Answer
$-3 \sqrt 2+i \ 3 \sqrt 2$
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 =6(\cos 135^{\circ} +i \sin 135^{\circ} ) ....(1) $
Now, $\cos 135^{\circ}=\cos (90^{\circ} +45^{\circ}) =-\sin 45^{\circ}=- \dfrac{\sqrt 2}{2}$ and $\sin 135^{\circ}=\sin (90^{\circ} +45^{\circ}) =\cos 45^{\circ}=\dfrac{\sqrt 2}{2}$
Therefore, the equation (1) becomes:
$z= 6( \dfrac{-\sqrt 2}{2} +i \dfrac{\sqrt 2}{2})=-3 \sqrt 2+i \ 3 \sqrt 2$
