Finding the exact value of $b$ when given the argument in $z=(b+i)^2$ 
In the answer to the above question, there are two methods. My method was that I expanded $(b+i)^2$ out and I do $\arctan \frac{2b}{b^2+1}$ and then solve quadratic equation.
However, my friend did it in this way: 

I don't understand why he can ignore the squared bit of $(b+i)$ and then take the complementary angle - $30^\circ$?
Please advise.
Thanks in advance and sorry for any wrong tags or title naming. 
 A: Its simple $arg$ shows similar properties  to logs so its $arg(b+i)+arg(b+i)=30+30$ thus $arg(b+i)=30$ and then normal procedure
A: HINT:
What your friend has used is basically this property:
$$arg(z_1z_2)=arg(z_1)+arg(z_2)+2k\pi$$
where $z_1=z_2$ and hence $k=0$.
Do you get now what your friend has done?
A: Notice:


*

*$$(a+bi)^2=(a+bi)(a+bi)=(a^2-b^2)+(2ab)i$$

*$$\arg(x+iy)=\begin{cases}
\arctan\left(\frac{y}{x}\right)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{if}\space x>0\\
\arctan\left(\frac{y}{x}\right)+\pi\space\space\space\space\space\space\space\space\text{if}\space x<0\space\text{and}\space y\ge0\\
\arctan\left(\frac{y}{x}\right)-\pi
\space\space\space\space\space\space\space\space\text{if}\space x<0\space\text{and}\space y<0\end{cases}$$

So, given:


*

*$$z=(b+i)^2\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{with}\space\space b\in\mathbb{R^+}$$

*$$\arg((b+i)^2)=\frac{\pi}{3}\space\space\space\space\space\space\text{with}\space\space b\in\mathbb{R^+}$$


So, solving the problem:
$$\arctan\left(\frac{2\cdot b\cdot1}{b^2-1^2}\right)=\frac{\pi}{3}\Longleftrightarrow$$
$$\arctan\left(\frac{2b}{b^2-1}\right)=\frac{\pi}{3}\Longleftrightarrow$$
$$\frac{2b}{b^2-1}=\tan\left(\frac{\pi}{3}\right)\Longleftrightarrow$$
$$\frac{2b}{b^2-1}=\sqrt{3}\Longleftrightarrow$$
$$\frac{b}{b^2-1}=\frac{\sqrt{3}}{2}\Longleftrightarrow$$
$$\frac{1}{b-\frac{1}{b}}=\frac{\sqrt{3}}{2}\Longleftrightarrow$$
$$b-\frac{1}{b}=\frac{2}{\sqrt{3}}\Longleftrightarrow$$
$$\frac{b^2-1}{b}=\frac{2}{\sqrt{3}}\Longleftrightarrow$$
$$\sqrt{3}(b^2-1)=2b\Longleftrightarrow$$
$$\sqrt{3}b^2-2b-\sqrt{3}=0\Longleftrightarrow$$
Solving $b$ with the quadratic formula gives:
$$b=\sqrt{3}\space\space\space\vee\space\space\space b=-\frac{1}{\sqrt{3}}$$
So, $b$ has to be positive:
$$z=\left(\sqrt{3}+i\right)^2$$
