Answer
$-1-i $ or, $-(1+i)$
Work Step by Step
The given complex number can be written in the rectangular form as follows: $r (\cos \theta +i \sin \theta) = a+i b$
Where, $a=r \cos \theta ; b =r \sin \theta $
Let us suppose that $z =\sqrt 2(\cos 225^{\circ} +i \sin 225^{\circ} ) ....(1) $
Now, $\cos 225^{\circ}=\cos (270^{\circ} -45^{\circ}) =-\cos 45^{\circ}=- \dfrac{\sqrt 2}{2}$ and $\sin 225^{\circ}=\sin (270^{\circ} - 45^{\circ}) =-\cos 45^{\circ}= - \dfrac{\sqrt 2}{2}$
Therefore, the equation (1) becomes:
$z= \sqrt 2( \dfrac{-\sqrt 2}{2} +i \dfrac{-\sqrt 2}{2})=-1-i $ or, $-(1+i)$
