Question III.d: Find the resultant of (8 +×j6)× (−10 −j7 . 5)

Resultant of (8 +j6)( -10 −j7.5)
Let Z is the resultant, then

=10|\underset{}{\underline{tan}^{-1}} 0.75×12.56|\underset{}{\underline{180°}} + tan^{-1} 0.75=10|\underset{}{\underline{37°}} ×12.56|\underset{}{\underline{180°}} +37°

=125.6 |\underset{}{\underline{37°+180°+37°}} =125.6 |\underset{}{\underline{254°}}

The location of the vector is shown in complex plane (see Fig. 39)

\left|OA\right| =125.6

θ=254°.

=125.6  cos (180° +74°) +j  125.6  sin (270°-16°)