Ta có
$1=\left ( \cos 0+i\sin0\right )$
$1+ i=\sqrt 2\left ( \cos \frac{\pi}{4}+i\sin\frac\pi4\right )$
Suy ra
$\dfrac{1}{2(1+i)} = \frac1{2\sqrt 2}\left ( \cos \left (0- \frac{\pi}{4} \right )+i\sin\left ( 0- \frac{\pi}{4} \right ) \right )$
$\dfrac{1}{2(1+i)} = \frac1{2\sqrt 2} \left ( \cos \left (- \frac{\pi}{4} \right )+i\sin\left ( - \frac{\pi}{4} \right )\right )$