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

352 Chapter 9 Complex numbers
y
y
z
P
z such that |z – 2| = 3
1 2 3 4 5
x
A
0 1 2
x
Figure9.23
Pointszand2 +0j.
Figure9.24
Locusofpointssatisfying |z −2| = 3.
Recall from Section 9.6 that the graphical representation of the sum and difference of
vectors in the plane, and the sum and difference of complex numbers, are equivalent.
Sincevector ⃗ OPrepresentsthecomplexnumberz,andvector ⃗ OArepresentsthecomplex
number2, ⃗ AP = ⃗ OP− ⃗ OAwillrepresentz−2.Therefore |z−2|representsthedistance
betweenAandP.Wearegiventhat |z−2| = 3,whichthereforemeansthatPcanbeany
point such that its distance from A is 3. This means that P can be any point on a circle
ofradius3centredatA(2,0).ThelocusisshowninFigure9.24. |z −2| < 3represents
the interior of the circle while |z −2| > 3 represents the exterior. Alternatively we can
obtain the same result algebraically: given |z −2| = 3 and also thatz =x +jy, we can
write
|z−2|=|(x−2)+jy|=3
that is
√
(x−2)2 +y 2 =3
or
(x−2) 2 +y 2 =9
Generally,theequation (x−a) 2 +(y−b) 2 =r 2 representsacircleofradiusrcentredat
(a,b), soweseethat (x −2) 2 +y 2 = 9representsacircleofradius3centredat (2,0),
as before.
Example9.25 Usethe algebraicapproachtofind the locusofthe pointzwhichsatisfies
|z−1|= 1 2 |z−j|
Solution Ifz =x +jy, then we have
|(x−1)+jy|= 1 |x +j(y −1)|
2
Therefore,
(x−1) 2 +y 2 = 1 4 {x2 + (y−1) 2 }