring to when you write $\sqrt{a+ib}$").
There is a formula for the square root of a complex number (see Wikipedia):$$\sqrt{a+ib}=\pm\left(\sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}}+i\operatorname{sgn}(b) \sqrt{\frac{-a + \sqrt{a^2 + b^2}}{2}}\right)$$So if $b\mapsto -b$, assuming $b$ to be strictly positive, then$$\sqrt{a-ib}=\pm\left(\sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}}-i\sqrt{\frac{-a + \sqrt{a^2 + b^2}}{2}}\right)$$Then, indeed, $\sqrt{a-ib}=\overline{\sqrt{a+ib}}$ if $b\neq 0$.
$\endgroup$ 7 $\begingroup$Consider
$$\sqrt{a+ib}=\exp\left(\frac{1}{2}\log(a+ib)\right)=\exp\left(\frac{1}{2}\log\left|a+ib\right|+i\theta_n\right),$$ where $\theta_n=\phi+\pi n$ for $n\in\mathbb{Z}$, and $\phi= \text{Arg}(a+ib)/2$.
We then have $$\sqrt{a+ib} = \sqrt{\left|a+ib\right|}\left(\cos\theta_n+i\sin\theta_n\right).$$ This expands to get $$\sqrt{\left|a+ib\right|}\left(\cos(\phi+\pi n)+i\sin(\phi+\pi n)\right),$$ and then $$\sqrt{\left|a+ib\right|}\left((-1)^n\cos(\phi)+i(-1)^n\sin(\phi)\right).$$
Hence, we have the complex numbers$$\sqrt{a+ib} = \pm\sqrt{\left|a+ib\right|}\left(\cos\phi+i\sin\phi\right).$$ So the symbol $\sqrt{a+ib}$ produces two complex numbers, not just the one, which is why your teacher might say $\sqrt{a+ib}$ is not a complex number.
$\endgroup$ 2 $\begingroup$Say $$a+ib=z$$ and
$$\sqrt{z}= \begin{cases} \sqrt{|z|}e^{\frac{i\phi}{2}} \\ \sqrt{|z|}e^{\frac{i\phi}{2 + \pi}} \end{cases}$$
for a chosen branch of $\sqrt{}$ - especially $\sqrt{z}$ should be well defined by the choice of the branch (this is important to mention, because for defining $\sqrt{}$ we have always to cut out a line including $0$ out of the $\Bbb C$ plane, f.i. for choosing $\phi\in (-\pi,\pi)$ it has to be for $z:=a+ib$ that $a>0$ or $b\neq 0$ - for other choices of the range of $\phi$ there apply corresponding restrictions to what $z$ can be).
Thereby we have these two values for the conjugate of the root $$\overline{\sqrt{z}}= \begin{cases} \sqrt{|z|}e^{\frac{-i\phi}{2}} \\ \sqrt{|z|}e^{\frac{-i\phi}{2 + \pi}} \end{cases}$$
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