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lec31.ppt
1. NPTEL β Physics β Mathematical Physics - 1
Module 6
Lecture 31
Complex analysis
Complex Numbers
Consider an equation, π₯2 + 1 = 0. No real number satisfies this equation. To allow for a
solution of this equation, complex numbers can be introduced. They are not only
confined to the real axis. This complex numbers are pairs of numbers that denote
coordinates of points in the complex plane.
Real numbers, including zero and negative numbers, integers or fractions, rational and
irrational numbers can be represented on a line called the real axis as shown below.
Thus, conversely corresponding to each point on the line, there is a real number.
The coordinates of A represent a complex number, (π₯, π¦). Since B lies on the real axis,
the coordinate of B is represented by a real number and for a point C, it is
purely imaginary.
Thus a complex no. is defined as π§ = π₯ + ππ¦ where x and y are real and i is an
imaginary quantity which has a value ββ1 .
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2. NPTEL β Physics β Mathematical Physics - 1
Properties of Complex numbers
1.Complex numbers, π§1 = π₯1 + ππ¦1 and π§2 = π₯2 + ππ¦2 are added as
π§1 + π§2 = (π₯1 + π₯2 ) + π(π¦1 + π¦2 )
2.Two complex numbers, π§1 = π₯1 + ππ¦1 and π§2 = π₯2 + ππ¦2 when multiplied yields,
π§1π§2 = (π₯1 + ππ¦1)(π₯2 + ππ¦2) = (π₯1π₯2 β π¦1π¦2) + π(π¦1π₯2 + π₯1π¦2)
3.Inverse of a complex number is found as in the following,
Let π§β1 = π’ + ππ£ such that
(π’ + ππ£)(π₯ + ππ¦) = 1
π₯π’ β π¦π£ = 1
π¦π’ + π₯π£ = 0} π’ =
1 + π¦π£
π₯
π¦ (
1 + π¦π£
π₯
) + π₯π£ = 0
β + π£ + π₯π£ = 0
π¦
π₯ π₯
π¦2
β π¦ + (π¦2 + π₯2)π£ = 0 β π£ = β
π¦
π₯2 + π¦2
Similarly π’ =
π₯
π₯2+π¦ 2
Thus, π§β1 = (
π₯ βπ¦
π₯2+π¦ 2 π₯2+π¦2
, ) (π§ β 0)
4. The binomial formula for complex numbers is
(π§1 + π§2)π = βπ
(π
)π§1
πβπ π§ π
π=0 π
π!
2 (π = 1,2β¦ . . )
where (π
) =
π
Also 0! = 1
π!(πβπ)!
π = 0,1,2 β¦ β¦ β¦ β¦ β¦ β¦ β¦ π
5. The equation |π§ β 1 + 3π| = 2 represents the circle whose center is π§0 = (1, β3)
and radius is π = 2 where |π§| denotes the magnitude and is defined as βπ₯2 + π¦2
|(π₯ β 1) + π(3 + π¦)| = 2
Thus, (π₯ β 1)2 + (π¦ + 3)2 = 22
So the center lies at (1, β3π) in the complex plane and the radius is 2.
6. |π§ + 4π| + |π§ β 4π| = 10 represents an ellipse with foci at (0, Β±4).
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3. NPTEL β Physics β Mathematical Physics - 1
|π₯ + (π¦ + 4)π| + |π₯ + (π¦ β 4)π| = 10
βπ₯2 + (π¦ + 4)2 + βπ₯2 + (π¦ β 4)2 = 10
β
π₯2 + (π¦ + 4)2
10
+ β
π₯2 + (π¦ β 4)2
10
= 1
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The foci of the ellipse are at (0,
Β±4)