Can i just check this which leads to my question: a square root of a number a is a number y such that $y^2$= a.
all a a=/=0 have 2 square roots so we choose a branch of this function so that it is single valued, called the principle branch.
in the case when a is non-negative real that is the positive square root, denoted by $\sqrt{x}$ so the principle square root function is f(x)= $\sqrt{x}$ x>=0.
if what i have said makes some sense then my question is how do we define the principle branch of the complex root function and is radical notation used to represent it.
so does f(x) = $\sqrt{x}$ x any complex number represent this function and how is it defined? From wikipedia, if x is in exponential form then $\sqrt{x}$ = $\sqrt{r}$e^(itheta/2) with theta between - pi and pi. Thanks for the help
Please don't attack me because to me this is a valid question.
For given $a\in\dot {\mathbb C}$ (meaning $a\ne 0$) the equation $z^2=a$ has two solutions "of equal rights". If $a>0$ is a positive real number the positive one of these two roots is called the square root of $a$, and is denoted by $\sqrt{\mathstrut a}$, and $\sqrt{0}=0$ is also o.k.
In all other contexts the $\sqrt{\mathstrut}$ sign may only be used in a colloquial sense or if accompagned by a comment.
Now the principal value of the square root, which I denote by $\>{\rm pv}\sqrt{\cdot}\>$. This is an analytic function defined in the slit complex plane $\Omega:=\bigl\{z\in{\mathbb C}\bigm| z\ne-|z|\bigr\}$ (meaning: the negative real axis is removed). The function $\>{\rm pv}\sqrt{\cdot}\>$ continues, or: extends, the square root function $x\mapsto\sqrt{\mathstrut x}$ from the positive real axis into the complex domain. It is defined as follows: We begin with the principal value of the logarithm $${\rm Log}:\quad\Omega\to{\mathbb C},\qquad z\mapsto\log|z|+ i{\rm Arg}(z)\ ,$$ whereby $${\rm Arg}(z):={\rm the}\bigl({\rm arg}(z)\,\cap\>]{-\pi},\pi[\,\bigr)\qquad(z\in\Omega)$$ is the principal value of the polar angle of $z$. One has $$\exp\bigl({\rm Log}(z)\bigr)\>\equiv\>z\qquad(z\in\Omega)\ .$$ Using this we put $${\rm pv}\sqrt{z}\>:=\>\exp\left({1\over2}{\rm Log}(z)\right)=\sqrt{|z|}\>e^{i\,{\rm Arg}(z)/2}\qquad(z\in\Omega)\ .$$ This function behaves as expected: $$\eqalign{\left({\rm pv}\sqrt{z}\right)^2&=\exp\bigl({\rm Log}(z)\bigr)=z\qquad(z\in\Omega)\>,\cr {\rm pv}\sqrt{x}&=\sqrt{x}\qquad(x>0)\>, \cr {\rm pv}\sqrt{\bar z}&=\overline{{\rm pv}\sqrt{z}}\>,\cr}$$ and one even has $${\rm pv}\sqrt{a\,b}={\rm pv}\sqrt{a}\cdot{\rm pv}\sqrt{b}\qquad\bigl({\rm Re}(a)>0, \ {\rm Re}(b)>0\bigr)\ .$$ The price to pay for these goodies is that ${\rm pv}\sqrt{z}$ remains undefined on the negative real axis. Crossing this slit multiplies ${\rm pv}\sqrt{z}$ by $-1$. Now, if it so happens that $z_0=-1$ is your working point, you can always produce a representant $r(z)$ of the square root function which is analytic in a full neighborhood of $z_0$ by writing $r(z):=i\, {\rm pv}\sqrt{-z}$. This $r(z)$ coincides with ${\rm pv}\sqrt{z}$ in the upper half plane, is $\ =-{\rm pv}\sqrt{z}$ in the lower half plane, and is undefined on the positive real axis.