Question B.4: Evaluate: (a) (2 + j5)(8e^j10°)/2 + j4 + 2 − 40° (b) j(3 − j...
Evaluate:
\text { (a) } \frac{(2+j 5)\left(8 e^{j 10^{\circ}}\right)}{2+j 4+2 \underline{/-40^{\circ}} } \text { (b) } \frac{j(3-j 4)^{*}}{(-1+j 6)(2+j)^{2}}
(a) Since there are terms in polar and exponential forms, it may be best to express all terms in polar form:
2+j 5=\sqrt{2^{2}+5^{2}} \underline{/ \tan ^{-1} 5 / 2} =5.385\underline{ / 68.2^{\circ}}
(2+j 5)\left(8 e^{j 10^{\circ}}\right)=\left(5.385\underline{ / 68.2^{\circ}} \right)\left(8\underline{ / 10^{\circ}} \right)=43.08 \underline{/ 78.2^{\circ}}
2+j 4+2 \underline{/-40^{\circ}} =2+j 4+2 \cos \left(-40^{\circ}\right)+j 2 \sin \left(-40^{\circ}\right)
=3.532+j 2.714=4.454\underline{ / 37.54^{\circ}}
Thus,
\frac{(2+j 5)\left(8 e^{j 10^{\circ}}\right)}{2+j 4+2\underline{ /-40^{\circ}} }=\frac{43.08 \underline{/ 78.2^{\circ}} }{4.454\underline{ / 37.54^{\circ}} }=9.672\underline{ / 40.66^{\circ}}(b) We can evaluate this in rectangular form, since all terms are in that form. But
j(3-j 4)^{*} = j(3 + j4) = −4 + j3
(2 + j)² = 4 + j4 − 1 = 3 + j4
(−1 + j6)(2 + j)² = (−1 + j6)(3 + j4) = −3 − 4j + j18 − 24
= −27 + j14
Hence,
\frac{j(3-j 4)^{*}}{(-1+j 6)(2+j)^{2}}=\frac{-4+j 3}{-27+j 14}=\frac{(-4+j 3)(-27-j 14)}{27^{2}+14^{2}}
=\frac{108+j 56-j 81+42}{925}=0.1622-j 0.027