How to find $\arg(z)$ and $|z|$? How to find $|z|$ and $\arg (z)$
$z$ is complex number and
$z$ is defined by $$z=\left(\cos\frac\pi5+i\sin\frac\pi5\right)^{15}\cdot(3-3i)^{20}$$
I`ve tried to behave it like $$e^{i15\frac\pi5}\cdot e^{i20\frac\pi4}$$ and got in result = $$e^{i3\pi}\cdot e^{i5\pi}=e^{i8\pi}$$ which gives me $$(-1)^8=1$$ and if $$+z=|z|$$ so $$|z|=1$$ So is it my answer and how to find? $$Arg(z)$$
 A: This is just a starter to actually answer the question, I'd be willing to have a longer discussion in comments or chatroom;
$\left(\cos(\frac{\pi}{5})+i\sin(\frac{\pi}{5})\right)^{15}(3-3i)^{20}=(e^{i\frac\pi5})^{15}\left(3\sqrt{2}e^{-i\frac\pi4}\right)^{20}=e^{3\pi i}(3\sqrt{2})^{20}\left(e^{-i\frac\pi4}\right)^{20}=e^{3\pi i}\cdot3^{20}\cdot\sqrt{2}^{20}\left(e^{20\cdot-i\frac\pi4}\right)=e^{3\pi i}\cdot3^{20}\cdot\sqrt{2}^{20}\left(e^{-5\pi i}\right)=3^{20}\cdot\sqrt{2}^{20}e^{-2\pi i}$
but note that $3^{20}\cdot\sqrt{2}^{20}$, but again by laws of exponents you can simplify further because $\sqrt{2}=(2)^{\frac12}$, so $3^{20}\cdot\sqrt{2}^{20}=3^{20}\cdot\left((2)^{\frac12}\right)^{20}=3^{20}\cdot2^{\left(20\cdot\frac12\right)}=3^{20}\cdot2^{10}$
so the final answer is $3^{20}\cdot2^{10}$, or 3570467226624. This is a pure real number with imaginary part 0. It's so ludicrously large, but this actually makes sense conceptually because the modulus of $3-3i$ (which for goodness sake you're taking a 20th power of) is $\sqrt{3^2+3^2}=3\sqrt{2}\approx4.2426$, and so with each multiplication you increase the modulus by a constant factor, essentially blowing it up exponentially.
Also this is a cool video: https://www.youtube.com/watch?v=-dhHrg-KbJ0&t=531s
You'll notice that part of the entire point of Euler's Formula is that it's one possible parametrization of the unit circle. There is further intuition on Euler's Formula in this video too, although its a bit long winded and I haven't watched it in a while.
To me one of the craziest parts about Euler's formula is the seemingly ridiculus connection between rotation, and taking exponents. Like literally, that's just straight up crazy.
Also, this might be of interest to you:
$\left(re^{i\theta}\right)^n=r^ne^{in\theta}$
From this, we can draw some general conclusions:

*

*Just as is the case with real numbers, if $r<1$, then $r^n$ (for high powers of $n$) become very tiny. If $r=1$ then everything about $r^n$ stays fixed, and you have a pure rotation. Finally if $r$ where greater than 1 - even if $r$ is only slightly bigger like 1.000000001 with lots of zero's, with a high enough power of $n$ you'd expect to get complex numbers with a very large modulus.


*$e^{it}$ is periodic, so the result can point in any direction depending purely on the value of $t$
Interestingly, Euler's Theorem is also super handy for taking square roots of complex numbers too, since square roots are basically just another name for raising to the one-halfth power, after all;
$\left(re^{i\theta}\right)^\frac12=r^\frac12e^{i(\theta/2)}$
which also pretty ridiculous. I've made you a Desmos thing to play around with: https://www.desmos.com/calculator/dvfoycjgz8 . The red circle to pick a $z$, and the slider $s$ is the exponent. As you can see you can always pick values of the power that loop you all the way back to numbers where the real part is the only part. It just so happens to be that in a lot of the questions you see in school they give you problems that are more likely to work out nicely, although this is not generally the case.
A: Just remember some basic properies of $\arg$ and absolue value: If a value is given in polar form, then the polar coordinates are obtained.  With $r, \varphi\in \Bbb R$ and $r\geqslant 0$:
$$\begin{align*}
|r\cdot e^{i\varphi}| &= r \\
\arg (r\cdot e^{i\varphi}) &\equiv \varphi 
\end{align*}$$
Where $\equiv$ means that the value is only determined up to an integer multiple of $2\pi$.  Common choices for $\arg$ are $0\leqslant\arg<2\pi$  or $-\pi < \arg\leqslant \pi$.
Moreover, for a product of two complex numbers we have:
$$\begin{align}
|z\cdot w| &= |z|\cdot|w| \\
\arg(z\cdot w) &\equiv \arg z + \arg w \\
\end{align}$$
which implies for $z\neq 0$ and real exponents $p\neq 0$
$$\begin{align}
|z^p| &= |z|^p \\
\arg(z^p) &\equiv p\cdot\arg z\\
\end{align}$$
So the first lines for your $\arg z$ would prepare for easy computation and read:
$$\begin{align}
z &= \left(\cos\frac\pi5+i\sin\frac\pi5\right)^{15}\!\!\cdot\,(3-3i)^{20} \\
&= \left(\exp\frac{\pi i}5\right)^{15}\!\!\cdot\,3^{20}\cdot(1-i)^{20} \\
\end{align}$$
Hence:
$$\begin{align}
\arg z &\equiv 15\cdot \frac\pi5 + 20\cdot\underbrace{\arg(3)}_{\textstyle=0} + 20\cdot\underbrace{\arg(1-i)}_{\textstyle=-\pi/4} \\
&\equiv 3\pi - 5\pi \equiv 0 \mod 2\pi
\end{align}$$
