(a) (5-j 6)-(2+j 8)/(-3+j 4)(5-j)+(4-j 6) (b) (240 ∠75° +160 ∠-30°)(60-j 80)/(67+j 84) (20 ∠32°) (c) (10+j 20/3+j 4)^2 √(10+j 5)(16-j 120)

Question

(a) \frac{(5-j 6)-(2+j 8)}{(-3+j 4)(5-j)+(4-j 6)}

(b) \frac{\left(240 \angle 75^{\circ}+160 \angle-30^{\circ}\right)(60-j 80)}{(67+j 84)\left(20 \angle 32^{\circ}\right)}

(c) \left(\frac{10+j 20}{3+j 4}\right)^{2} \sqrt{(10+j 5)(16-j 120)}

Accepted Answer

(a) [latex]\frac{3-\mathrm{jl} 4}{-7+\mathrm{jl} 7}=\frac{14.318 \angle-77.91^{\circ}}{18.385 \angle 112.38^{\circ}}=0.7788 \angle 169.71^{\circ}=\frac{-0.7663+\mathrm{j} 0.13912}{-10}[/latex]

(b) [latex]\frac{(62.116+j 231.82+138.56-j 80)(60-j 80)}{(67+j 84)(16.96+j 10.5983)}=\frac{24186-j6944.9}{246.06+j 2134.7}=-1.922-j 11.55[/latex]

(c) [latex](-2+j 4)^{2} \sqrt{(260-j 120)}=\left(20 \angle-126.86^{\circ}\right)\left(16.923 \angle-12.38^{\circ}\right)[/latex]
[latex]=338.46 \angle-139.24^{\circ}=-256.4-j 221[/latex]

Question : (a) (5-j 6)-(2+j 8)/(-3+j 4)(5-j)+(4-j 6) (b) (240 ∠75° +160...