Answer
$[2(cos~45^{\circ}+i~sin~45^{\circ})]\cdot[5(cos~90^{\circ}+i~sin~90^{\circ} )] = [2(cos~(-315)^{\circ}+i~sin~(-315)^{\circ})]\cdot[5(cos~(-270)^{\circ}+i~sin~(-270)^{\circ} )]$
Work Step by Step
$45^{\circ} = -315^{\circ}+360^{\circ}$
Therefore, $45^{\circ}$ and $-315^{\circ}$ are co-terminal angles.
$cos~45^{\circ}+i~sin~45^{\circ} = cos~(-315^{\circ})+i~sin~(-315^{\circ})$
Similarly:
$90^{\circ} = -270^{\circ}+360^{\circ}$
Therefore, $90^{\circ}$ and $-270^{\circ}$ are co-terminal angles.
$cos~90^{\circ}+i~sin~90^{\circ} = cos~(-270^{\circ})+i~sin~(-270^{\circ})$
Therefore:
$[2(cos~45^{\circ}+i~sin~45^{\circ})]\cdot[5(cos~90^{\circ}+i~sin~90^{\circ} )] = [2(cos~(-315)^{\circ}+i~sin~(-315)^{\circ})]\cdot[5(cos~(-270)^{\circ}+i~sin~(-270)^{\circ} )]$
