082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

Similarly, wefind, by equating imaginary parts
6=−y
so thaty = −6.
9.3 Operations with complex numbers 329
Theoperationsofaddition,subtraction,multiplicationanddivisioncanallbeperformed
on complex numbers.
9.3.1 Additionandsubtraction
Toaddtwocomplexnumberswesimplyaddtherealpartsandaddtheimaginaryparts;
tosubtractacomplexnumberfromanotherwesubtractthecorrespondingrealpartsand
subtractthe corresponding imaginary parts as shown inExample 9.6.
Example9.6 Ifz 1
= 3 −4jandz 2
= 4 +2jfindz 1
+z 2
andz 1
−z 2
.
Solution
z 1
+z 2
=(3−4j)+(4+2j)
=(3+4)+(−4+2)j
=7−2j
z 1
−z 2
=(3−4j)−(4+2j)
=(3−4)+(−4−2)j
=−1−6j
9.3.2 Multiplication
Wecanmultiplyacomplexnumberbyarealnumber.Boththerealandimaginaryparts
of the complex number aremultiplied by the real number. Thus 3(4 −6j) = 12 −18j.
To multiply two complex numbers we use the factthat j 2 = −1.
Example9.7 Ifz 1
= 2 −2jandz 2
= 3 +4j, findz 1
z 2
.
Solution
z 1
z 2
= (2 −2j)(3 +4j)
Removing brackets wefind
z 1
z 2
=6−6j+8j−8j 2
=6−6j+8j+8 usingj 2 =−1
=14+2j
Example9.8 Ifz = 3 −2jfindzz.
Solution Ifz = 3 −2jthen itsconjugate isz = 3 +2j. Therefore,
zz = (3 −2j)(3 +2j)
=9−6j+6j−4j 2
=9−4j 2
= 13
We see thatthe answer isareal number.