Ta có
$1+\sqrt 3 i=2\left ( \cos \frac{\pi}{3}+i\sin\frac\pi3\right )$
$1+ i=\sqrt 2\left ( \cos \frac{\pi}{4}+i\sin\frac\pi4\right )$
Suy ra
$\dfrac{1+i\sqrt3}{1+i} = \sqrt 2\cos \left ( \frac{\pi}{3}- \frac{\pi}{4} \right )+i\sin\left ( \frac{\pi}{3}- \frac{\pi}{4} \right )$
$\dfrac{1+i\sqrt3}{1+i} = \sqrt 2 \left ( \cos \left ( \frac{\pi}{3}- \frac{\pi}{4} \right )+i\sin\left ( \frac{\pi}{3}- \frac{\pi}{4} \right )\right )$
$\dfrac{1+i\sqrt3}{1+i} =\sqrt 2\left ( \cos \frac{\pi}{12}+i\sin\frac\pi{12}\right )$