Mathematical Physics 1 2 ae

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1 
 Prof. A.F.Guimarães 
Mathematical Physics 1 – Problems 2a 
 Problem 1
Verify the followings rules for scalar and vector products. 
( ! ∙ #) = $%(&*̅&+) = $%(&*&+̅), [ ! × #] = ,-(&*̅&+) = −,-(&*&+̅). 
Solution: 
Let ! and # be the complex numbers given by: 
 ! = 0*(cos 1 + 3 sin 1) # = 0+(cos 4 + 3 sin 4) 
(1.1) 
Let us calculate &*̅&+, 
&*̅&+ = 0*0+(cos 1 + 3 sin 1)(cos 4 − 3 sin 4) &*̅&+ = 0*0+[cos 4 cos 1 + sin 4 sin 1 +3(sin 4 cos 1 − sin 4 cos 1)] ∴ &*̅&+ = 0*0+[cos(4 − 1) + 3 sin(4 − 1)] 
(1.2) 
Where | !| = 0* and | #| = 0*. Another product (&*&+̅) is 
given by: 
&*&+̅ = 0*0+[cos(4 − 1) − 3 sin(4 − 1)] 
(1.3) 
The scalar product is given by: 
( ! ∙ #) = |&*||&+| cos(4 − 1) 
 (1.4) 
By the use of the result from (1.2) or (1.3) we can write, 
( ! ∙ #) = $%(&*̅&+) = $%(&*&+̅) 
(1.5) 
And the product of vectors, 
[ ! × #] = |&*||&+| sin(4 − 1) 
(1.6) 
Hence, 
[ ! × #] = ,-(&*̅&+) = −,-(&*&+̅) 
(1.7) 
 Problem 2
Show that: $% &* $% &+ = *+ $%(&*&+) + *+ $%(&*&+̅). 
Solution: 
Consider &* = 6 + 37 and &+ = 8 + 39. Then: 
$% &* $% &+ = 68 
(2.1) 
Let us now calculate the following products, &*&+ and &*&+̅. 
So: 
 &*&+ = 68 − 79 + 3(69 + 87) &*&+̅ = 68 + 79 + 3(87 − 69) 
(2.2) 
 
By the use of (2.2), we have 
 $%(&*&+) + $%(&*&+̅) = 268 = 2($% &* $% &+) 
∴ $% &* $% &+ = 12 [$%(&*&+) + $%(&*&+̅)] 
(2.3) 
 
 Problem 3
 
Prove that, if &(3 − 1) = −&̅(3 + 1), then arg & is either <
> or − ?<> . 
Solution: 
Let us choose & = 6 + 37, then, 
 (6 + 37)(3 − 1) = −(6 − 37)(3 + 1) 63 − 6 − 7 − 37 = −63 − 6 − 7 + 37 23(6 − 7) = 0 ∴ 6 = 7 
(3.1) 
 
The argument of & is given by: 
 
arg & = tgA* 76 ∴ arg & = tgA* 1 
(3.2) 
 
But −B < arg & ≤ B, then arg & = <> or arg & = − ?<> . 
 
 Problem 4
 
If 6 and 7 are two complex numbers and E is a real 
parameter, then the expression & = 6 + E(7 − 6) represents 
a curve in complex plane. Set & = F + 3G and determine the 
parametric equations F = F(E) e G = G(E) of this curve. 
What kind of curve is it? 
Solution 
The numbers 6 and 7 are given by 6 = 6* + 36+ and 7 =7* + 37+. Here, 7* − 6* ≠ 0 and 7+ − 6+ ≠ 0. So we can 
write: 
 & = 6* + 36+ + E(7* + 37+ − 6* − 36+) ∴ & = 6* + E(7* − 6*) + 3[6+ + E(7+ − 6+)] 
(4.1) 
 
As & = F + 3G we have: 
 F = 6* + E(7* − 6*) G = 6+ + E(7+ − 6+) 
(4.2) 
 
