If we let a= 4, b= 5, c= 3, and d=−8, then we havec^2 +d^2 = 32 + (−8)^2= 9 + 64
= 73
and therefore(4+j5) / (3 −j8)
= [4 × 3 + 5 × (−8)] / 73 +j{5× 3 − [4 × (−8)]} / 73
= (12 − 40) / 73 +j(15+ 32) / 73
=−28/73+j(47/73)- To find (a+jb)− (a−jb), we multiply the second complex number through by −1, and
then add the real parts and the imaginary parts separately, getting
(a+jb)− (a−jb)= (a+jb)+ [−1(a−jb)]= (a+jb)+ (−a+jb)
=a+ (−a)+jb+jb
=j 2 b
To find (a−jb)− (a+jb), we again multiply the second complex number through by −1,
and then add the real parts and the imaginary parts separately, getting(a−jb)− (a+jb)= (a−jb)+ [−1(a+jb)]= (a−jb)+ (−a−jb)
=a+ (−a)+ (−jb)+ (−jb)
=−j 2 b
Note that in these answers, the numerals 2 are not exponents! We have j times the quan-
tity 2b in the first case, and −j times the quantity 2b in the second case.- To find (a+jb) / (a−jb), let’s first change the subtraction in the denominator to
negative addition. That will give us the expression
(a+jb) / [a+j(−b)]Now we can use the quotient formula for complex numbers. Let’s state it again for
reference. If a,b,c, and d are real numbers, and as long as c and d aren’t both equal
to 0, then(a+jb) / (c+jd)
= (ac+bd) / (c^2 +d^2 )+j(bc−ad) / (c^2 +d^2 )Chapter 21 667