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?
Best Answer
Note, we have $|w| = 5$. Let $\theta \in Arg(w)$ and then from your corresponding diagram of the triangle form my $w$, $\cos(\theta) = \frac{3}{5}$ and $\sin(\theta) = \frac{4}{5}$. Therefore, from $\sqrt{z} = \sqrt{z}\left( \cos(\frac{\theta}{2}) + i\sin(\frac{\theta}{2})\right )$, we essentially arrive at our answer.
Recall the half-angle identities of both cosine and sine. i.e.,
$$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1}{2}(1 + \cos(\theta))}$$
$$\sin \left (\frac{\theta}{2} \right) = \sqrt{\frac{1}{2}(1 - \cos(\theta))}$$
From plugging in the corresponding values into the above equations, we find that $\cos(\frac{\theta}{2}) = \frac{2}{\sqrt{5}}$ and $\sin(\frac{\theta}{2}) = \frac{1}{\sqrt{5}}$.
Then we obtain $\boxed{\sqrt{3 + 4i} = \pm (2 + i)}$