Write $ z = \frac{(1-i)^3(√3+i)}{4i}$ to polar form Write the complex number in polar form:
$$ z = \frac{(1-i)^3(\sqrt 3+i)}{4i}$$
So my try goes as follows:
\begin{align}
\frac{(1−i)^3(\sqrt 3+i)}{4i} &= 
\frac{(1−i)^3(\sqrt 3+i) \times -4i}{16}\\& =
\frac{(1-3i-3+i)(\sqrt 3+i)\times-4i}{16}\\& =
\frac{(-2-2i)(\sqrt 3+i)\times-4i}{16} \\&=
\frac{(-1-i)(\sqrt 3+i)\times-4i}{8}\\& =
\frac{(4i - 4)(\sqrt 3+i)}{8} \\&=
\frac{(i-1)(\sqrt 3+i)}{2} \\&=
\frac{(\sqrt 3 \times i-1-\sqrt 3-i)}{2}\\& =
\frac{-(\sqrt 3+1)}{2} + \frac{(\sqrt 3-1)i} {2}\end{align}
Polar form:
$$r =\sqrt {\left(\frac {-1+\sqrt 3)}{2}\right)^2 + \left(\frac{\sqrt 3-1}{2}\right)^2}  = \sqrt{2} $$
$$\tan(v) = \frac{(\sqrt 3-1)} { -(1+\sqrt 3)} \iff v = \arctan(\sqrt 3-2) $$
put it equal to $K$
$$z = \sqrt{2}(\cos K, i \sin K), \\ K =  \arctan(\sqrt 3-2)$$
To me, this doesn’t seem like a clean answer, since this question could potentially be on an upcoming exam. Is there something I am missing, like a better approach?

The beat way to change to polar form is to change individual terms to polar form first
So
$$(1-i)^3=\left(\sqrt{2}e^{-\frac{\pi i}{4}}\right)^3=2\sqrt{2}e^{-\frac{3\pi i}{4}}$$
$$\sqrt{3}+i=2e^{\frac{\pi i}{6}}$$
and
$$4i=4e^{\frac{\pi i}{2}}$$
Which means that
$$z=\sqrt{2}e^{-\frac{13\pi i}{12}}$$
