$$Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2$$
$$Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$$
I got that $Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 1 \ne 2$
And, that $\Big(({\frac{1+i}{1-i})^5\Big)} = i $ , which means that $Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$
Can you guys confirm that it's true? Thanks in advance!
I looked at your link for the solution: first one is actually easier if you just use the Binomial theorem on both the numerator and denominator and then just simplify. The second one, the way you have done it is fine.
First notice that $(1-i)^4=1-4i+6i^2-4i^3+i^4=1-6+1-4i+4i=-4$. So if we only find the real part of the numerator we are done. Now $Re(1+i\sqrt{3})^4=Re(1+4i\sqrt{3}+6i^2(3)+4i^3(3\sqrt{3})+i^43^2)=1-18+9=-8$, and then $\frac{-8}{-4}=2$ as required.
The second one, as mentioned, your method is good...
You can write $$ 1+i\sqrt{3}=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)= 2(\cos(\pi/3)+i\sin(\pi/3)) $$ so $$ (1+i\sqrt{3})^4=16\cos(4\pi/3)+i\sin(4\pi/3)= 16\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right). $$ On the other hand $$ 1-i=\sqrt{2}\left(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)= \sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4)) $$ so $$ (1-i)^4=4(\cos(-\pi)+i\sin(-\pi))=-4. $$ Therefore $$ \left(\frac{1+i\sqrt{3}}{1-i}\right)^4= \frac{8(-1-i\sqrt{3})}{-4}= 2+2i\sqrt{3}. $$ Similarly, $1+i=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))$, so $$ \frac{1+i}{1-i}= \frac{\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))}{\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4))}= \cos(\pi/2)+i\sin(\pi/2)=i $$ and so its fifth power is again $i$. Perhaps more easy, in this case, is $$ \frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+2i+i^2}{1+1}=i. $$
