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?
 A: 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
$$z=\sqrt{2}e^{-\frac{13\pi i}{12}}$$
A: Problems
The absolute value $r$ is correct, but the angle is not!
You used the wrong formula for the angle...
If you got a complex number $z = a + b \cdot \mathrm{i} = r \cdot \operatorname{cis}(\theta)$ then your angle is $\theta = \operatorname{arctan2}(b, ~a)$. But you used another formula.
If you use $\theta = \operatorname{arctan2}(b, ~a)$ you will get: $\theta = \operatorname{arctan2}\left(\Im\left(\frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\right), ~\Re\left(\frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\right)\right) = \arctan\left(\frac{-1 + \sqrt{3}}{-1 - \sqrt{3}}\right) + \pi = \frac{11}{12} \cdot \pi + 2 \cdot k \cdot \pi$.
better way to approach this

Is there something I am missing, like a better way to approach this?

Yes there is.
But you could also write the individual terms that are multiplied or divided with each other in polar form and then summarize them using the power laws.
If you don't whant to this 'cause you don't like the polar form.
You just can say/use (algebraic):
$$
\begin{align*}
z &= a + b \cdot \mathrm{i}\\
z &= r \cdot \operatorname{cis}(\theta)\\
\\
r &= |z| = \sqrt{a^{2} + b^{2}}\\
\theta &= \arg(z) = \operatorname{arctan2}\left( b, ~a \right)\\
\\&\text{with}\\\\
\operatorname{arctan2}\left( b, ~a \right) &=
\begin{cases}
\arctan\left({\frac {b}{a}}\right) \qquad\quad~~~ \text{for } a > 0\\
\arctan\left({\frac {b}{a}}\right) + \pi \qquad \text{for } a < 0 \quad y > 0\\
\pi \qquad\qquad\qquad\quad~~~ \text{for } a < 0 \quad y = 0\\
\arctan\left({\frac {b}{a}}\right) - \pi \qquad \text{for } a < 0 \quad y < 0\\
\frac{\pi}{2} \qquad\qquad\qquad\quad~~ \text{for } a  = 0 \quad y > 0\\
-\frac{\pi}{2} \qquad\qquad\qquad~~~ \text{for } a  = 0 \quad y < 0\\
\end{cases} + 2 \cdot k \cdot \pi
\end{align*}
$$
If you use that you'll get:
$$
\begin{align*}
z &= \frac{\left(1 - \mathrm{i}\right)^{3} \cdot \left(\sqrt{3} + \mathrm{i}\right)}{4 \cdot \mathrm{i}}\\
z &= -\frac{1 + \sqrt{3}}{2} + \left( \frac{-1 + \sqrt{3}}{2} \right) \cdot \mathrm{i}\\
z &= \sqrt{\left( -\frac{1 + \sqrt{3}}{2} \right)^{2} + \left( \frac{-1 + \sqrt{3}}{2} \right)^{2}} \cdot \operatorname{cis}\left(\operatorname{arctan2}\left( -\frac{1 + \sqrt{3}}{2}, ~\frac{-1 + \sqrt{3}}{2} \right)\right)\\
z &= \sqrt{2} \cdot \operatorname{cis}\left(\arctan\left(\frac{-1 + \sqrt{3}}{-1 - \sqrt{3}}\right) + \pi\right)\\
z &= \sqrt{2} \cdot \operatorname{cis}\left(\frac{11}{12} \cdot \pi + 2 \cdot k \cdot \pi\right)\\
z &= \sqrt{2} \cdot \operatorname{cis}\left(\frac{11}{12} \cdot \pi\right)\\
z &= \sqrt{2} \cdot \left(\cos\left(\frac{11}{12} \cdot \pi\right) + \sin\left(\frac{11}{12} \cdot \pi\right) \cdot \mathrm{i} \right)\\
\end{align*}
$$
