How do I find real Re(z) and imaginary Im(z) part of $z = (1-i)^{2019} $

The complex number started out as $z=\left(\frac{2+4i}{-1+3i}\right)^{2019}$ which I then simplified to said $z = (1-i)^{2019} $ . I am very new to complex numbers and so far haven't had success in finding a general principle of calculating Re(z) and Im(z), which is why I am looking for insight on how this could be solved.

You question is already answered in detail with the relevant theory and hope you got it. Now I just want to add a complementary solution to it.

$(1-i)^{2019}=(1-i)(1-i)^{2018}=(1-i)((1-i)^2)^{1009}=(1-i)(-2i)^{1009}=(1-i)(-2)^{1009}(i)^{1009}=(1-i)(-2)^{1009}.i.(i)^{1008}=(1-i)(-2)^{1009}.i.(i^2)^{504}=(1-i)(-2)^{1009}.i.(-1)^{504}=(1-i)(-2)^{1009}.i=(i-i^2)(-2^{1009})=-(1+i).2^{1009}$

This is practically impossible without knowing about exponential forms of complex numbers.

If you haven't learned this. Consider that for any complex number $z \ne 0$ and $z = a + bi$. Then $|z|=\sqrt{a^2 + b^2}> 0$. Let's call $r:=|z|=\sqrt{a^2+ b^2}$. Then $z = r(\frac ar + \frac bri)$. If call $c=\frac ar$ and $s=\frac br$ we have $z = r(c + si)$.

Why on earth would we want to do that?

Well, $c,s$ are real numbers with the conditions that $$c^2 + s^2 = \frac {a^2}{r^2} + \frac {b^2}{r^2}=\frac {a^2}{a^2 + b^2} + \frac {b^2}{a^2 + b^2} = \frac {a^2 + b^2}{a^2 + b^2}= 1$$ So $c^2 + s^2 =1$. The only pairs of real numbers where that is true, and for every pair of real numbers where that is true, are $\cos$ and $\sin$ pairs of some angle. And that angle is $\tan^{-1} = \frac sc =\tan^{-1} \frac {\frac br}{\frac ar}=\frac ba$.

So if we call $\theta = \arg z = \tan^{-1}\frac ba$ then we can write any (non-zero) complex number $z = a+bi$ as $z = r(\cos \theta + i\sin \theta)$ where $r=|z|=\sqrt{a^2 + b^2}$ and $\theta = \arg z =\tan^{-1}\frac ba$.

Note: if we were to "plot" the complex number on Cartesian plane as the point $(a,b)$ where we plot $a$ on the real (horizontal) axis and $y$ on the imaginary (vertical) axis then $r =|z|$ would be the magnitude (distance from the origin) of the point and $\theta = \arg z$ would be the angle of the point-to the origin-to the (positive) $x$-axis.

So for any non-negative complex number $z$ we can write this as $$z = r(\cos \theta + i\sin\theta )$$ where $r=|z|$ and $\theta = \arg z$.

So why on earth would we want to do that?

Well, this makes a nearly miraculous change in how we multiply (and how we visualize multiplying) complex numbers together!

Consider $$z = r(\cos \theta + i\sin \theta)\space \text{and}\space w = s(\cos \phi + i\sin \phi)$$ Then \begin{align} zw &= r(\cos \theta + i\sin \theta)\cdot s(\cos \phi + i\sin \phi)\\ &=[rs][\cos\theta\cos \phi + i(\sin\theta\cos\phi + \cos\theta\sin\phi) - \sin\theta \sin\phi]\\ &=rs([\cos\theta\cos\phi -\sin\theta\sin\phi] + i[\sin\theta\cos \phi + \cos\theta\sin\phi])\\ &=rs(\cos(\theta +\phi) + i\sin(\theta + \phi))\\ &=|z||w|(\cos(\arg z + \arg w) + i\sin(\arg z + \arg w))\\ \end{align} This is amazing!

If we consider a complex number to be a positive real distance from the origin and to have an angle in the complex plane... multiplying two complex numbers is simply a matter of multiplying their distances and adding their angles!.

So that would mean if

$$z = r(\cos\theta + i\sin \theta)$$ then

$$z^n = r^n(\cos(n\theta) + i\sin(n\theta))$$

Now we can do your question "easily". \begin{align} 1 - i &= \sqrt{1^2 + 1^2}(\frac 1{\sqrt{1^2 + 1^2}} + \frac {-1}{\sqrt{1^2+1^2}})\\ &=\sqrt 2\left(\frac 1{\sqrt 2} + \frac {-1}{\sqrt 2}\right)\\ &=\sqrt 2\left(\cos [-\frac \pi4] + i\sin[-\frac \pi 4]\right)\\ \end{align}

And so \begin{align}(1-i)^{2019} &= (\sqrt 2)^{2019}\left(\cos[-\frac {2019}4\pi] + i\sin[-\frac {2019}4\pi]\right)\\ &=2^{1009}\sqrt 2\left(\cos (-\frac 34\pi) + i\sin(-\frac 34\pi)\right)\\ &=2^{1009}\sqrt 2\left(-\frac 1{\sqrt 2} - i\frac 1{\sqrt 2}\right)\\ &=-2^{1009} - 2^{1009}i=\\ &=-2^{1009}(1+i)\\ \end{align}

Alternatively...

$(1-i)^2 = 1^2 -2i + i^2 =(1+(-1)) -2i = 0-2i =-2i$.

$(1-i)^3 =(1-i)^2\cdot (1-i)=-2i(1-i) =-2i+2i^2=-2-2i$.

$(1-i)^4=\begin{cases} (1-i)^2=(-2i)^2=4i^2 =-4\\(1-i)^3(1-i)= (-2-2i)(1-i)= -2+(-2i+2i) +2i^2= -2+0-2=-4\end{cases}\qquad\space\quad=-4$

So $(1-i)^{4k} = (-4)^k$

So $(1-i)^{2019}=(1-i)^{2016}(1-i)^3=(-4)^{504}(-2-2i)=$

$2^{1008}\cdot2(-1-i)= -2^{1009}(1+i)$

Note that $\require{cancel}1-i=\sqrt2\left(\cos\left(-\frac\pi4\right)+\sin\left(-\frac\pi4\right)i\right)$ and that therefore\begin{align}(1-i)^{2019}&=\sqrt2^{2019}\left(\cos\left(-2019\frac\pi4\right)+\sin\left(-2019\frac\pi4\right)i\right)\\&=\sqrt2\,2^{1009}\left(\cos\left(\frac{5\pi}4\right)+\sin\left(\frac{5\pi}4\right)i\right)\\&=\cancel{\sqrt2}\,2^{1009}\left(-\frac1{\cancel{\sqrt2}}-\frac i{\cancel{\sqrt2}}\right).\end{align}