Complex Equations

If two complex numbers are equal, then their real parts are equal and their imaginary parts are equal. Hence, if a+jb=c+jd, then a=c and b=d. This is a useful property, since equations having two unknown quantities can be solved from one equation. Complex equations are used when deriving balance equations with AC bridges.

Example 7.3

Solve the following complex equations:

(a) 3(a+jb)=9 –j2

(b) (2 1 j)(–2+j)=x+jy

(c) (a 2 j2b)+(b –j3a)=5+j2

Solution

.

(a) 3(a+jb)=9 –j2. Thus, 3a+j3b=9 –j2

Equating real parts gives: 3a=9, i.e., a =3

Equating imaginary parts gives:

(b) (2 1 j)(–2+j)=x+jy

Thus, 

Equating real and imaginary parts gives: x =–5, y =0

(c) (a 2 j2b)+(b –j3a)=5+j2

Thus, (a+b)+j(–2b – 3a)=5 +j2

(7.1)

(7.2)

We have two simultaneous equations to solve. Multiplying equation (7.1) by 2 gives:

(7.3)

Adding equations (7.2) and (7.3) gives –a=12, i.e., a =–12

From equation (7.1), b =17

Example 7.4

An equation derived from an AC bridge network is given by:

R₁, R₃, R₄ and C₄ are known values. Determine expressions for R₂ and L₂ in terms of the known components.

Solution

Multiplying both sides of the equation by (1/R₄+jωC₄) gives:

i.e., 

Equating the real parts gives: R₂=R₁R₃/R₄

Equating the imaginary parts gives:

ωL₂=R₁R₃ωC₄, from which, L₂=R₁R₃C₄