(a) 化簡 (3+2i)(1-2i)/(4-i)(1+2i),並以a+bi表示答案 (b) 若(3+2i)(1-2i)/(4-i)(1+2i)=(7/4)x-5yi,求數式中x和y的值。?

(a)化簡 (3+2i)(1－2i)/[(4－i)(1+2i)]，並以a+bi表示答案
(3+2i)(1－2i)/[(4－i)(1+2i)]
=(3－6i+2i－4i^2)/(4+8i－i－2i^2)
=(3－6i+2i+4)/(4+8i－i+2)
=(7－4i)/(6+7i)
=(7－4i)(6－7i)/(36+49)
=(42－49i－24i－28)/(36+49)
=(14－73i)/85
=(14/85)－(73/85)i
or
(3+2i)(1－2i)/[(4－i)(1+2i)]
=(3+2i)(1－2i)(4+i)(1－2i)/[(16+1)*(1+4)]
=(3+2i)(4+i)(1－2i)(1－2i)/85
=(12+3i+8i－2)(1－2i－2i－4)/85
=(10+11i)(－3－4i)/85
=(－30－40i－33i+44)/85
=(14－73i)/85
=(14/85)－(73/85)i
(b)若(3+2i)(1-2i)/[(4-i)(1+2i)]=(7/4)x-5yi,求數式中x和y的值?
Sol
x=(14/85)/(7/4)=(14/85)*(4/7)=8/85
y=(－73/85)/(－5)=73/425

a)
(3+2i)(1-2i)/(4-i)(1+2i)
=(14/85)-(73/85)i

b)
(3+2i)(1-2i)/(4-i)(1+2i)=(7/4)x-5yi
(14/85)-(73/85)i=(7/4)x-5yi
14/85=(7/4)x
x=8/85
-73/85=-5y
y=73/425