Argument of $3+4i$ for $z = \sqrt{3 + 4i}$, I am trying to put this in Standard form, where z is complex. I let $w = 3+4i$ and find that the modulus, $|w|=r$, is 5. I am having trouble solving for arg(w).
I find that $\tan^{-1}{\theta} = \frac{4}{3}$. However, this is not an angle well known. How do I find it?

Suppose $\sqrt{3+4i}$ were in standard form, say $x+yi$.  Then we would have $$\begin{align}
x+yi & = \sqrt{3+4i}\\
(x+yi)^2 & = 3+4i\\
(x^2-y^2) + 2xyi & = 3+4i
\end{align}
$$
so
$$\begin{align}
x^2 -y^2 &= 3 \\
2xy &= 4 \\
\end{align}
$$
From the second equation we have $y = \frac2x$.  Putting this into the first equation we obtain $$x^2 - \frac4{x^2}  = 3.$$ Multiplying through by $x^2$, then setting $z=x^2$ we obtain the quadratic equation $$z^2 -3z -4 = 0$$ which we can easily solve to obtain $z=4$.  (The other root, $z=-1$, is spurious since $z = x^2$ and $x$ is real.)  Then since $x^2=z$ and $y=\frac2x$ we get $\color{darkblue}{x=2, y=1}$ and $\color{darkred}{x=-2, y=-1}$.
