Understanding Engineering Mathematics

Problem 12.3

Multiply (3−j)and(2Y 3 j).

( 3 −j)( 2 + 3 j)= 6 + 9 j− 2 j− 3 j^2 = 6 + 7 j− 3 (− 1 )= 9 + 7 j

Given a complex numberz=a+jbthere is a very useful operation we can perform,
that converts the number to its ‘conjugate’. Thus, thecomplex conjugateof a complex
numberz=a+ibis defined as

̄z=a+(−j)b (alternative notationz∗)

NB:In general the complex conjugate of a complex expression is obtained by changing
the sign ofjeverywhere it occurs.For examples of this see Reinforcement Exercise 5,
and 9.

Problem 12.4

Show thatz ̄z is a positive real number which is only zero ifa=b=0,

i.e. ifz=0. Also show thatzYz ̄is always a real number.

This is the crux of the importance ofz ̄.Wehave

̄zz=(a−jb)(a+jb)=a^2 −(jb)^2 =a^2 −j^2 b^2 =a^2 +b^2

on using the difference of two squares (42
➤
). Nowa^2 andb^2 are both non-negative
numbers and sozz ̄is clearly positive. Also, the result can only add up to zero if botha
andbare zero. Sozz ̄ =0 only ifa=0andb=0, i.e. ifz=0. We will see that these
properties ofz ̄zare very important when we come to dividing complex numbers.
Also, we have directly

z+ ̄z=a+jb+a−jb= 2 a

which is a real number.
The complex conjugate is also useful for characterising the real and imaginary parts of
a complex number. Thus, as you can easily check, ifzis purelyreal,thenz ̄=zwhile if
zis purelyimaginary,thenz ̄=−z. In general, you can see that (see Problem 12.4)

Rez=

z+ ̄z

2

Imz=

z− ̄z

2 j

Problem 12.5

Forz= 2 Y 3 jevaluatez ̄z,Rezand Imz.

Forz= 2 + 3 jwe havez ̄= 2 − 3 jand so

z ̄z= 22 + 32 = 13

Rez=

2 + 3 j+ 2 − 3 j

2

= 2 , Imz=

2 + 3 j−( 2 − 3 j)

2 j

= 3

which last results are obvious by inspection ofz.