get the argument for this complex number

Question

I rationalized and I obtained:

I get: $|z| = 2^6 = 64$

But $\arg(z) = 2 + \sqrt{3}$ ?

Answer 1

Since $\sqrt{3}-i=2e^{-i\pi/6}$ and $1+i=\sqrt{2}e^{i\pi/4}$, $\frac{\sqrt{3}-i}{1+i}=\sqrt{2}e^{-5i\pi/12}$ and $z=2^6e^{-5i\pi}=-64$, which has modulus $64$, argument $\pi$.