$$Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2$$
$$Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$$
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!
In this image I got to$(\frac{1−\sqrt3+i+i\sqrt3}{2})^4$ and then, I'm not sure how to continue.
Hint: Brute force works: $$ \begin{align} &\frac1{16}\mathrm{Re}\left[((1-\sqrt3)+i(1+\sqrt3))^4\right]\tag{1}\\ &=\frac1{16}\left(\color{#C00000}{(1-\sqrt3)^4}-\color{#00A000}{6(1-\sqrt3)^2(1+\sqrt3)^2}+\color{#C00000}{(1+\sqrt3)^4}\right)\tag{2}\\ &=\frac1{16}\left(\color{#C00000}{2(1+6\cdot3+9)}-\color{#00A000}{6(-2)^2}\right)\tag{3}\\[6pt] &=2\tag{4} \end{align} $$ where we have used the binomial theorem a few times $$ (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\tag{5} $$ Explanation:
$(1)$: copy problem
$(2)$: expand using $(5)$ and drop the terms with an $i$
$(3)$: expand the red terms using $(5)$ and drop the odd exponents ($-\sqrt3$ will cancel $+\sqrt3$)
$(3)$: also use that $(1-\sqrt3)(1+\sqrt3)=1-3=-2$
$(4)$: arithmetic