About the argument of a complex number I know that the square root of complex number is a multivalued function, and by definition:
$\sqrt{z}=w\iff w^2=z$ i.e. the square root of z are the solutions of complex equation $w^2=z$.
Problem
I would like to evaluate the argument(s) of
$$w=\sqrt{1+i}$$
using the properties of the principal value of $\arg(\cdot)\in (-\pi, \pi]$.
By definition $$w=\sqrt{1+i}\implies w^2=1+i$$ so $\arg(w^2)=\arg(1+i)$. It's easy to show that $\arg(1+i)=\frac{\pi}{4}$ hence $$\arg(w^2)=\arg(1+i)\iff \arg(w^2)=\frac{\pi}{4}$$
Now I use the property $\arg(w^n)=n\arg(w)$ so the equation becomes
$$2\arg(w)=\frac{\pi}{4}\implies \arg(w)=\frac{\pi}{8}.$$ The problem here is that $\frac{\pi}{8}$ is the argument of only one square root of $\sqrt{1+i}$. I lost the argument of the other one... but where? 
 A: $$\arg(w^2)=\arg(1+i) \implies \arg(w^2)=\frac{\pi}{4}+2k\pi$$
$$\arg(w^n)= n\arg(w)$$
$$\implies \arg(w)=\frac{\pi}{8}+k\pi$$
A: 
PRIMER:  In general, $\displaystyle \arg(z^n)\ne n\arg(z)$

While it is true that  
$$\arg(z_1z_2)=\arg(z_1)+\arg(z_2)\tag 1$$
the equality in $(1)$ is interpreted as a set equivalence.  It means that any value of $\arg(z_1z_2)$ can be expressed as the sum of some value of $\arg(z_1)$ and some value of $\arg(z_2)$.  And conversely, it means that the sum of any value of $\arg(z_1)$ and any value of $\arg(z_2)$ can be expressed as some value of $\arg(z_1z_2)$.
However, it is not true in general, that $\arg(z^2)=2\arg(2)$.  
For example, take $z=i$.  Then, for $\arg(z^2)=\arg(-1)=\pi+2n\pi=3\pi$, for $n=1$, there is no value of $2\arg(i)=\pi+4n\pi$ that is equal $3\pi$.
Therefore, $\arg(z^n)\ne n\arg(z)$ in general.

Now, for $z=1+i=\sqrt{2}e^{i(\pi/4+2n\pi)}$, we see that $\arg(z)=\pi/4+2n\pi$ is mulita-valued.  And $\arg(\sqrt{z})=\pi/8+n\pi$ is also multi-valued.  
If one restricts the argument of $z$ such that $-\pi<\arg(z)\le \pi$, then $\arg(z)$ assumes the values $\pi/4$ and $-3\pi/4$.  Finally, the argument of the square root assumes, therefore, the two values $\pi/8$ and $-3\pi/8$.
A: Well, we have when $\text{z}\in\mathbb{C}$:
$$\text{z}=\left|\text{z}\right|\cdot\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)\tag1$$
Where we work in radians, $\left|\text{z}\right|=\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}$, $0\le\arg\left(\text{z}\right)<2\pi$ and $\text{k}\in\mathbb{Z}$
So, when we take the square root of $\text{z}$, we get:
$$\sqrt{\text{z}}=\sqrt{\left|\text{z}\right|\cdot\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)}=\left(\left|\text{z}\right|\cdot\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)\right)^\frac{1}{2}=$$
$$\sqrt{\left|\text{z}\right|}\cdot\left(\exp\left(\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i\right)\right)^\frac{1}{2}=\sqrt{\left|\text{z}\right|}\cdot\exp\left(\frac{\left(\arg\left(\text{z}\right)+2\pi\text{k}\right)i}{2}\right)\tag2$$
So, when $\text{z}=w=1+i$ we get:
$$\sqrt{1+i}=\sqrt{\left|1+i\right|}\cdot\exp\left(\frac{\left(\arg\left(1+i\right)+2\pi\text{k}\right)i}{2}\right)=$$
$$\sqrt{\sqrt{2}}\cdot\exp\left(\frac{\left(\frac{\pi}{4}+2\pi\text{k}\right)i}{2}\right)=\sqrt[4]{2}\cdot\exp\left(\left(\frac{\pi}{8}+\pi\text{k}\right)i\right)\tag3$$
And we know that:


*

*$$\cos\left(\frac{\pi}{8}+\pi\text{k}\right)=\pm\space\frac{\sqrt{2+\sqrt{2}}}{2}\tag4$$

*$$\sin\left(\frac{\pi}{8}+\pi\text{k}\right)=\pm\space\frac{\sqrt{2-\sqrt{2}}}{2}\tag5$$

