14.12.2012 Views

Complex Analysis - Maths KU

Complex Analysis - Maths KU

Complex Analysis - Maths KU

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

1.6 Applications 39<br />

c = 0.The latter quadratic equation is known as the auxiliary equation.<br />

Now when the coefficients of a polynomial equation are real, the equation cannot<br />

have just one complex root; that is, complex roots must always appear<br />

in conjugate pairs.Thus, if the auxiliary equation possesses complex roots<br />

α + iβ, α − iβ, β > 0, then two solutions of ay ′′ + by ′ + cy = 0 are complex<br />

exponential functions y = e (α+iβ)x and y = e (α−iβ)x .In order to obtain real<br />

solutions of the differential equation we use Euler’s formula<br />

e iθ = cos θ + i sin θ, (6)<br />

where θ is real.With θ replaced, in turn, by β and –β, we use (6) to write<br />

e (α+iβ)x = e αx e iβx = e αx (cos βx + i sin βx) and e (α−iβ)x = e αx e −iβx = e αx (cos βx − i sin βx).(7)<br />

Now since the differential equation is homogeneous, the linear combinations<br />

y1 = 1<br />

�<br />

e<br />

2<br />

(α+iβ)x + e (α−iβ)x�<br />

and y2 = 1<br />

�<br />

e<br />

2i<br />

(α+iβ)x − e (α−iβ)x�<br />

are also solutions.But in view of (7), both of the foregoing expressions are<br />

real functions<br />

y1 = e αx cos βx and y2 = e αx sin βx. (8)<br />

EXAMPLE 2 Solving a Differential Equation<br />

Solve the differential equation y ′′ +2y ′ +2y =0.<br />

Solution We apply the quadratic formula to the auxiliary equation m 2 +<br />

2m+2 = 0 and obtain the complex roots m1 = −1+i and m2 = m1 = −1−i.<br />

With the identifications α = −1 and β = 1, (8) gives us two solutions<br />

y1 = e −x cos x and y2 = e −x sin x.<br />

You may recall that the so-called general solution of a homogeneous linear<br />

nth-order differential equations consists of a linear combination of n linearly<br />

independent solutions.Thus in Example 2, the general solution of the<br />

given second-order differential equation is y1 = c1y1 + c2y2 = c1e −x cos x +<br />

c2e −x sin x, where c1 and c2 are arbitrary constants.<br />

Exponential Form of a <strong>Complex</strong> Number We hasten to point<br />

out that the results given in (6) and (7) were assumptions at that point<br />

because the complex exponential function has not been defined as yet.As a<br />

brief preview of the material in Sections 2.1 and 4.1, the complex exponential<br />

e z is the complex number defined by<br />

e z = e x+iy = e x (cos y + i sin y). (9)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!