 
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2 
The equations in (4.2) represent a straight line. 
 Problem 5
Show that the equation & = I%JK + L%AJK, where I and L are 
complex constants and E is a real parameter, represents an 
ellipse. Describe this ellipse (its semi axes, centre, 
orientation and etc.) in terms of I and L. 
Solution 
Consider the following complex constants: 
I = 6%JM; L = 7%JO 
(5.1) 
Here 6 = |I|, 7 = |L| and 4 and P are real constants. Hence: 
& = 6 8QR(E + 4) + 7 8QR(E − P) +3[6 R3S(E + 4) − 7 R3S (E − P)] 
(5.2) 
If we write & = F + 3G, then: 
F = 6 8QR(E + 4) + 7 8QR(E − P) G = 6 R3S(E + 4) − 7 R3S (E − P) 
(5.3) 
The expressions in (5.3) represent the parametric equations 
of the aforementioned ellipse, whose centre is at origin of 
complex plane. The magnitude of & is given by: 
|&| = T6+ + 7+ + 267 8QR(2E + 4 − P) 
(5.4) 
According of (5.4), we can conclude that the major semi axis 
of the ellipse is given by: 
|&|UVW = T6+ + 7+ + 267 = 6 + 7 ∴ |&|UVW = |I| + |L| 
(5.5) 
As long as cos(2E + 4 − P) = 1. In this case, the parameter E 
becomes: EUVW = OAM+ . And the minor semi axis is given by: 
|&|UJX = T6+ + 7+ − 267 = 6 − 7 ∴ |&|UJX = |I| − |L| 
(5.6) 
As long as cos(2E + 4 − P) = −1. In this case, the parameter 
E becomes: EUJX = OAMY<+ , once that −B < E ≤ B. The 
orientation of the major semi axis is given by the help of 
(5.3), with EUáW . Then: 
EZ1 = 6 R3S \
P + 42 ^ + 7 R3S \P + 42 ^
6 8QR \P + 42 ^ + 7 8QR \P + 42 ^
= EZ \P + 42 ^ 
∴ 1 = P + 42 
(5.7) 
We have the following expression for the ellipse’s 
eccentricity: 
 
% = T|&|UáW+ − |&|UíX+|&|UáW =
2√67
6 + 7 
∴ % = 2T|I||L||I| + |L| 
(5.8) 
 
By the help of (5.7) and (5.8), we can write the following 
expression for the focuses: 
 
*` = %|&|UVW%J(bA<) 
+` = %|&|UVW%Jb 
(5.9) 
 
 
For an example, let us take two complex numbers A and B: 
 
I = 2%J<?; L = %J<f 
(5.10) 
 
By the use of (5.3) we have: 
 
F = 2 8QR hE + B3k + 8QR hE −
B
6k 
G = 2 R3S hE + B3k − R3S hE −
B
6k 
(5.11) 
 
According of (5.5) and (5.6), the major semi axis has 
magnitude 3 and the minor semi axis has magnitude 1. The 
orientation of major semi axis, by the use of (5.7), is mp rad. 
The figure 5.1 shows the ellipse (blue) on the plane. 
 
Figure 5.1 – The ellipse (blue) resultants from the addition of Aeit with Be-it. 
Major axis (red) and minor axis (yellow). 
 
From figure 5.1 on we can associate the situation as being 
the addition of two vectors whose endpoints rotate around 
-2,5
-1,5
-0,5
0,5
1,5
2,5
-2,5 -1,5 -0,5 0,5 1,5 2,5
B 
A 
z 
q = uvwxy
 
