The answer is: $|z|=125$ and $Arg(z)=\pi/2+3\arctan (\frac{4}{3})+2\pi k$, $k=0,+-1,+-2,...$
I know how we get $|z|=125$, but the question is how we get that $Arg(z)$. By opening brackets and simplifying statement we get $z=44+117i$. Then $Arg(z)$ should be equal to $\arctan(117/44)+2\pi k$. I don't understand it, please explain
By your work $$(-4+3i)^3=125\left(\operatorname{cis}\left(\pi-\arctan\frac{3}{4}\right)\right)^3=$$ $$=125\left(\operatorname{cis}\left(\frac{\pi}{2}+\arctan\frac{4}{3}\right)\right)^3=125\operatorname{cis}\left(-\frac{\pi}{2}+3\arccos\frac{4}{3}\right)$$ and indeed $$\tan\left(-\frac{\pi}{2}+3\arctan\frac{4}{3 }\right)=-\frac{1-3\left(\frac{4}{3}\right)^2}{3\left(\frac{4}{3}\right)-\left(\frac{4}{3}\right)^3}=\frac{117}{44}.$$ We used $$\tan3\alpha=\frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}.$$
Note that the principal argument should be a single value in the appropriate range. There is generally no need to state the periodicity like that, although in some contexts the argument is considered to be multi-valued.
Second of all, I'm not sure why your answer key complicates things by taking $\arctan \frac 43$, when it should be the imaginary part divided by the real part ("rise over run"). They took it the other way around, which is the only reason they needed to introduce the $\frac{\pi} 2$ term (remember the identity $\cot x = \frac 1{\tan x} =\tan(\frac{\pi} 2 - x) $).
I would simply start with $\mathrm{arg(z)} = \pi - \arctan \frac 34$ since the point of $z$ lies in the second quadrant. Then multiply by $3$ modulo $2\pi$ to find the argument of the cube, i.e. $\mathrm{arg(z^3)} = 3\pi - 3\arctan\frac 34 \pmod{2\pi} = \pi - 3\arctan\frac 34$. Note that since $ 0< \arctan \frac 34 <\frac {\pi} {4}$ (by the increasing monotonicity of tangent within the first quadrant and the fact that $\tan \frac{\pi}4=1$, you're guaranteed to have a value for the final result in the range $(0,\pi)$, so you don't have to make any further adjustments to it to get the principal argument. However, if you require the full multivalued form, here is where you add $2k\pi, k\in \mathbb {Z}$.
Somewhat alternative approach that avoids the use of the tangent function.
The trick is to convert $x + iy$ to $r\times (\cos \theta + i\sin \theta),$
where $r > 0.$
Once this is done, you have, by De Moivre's theorem that
$[r\times (\cos \theta + i\sin \theta)]^3
~=~ r^3 \times (\cos 3\theta + i\sin 3\theta).$
This means that if $\theta$ is the Argument of $z$, then $3\times \theta$ is the Argument of $z^3$ (within modulus of $2\pi$).
Here, if $z = -4 + 3i$ and $\theta$ is the Argument of $z$,
then you have that $|z| = 5, \cos \theta = \dfrac{-4}{5},$
and $\sin\theta = \dfrac{3}{5}.$
Then, it is immediate that $|z|^3$ is $125$.
Further, https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Multiple-angle_formulae can be consulted.
This gives $\sin (3\theta) = 3\sin \theta - 4\sin^3 \theta$
and $\cos (3\theta) = 4 \cos^3 \theta - 3 \cos \theta.$
With $\sin \theta = \dfrac{3}{5}$ and $\cos \theta = \dfrac{-4}{5}$ we have
$$\sin(3\theta) = \frac{9}{5} - \frac{108}{125} = \frac{117}{125}$$ and
$$\cos(3\theta) = \frac{-256}{125} - \frac{-12}{5} = \frac{44}{125}$$
Given an arbitrary complex number $z = a+ib$, step one is to convert it to the form $|z|e^{i\phi}$. This is always possible due to Euler's theorem $e^{i\phi} = \cos(\phi) + i \sin(\phi)$. Now if you raise $z$ to power $n$ then we have $z^n =|z|^n \left ({e^{i \phi}}\right )^n = |z|^n e^{i \phi n}$. So the angle you are after is $3\phi + 2 \pi k$, because $e^{i 2 \pi k} = 1$ for all $k$. In your case $\phi \approx 2.5$ radians. I did that calculation using the Matlab command [phi,z] = cart2pol(-4,3). And yes $|z|^3 = 125$. Hopefully this answers your question.
