Let $x+iy=(1+i\sqrt{3})^{(1+i\sqrt{3})}$

Taking log on both sides and using

$1+i\sqrt{3}= 2e^{i\frac {\pi}{3}} (x+iy) = r (\cos \theta + isin \theta)=re^{i\theta}$

$\text{where }r=\sqrt{x^2 + y^2} \text{ and }\theta=\tan^{-1}\dfrac {y}{x}$

$\log (x+iy)=(1+i\sqrt{3})\log (1+i\sqrt{3}) = (1+i\sqrt{3}) \log (2e^{i \frac {\pi}{3}})$

$\text{Using }\tan^{-1}\sqrt{3} = \dfrac {\pi}{3}$

$\log (x+iy)= (1+i\sqrt{3}) \left [ \log (2)+\log e^{i\frac {\pi}{3}} \right ] $

$=(1+i\sqrt{3}) \left [\log (2)+ i \dfrac {\pi}{3} \right ]=\log (2) - \dfrac {\pi}{\sqrt{3}} + i \left ( \dfrac {\pi}{3}+ \sqrt{3}\log 2 \right )$

$\therefore x+iy=e^{\left ( \log 2-\frac {\pi}{\sqrt{3}} \right )} e^{i\left ( \frac {\pi}{3}+ \sqrt{3} \log 2 \right )}$

$ =e^(\log 2) e^{\left (- \frac {\pi}{\sqrt{3}} \right )} \left ( \cos \left ( \dfrac {\pi}{3}+ \sqrt{3}\log 2 \right ) + \sin \left ( \dfrac {\pi}{3}+ \sqrt{3}\log 2 \right ) \right )$

Hence, real part, $x=2e^{-\dfrac {\pi}{\sqrt{3}}} \left ( \cos \dfrac {\pi}{3}+ \sqrt{3} \log 2 \right )$