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the plane centre. The number A as the endpoint of a vector A 
(purple) that rotates in anti-clockwise and the number B, as 
the endpoint of a vector B (green) that rotates in clockwise. 
The result is a vector z (blue) whose endpoint rotates in 
anti-clockwise. In yellow, the minor axis with magnitude 1. 
In red, the major axis with magnitude 3. 
 Problem 6
Show that the transformation: 
& + 6 = z; (6 = complex constant) 
represents a translation in the complex plane. Show that the 
equation: 
(& + 1 − 3)(&̅ + 1 + 3) = 1 
represents a circumference in the complex plane. 
[Suggestion: Set & + 1 − 3 = z.] 
Solution 
Let us do & = F + 3G and 6 = 6W + 36. Then, 
& + 6 = F + 6W + 3‚G + 6ƒ 
(6.1) 
Now, let us write z = zW + 3z, so we can conclude that 
zW = F + 6W z = G + 6 
(6.2) 
The expressions in (6.2) represent the translation in both 
axes F and G. So we have a translation on the complex plane. 
Now, by the help of suggestion, we have 
(& + 1 − 3)(&̅ + 1 + 3) = z ∙ z ̅ = 1 
(6.3) 
 
But z ∙ z ̅ = |z|+. Then we can write 
zW+ + z+ = 1 
(6.4) 
The expression in (6.4) represents a circumference with 
radius 1. As soon as we use (6.2), we have the pair (−1,1) as 
the centre of the circumference. 
 Problem 7
Show that the equations: 
a) |& − 1| − |& + 1| = 1 
b) $%(1 − &) = |&| 
represent, respectively, a hyperbola and a parabola in the 
complex plane. Can these conclusions be drawn just by the 
inspection of the equations without any algebraic 
manipulations? [Suggestion: Remember of the basic 
properties of hyperbola and parabola] 
Solution: 
a) Let us calculate the modulus: 
 
|& − 1| = T(F − 1)+ + G+ 
(7.1) 
 
And 
 
|& + 1| = T(F + 1)+ + G+ 
(7.2) 
 
Then 
 
|& − 1| − |& + 1| = T(F − 1)+ + G+ − T(F + 1)+ + G+ = 1 
(7.3) 
 
After some manipulations, 
 
(F − 1)+ = 1 + (F + 1)+ + 2T(F + 1)+ + G+ 4F + 1 = 4[(F + 1)+ + G+] 12F+ − 4G+ = 3 
(7.4) 
 
The result from (7.4) can be written in the form of reduced 
equation as follows: 
 F+
1 4† −
G+
3 4†
= 1 
(7.5) 
 
The expression in (7.5) is represented in figure 7.1. 
Definition of hyperbola: For two given different points ‡* and ‡+, on a plane and 28 as the distance between them. ˆ 
is the medium point of the segment ‡*‡+‰‰‰‰‰‰. A hyperbola is 
defined as the set of points on plane whose the difference (in 
modulus) of the distances to ‡* and ‡+ is the 2a constant (0 < 26 < 28). So, by the inspection of the equation in “a”, 
we conclude that |& − 1| is the distance between the points z 
and 1, and |& + 1| is the distance between the points z and −1.In this case, there is a possibility to reach the conclusion 
about hyperbola by the inspection of the equation. 
b) The real part of equation is given by: 
 $%(1 − &) = 1 − F 
(7.6) 
i 
1 -1 Re 
z 
| − 1| 
| + 1| 
1
2 1 −
1
2 
Figure 7.1 – Hyperbola on complex plane. 
 
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Then 
1 − F = TF+ + G+ 
(7.7) 
After some manipulation we have 
G+ = −2 \F − 12^ 
(7.8) 
 
The expression in (7.8) represents a reduced equation of a 
parabola, with its focus at Ox axis (origin) as shown in the 
figure 7.2. 
Definition of parabola: For a given point ‡ and a straight line 9 on a plane, with ‡ ∉ 9, a parabola is defined as the set of 
points of the plane whose distance to point ‡ is the same as 
the distance to 9. 
In this case, there is also a possibility to reach the conclusion 
about parabola by the inspection of the equation. 
 Problem 8
Prove the inequalities below: 
a) |1 + &| ≤ 1 + |&| 
b) |1 − &| ≥ Œ1 − |&|Œ 
And then generalize them by setting & = &+ &*⁄ . (What if &* = 0?) 
Solution: 
Let us demonstrate the inequalities by a general way instead 
of as they have been proposed. 
a) Let the modulus given by: 
|&* + &+| 
(8.1) 
So the squared of (8.1) is given by: 
|&* + &+|+ = (&* + &+) ∙ (&* + &+‰‰‰‰‰‰‰‰‰) 
(8.2) 
Let the property given by: 
&* + &+‰‰‰‰‰‰‰‰‰ = &*̅ + &+̅ 
(8.3) 
 
