Complex Analysis - Maths KU

1.6 Applications 41
Electrical engineers often solve circuit problems such as this by using complex
analysis.First of all, to avoid confusion with the current i, an electrical
engineer will denote the imaginary unit i by the symbol j; in other words,
j 2 = −1. Since current i is related to charge q by i = dq/dt, the differential
equation (13) is the same as
L di
dt
+ Ri + 1
C q = E0 sin γt. (16)
Now in view of Euler’s formula (6), if θ is replaced by the symbol γ, then
the impressed voltage E0 sin γt is the same as Im(E0e jγt ).Because of this
last form, the method of undetermined coefficients suggests that we try a
solution of (16) in the form of a constant multiple of a complex exponential,
that is, ip(t) = Im(Ae jγt ). We substitute this last expression into equation
(16), assume that the complex exponential satisfies the “usual” differentiation
rules, use the fact that charge q is an antiderivative of the current i, and equate
coefficients of e jγt .The result is (jLγ + R +1/jCγ) A = E0 and from this
we obtain
E0
A = �
R + j Lγ − 1
� =
Cγ
E0
,
R + jX
(17)
where X is the reactance given in (14).The denominator of the last
expression, Zc = R + j(Lγ − 1/Cγ) = R + jX, is called the complex
impedance � of the circuit.Since the modulus of the complex impedance
is |Zc| = R2 +(Lγ − 1/Cγ) 2 , we see from (14) that the impedance Z and
the complex impedance Zc are related by Z = |Zc|.
Now from the exponential form of a complex number given in (11), we
can write the complex impedance as
Zc = |Zc| e jθ = Ze jθ where
Lγ −
tan θ =
1
Cγ
.
R
Hence (17) becomes A = E0/Zc = E0/(Zejθ ), and so the steady-state current
is given by
�
E0
ip(t) =Im
Z e−jθe jγt
�
. (18)
You are encouraged to verify that the expression in (18) is the same as that
given in (15).
Remarks Comparison with Real Analysis
We have seen in this section that if z1 is a complex root of a polynomial
equation, then z2 =¯z1 is another root whenever all the coefficients of the
polynomial are real, but that ¯z1 is not necessarily a root of the equation
when at least one coefficient is not real.In the latter case, we can obtain
another factor of the polynomial by dividing it by z − z1.We note that
synthetic division is valid in complex analysis.See Problems 25 and 26
in Exercises 1.6.