Complex Analysis - Maths KU

40 Chapter 1 Complex Numbers and the Complex Plane
Although the proof is postponed until Section 4.1, (9) can be used to show
that the familiar law of exponents holds for complex numbers z1 and z2:
e z1 e z2 = e z1+z2 . (10)
Because of (10), the results in (7) are valid.Moreover, note that Euler’s formula
(4) is a special case of (9) when z is a pure imaginary number, that is,
with x = 0 and y replaced by θ.Euler’s formula provides a notational convenience
for several concepts considered earlier in this chapter.The polar form
of a complex number z, z = r(cos θ + i sin θ), can now be written compactly
as
z = re iθ . (11)
This convenient form is called the exponential form of a complex number z.
For example, i = e πi/2 and 1 + i = √ 2e πi/4 . Also, the formula for the n nth
roots of a complex number, (4) of Section 1.4, becomes
z 1/n = n√ re i(θ+2kπ)/n ,k =0, 1, 2,... ,n− 1. (12)
Electrical Engineering In applying mathematics to physical situations,
engineers and mathematicians often approach the same problem in
completely different ways.For example, consider the problem of finding the
steady-state current ip(t) inanLRC -series circuit in which the charge q(t) on
the capacitor for time t>0 is described by the differential equation
L d2q dq
+ R
dt2 dt
+ 1
C q = E0 sin γt (13)
where the positive constants L, R, and C are, in turn, the inductance, resistance,
and capacitance.Now to find the steady-state current ip(t), we
first find the steady-state charge on the capacitor by finding a particular
solution qp(t) of (13).Proceeding as we would in a course in differential equations,
we will use the method of undetermined coefficients to find qp(t).This
entails assuming a particular solution of the form qp(t) =A sin γt + B cos γt,
substituting this expression into the differential equation, simplifying, equating
coefficients, and solving for the unknown coefficients A and B.It is left
as an exercise to show that A = E0X/(−γZ 2 ) and B = E0R/(−γZ 2 ), where
the quantities
X = Lγ − 1/Cγ and Z = � X 2 + R 2 (14)
are called, respectively, the reactance and impedance of the circuit.Thus,
the steady-state solution or steady-state charge in the circuit is
qp(t) =− E0X E0R
sin γt − cos γt.
γZ2 γZ2 From this solution and ip(t) =q ′ p(t) we obtain the steady-state current:
ip(t) = E0
� �
R X
sin γt − cos γt . (15)
Z Z Z