So, the expression (8.2) becomes 
 |&* + &+|+ = (&* + &+) ∙ (&*̅ + &+̅) 
(8.4) 
 
Let us calculate the left side product. Then 
 |&* + &+|+ = &*&*̅ + &+&+̅ + &*&+̅ + &+&*̅ 
(8.5) 
 
We can write the following relationship for the last two 
terms of the addition: 
 &*&+̅ + &+&*̅ = &*&+̅ + &*&+̅‰‰‰‰‰‰ = 2$%&*&+̅ 
(8.6) 
 
And let us use the following property 
 $% & ≤ |$% &| ≤ |&| 
(8.7) 
 
So, the expression in (8.6) can be written as: 
 &*&+̅ + &+&*̅ = &*&+̅ + &*&+̅‰‰‰‰‰‰ = 2$%&*&+̅ ≤ 2|&*&+̅| = 2|&*||&+̅| 
(8.8) 
 
Where |&+̅| = |&+|. The expression in (8.5) becomes 
 |&* + &+|+ ≤ |&*|+ + |&+|+ + 2|&*||&+| 
(8.9) 
 
Or, 
 |&* + &+|+ ≤ (|&*| + |&+|)+ 
(8.10) 
 
We can calculate the square root in both sides of (8.10) 
because both are positives. Then 
 |&* + &+| ≤ |&*| + |&+| 
(8.11) 
 
In (8.11), if &* = 0, we can consider only the equality. 
 
b) Let us follow the same way as we have done in ‘a’. So 
 |&* − &+|+ = (&* − &+) ∙ (&*̅ − &+̅) 
(8.12) 
 
This leads us to 
 |&* − &+|+ = |&*|+ + |&+|+ − (&*&+̅ + &+&*̅) 
(8.13) 
 
From (8.8), we can write 
 −(&*&+̅ + &+&*̅) = −(&*&+̅ + &*&+̅‰‰‰‰‰‰) = −2$%&*&+̅ ≥ −2|&*||&+| 
(8.14) 
i 
Re 
1 
z 
| | !"(1 − ) 
1 2† 0 
Figure 7.2 – Parabola on complex plane. 
 
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Then the expression in (8.13) becomes 
|&* − &+|+ ≥ (|&*| − |&+|)+ 
(8.15) 
In this case, the square root is given by: 
|&* − &+| ≥ T(|&*| − |&+|)+ = Œ|&*| − |&+|Œ 
(8.16) 
For a better understanding of the properties used here, see, 
e.g. Complex Variables and Applications, BROWN, J. W. e 
CHURCHILL, R. V., 9th edition, McGraw Hill, New York, 2014. 
 Problem 9
Use the DeMoivre’s formula to show that: 
R3S41 = 48QR?1R3S1 − 48QR1R3S?1 
 
What is the corresponding expression for 8QR41? 
Solution: 
The DeMoivre’s formula is given by: 
(8QR1 + 3R3S1)X = 8QRS1 + 3R3S S1 
(9.1) 
Let us do S = 4. Then, from (9.1) we have 
(8QR1 + 3R3S1)> = 8QR41 + 3R3S41 
(9.2) 
The power of four is given by: 
(8QR1 + 3R%S1)> = 8QR>1 + 48QR?1 ∙ 3R%S1 − 68QR+1R%S+1− 438QR1R%S?1 + R%S>1 
(9.3) 
Then from (9.2) and (9.3), we can write 
R%S41 = 48QR?1R%S1 − 48QR1R%S?1 
(9.4) 
And for 8QR41: 
8QR41 = 8QR>1 − 68QR+1R%S+1 + R%S>1 
(9.5) 
 Problem 10
Express the following complex numbers in the trigonometric 
form. 
a) & = √3 − 1Ž ; 
b) & = (−1 − 3)p 
Solution: 
a) First of all, let us write 3 − 1 in its polar form. Then 
3 − 1 = √2%J?<> 
(10.1) 
Here, √2 and ?<> are, the modulus (magnitude) and the main 
argument of 3 − 1, respectively. Thus: 
 
√3 − 1Ž = √2%J?<> ‘
*?
 
