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z −1 = 1 = 1   a − jb  = a − jb = z 2 (6.10)
2
z ( a + jb  ) a − jb a + b 2 z
from which we deduce that:

 1 Re( ) z
Re = (6.11)
 z  [Re( )] + [Im( )] 2
2
z
z
and
1

Im  = − Im( ) z (6.12)
 z  [Re( )] + [Im( )] 2
2
z
z
To summarize the above results, and to help you build your syntax for the
quantities defined in this section, edit the following script M-file and execute it:
z=3+4*j
zbar=conj(z)
modulz=abs(z)
modul2z=z*conj(z)
invz=1/z
reinvz=real(1/z)
iminvz=imag(1/z)

In-Class Exercises
Pb. 6.8 Analytically and numerically, obtain in the standard form an
expression for each of the following quantities:

34+ j 3 + j  12− j 3 +  j
a. b. c.  − 
25+ j ( 1− j)( 3 + j)  23+ j 2j 

Pb. 6.9 For any pair of complex numbers z and z , show that:
2
1
z + z = z + z
1 2 1 2
z − z = z − z
1 2 1 2
zz = zz
1 2 1 2
z (/ z ) = z / z
1 2 1 2
z = z

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