Determine all the possible values of $(\sqrt{3}+i)^i$ and specify which quadrant(s) of the plane contains these values. $$ \mathrm{used}: a^z = \exp(z\log a)$$
$$(\sqrt{3}+i)^i = \exp(i\log((\sqrt{3}+i))$$
$$(\sqrt{3}+i)^i = \exp(i(\ln2+i(\frac\pi6+2k\pi)))$$
$$\mathrm{use}: e^z = e^x(\cos y+i\sin y)$$
$$ = (\exp(\frac \pi6+2k\pi))(\cos(\ln2)+i\sin(\ln2))$$
After this I am not sure how to tell which quadrants contain these values? 
 A: You have the result in polar form so you can read it right off your expression.  The complex factor in your polar form shows that the points are at an angle of ln(2) radians counterclockwise from the positive real axis.  This number lies between 0 and $\pi/2$ meaning the points are in the first quadrant.
A: $$\left(\sqrt{3}+i\right)^i=\left(2e^{\frac{\pi i}{6}}\right)^i=2^ie^{i\cdot\frac{\pi i}{6}}=2^ie^{-\frac{\pi}{6}}=2^ie^{-\frac{\pi}{6}}=$$

Now, use that $n^i=e^{\ln(n)i}$:

$$e^{\ln(2)i}e^{-\frac{\pi}{6}}=\frac{\cos(\ln(2))+\sin(\ln(2))i}{e^{\frac{\pi}{6}}}$$
To find the quadrant, use the argument:
$$\arg\left[\frac{\cos(\ln(2))+\sin(\ln(2))i}{e^{\frac{\pi}{6}}}\right]=$$
$$\arg\left[\cos(\ln(2))+\sin(\ln(2))i\right]=$$
$$\arctan\left(\frac{\sin(\ln(2))}{\cos(\ln(2))}\right)=$$
$$\arctan\left(\tan(\ln(2))\right)=\ln(2)\approx0.693147$$
So your number is in the first quadrant, because if $x$ lay in the first quadrant, then the argument has to be:
$$0<\arg\left[x\right]<\frac{\pi}{2}$$
