For the numbers in [tex]a+bi[/tex] form, convert to polar form:
[tex]1+i=\sqrt2\dfrac{1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)[/tex]
By DeMoivre's theorem,
[tex](1+i)^5=(\sqrt2)^5\left(\cos\dfrac{5\pi}4+i\sin\dfrac{5\pi}4\right)=4\sqrt2\dfrac{-1-i}{\sqrt2}=-4-4i[/tex]
[tex]-1+i=\sqrt2\dfrac{-1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac{3\pi}4+i\sin\dfrac{3\pi}4}\right)[/tex]
[tex]\implies(-1+i)^6=(\sqrt2)^6\left(\cos\dfrac{18\pi}4+i\sin\dfrac{18\pi}4\right)=8i[/tex]
[tex]\sqrt3+i=2\dfrac{\sqrt3+i}2=2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)[/tex]
[tex]\implies2(\sqrt3+i)^{10}=2^{11}\left(\cos\dfrac{10\pi}6+i\sin\dfrac{10\pi}6\right)=2^{11}\dfrac{1-i\sqrt3}2=2^{10}(1-i\sqrt3)[/tex]
For the numbers already in polar form, DeMoivre's theorem can be applied directly:
[tex]2\left(\cos20^\circ+i\sin20^\circ\right)^3=2\left(\cos60^\circ+i\sin60^\circ\right)=2\dfrac{1+i\sqrt3}2=1+i\sqrt3[/tex]
[tex]2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)^4=2(\cos\pi+i\sin\pi)=-2[/tex]
At second glance, I think the 2s in the last two numbers should also be getting raised to the 3rd and 4th powers:
[tex]\left(2(\cos20^\circ+i\sin20^\circ)\right)^3=8\left(\cos60^\circ+i\sin60^\circ\right)=4+4\sqrt3[/tex]
[tex]\left(2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)\right)^4=16(\cos\pi+i\sin\pi)=-16[/tex]