∴ √3 − 1Ž = √2’ %J<> 
(10.2) 
 
b) Let us do as we have done in “a”. Then 
 
−1 − 3 = √2%J“<> 
(10.3) 
 
Here √2 e “<> are, the modulus and the argument of −1 − 3, 
respectively. However, mp does not represent the main 
argument, because 
“<
> > B. The main argument of a complex 
number must be in the interval −B < 1 ≤ B. Then, the main 
argument of −1 − 3 is given by: “<> − 2B = − ?<> . So, the 
expression in (10.3) becomes 
 
−1 − 3 = √2%AJ?<> 
(10.4) 
 
Then: 
 
(−1 − 3)>“ = √2%AJ?<> ‘
>“
 
∴ (−1 − 3)>“ = 2+“%AJ?<“ 
(10.5) 
 
 Problem 11
 
Show that 3J = %Ahm•Y+X<k (n=integer). 
Suggestion: [Calculate LQZ 3]. 
Solution: 
Let us write i in the polar form 
 
3 = %J∙<+ 
(11.1) 
 
Now, let us set Log I. 
 
–QZ 3 = 3 hB2 + 2SBk 
(11.2) 
 
Finally, let us take the expression (11.2) and multiply it for i: 
 
3 ∙ –QZ 3 = − hB2 + 2SBk 
–QZ 3J = − hB2 + 2SBk 
∴ 3J = %Ah<+Y+X<k 
(11.3) 
 
 
 
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 Problem 12
By means of the definitions of the complex functions R3S & 
and R3Sℎ &, show that: 
a) R3S(&* ± &+) = R3S&*8QR&+ ± 8QR&*R3S&+ 
b) R3Sℎ(&* ± &+) = R3Sℎ&*8QRℎ&+ ± 8QRℎ&*R3Sℎ&+ 
What are the corresponding relations for 8QR(&* ± &+) and 8QRℎ(&* ± &+)? 
Solution: 
The expression for R3S & is given by: 
R3S & = 123 (%J˜ − %AJ˜) 
(12.1) 
So, we can write 
R3S (&* ± &+) = 123 ‚%J(˜™±˜•) − %AJ(˜™±˜•)ƒ 
(12.2) 
The Euler’s formula is given by: 
%J˜ = 8QR & + 3R3S & 
(12.3) 
And 
%AJ˜ = 8QR & − 3R3S & 
(12.4) 
Let us use the expressions (12.2), (12.3) and (12.4), then 
R3S (&* ± &+) = 123 [(8QR&* + 3R3S&*)(8QR&+ ± 3R3S&+) −(8QR&* + 3R3S&*)(8QR&+ ∓ 3R3S&+)] 
(12.5) 
After algebraic manipulations, 
R3S (&* ± &+) = R3S&*8QR&+ ± 8QR&*R3S&+ 
(12.6) 
The expression for R3Sℎ(&* ± &+) is given by: 
R3Sℎ (&* ± &+) = 12 ‚%(˜™±˜•) − %A(˜™±˜•)ƒ 
(12.7) 
Let us use the expression (12.7) and the following 
expressions 
%˜ = 8QRℎ & + R%Sℎ & 
(12.8) 
And 
%A˜ = 8QRℎ & − R%Sℎ & 
(12.9) 
Then: 
 
R3Sℎ (&* ± &+) = 12 [(8QRℎ&* + R3Sℎ&*)(8QRℎ&+ ± R3Sℎ&+) −(8QRℎ&* + R3Sℎ&*)(8QRℎ&+ ∓ R3Sℎ&+)] 
(12.10) 
 
After algebraic manipulations, 
 R3Sℎ (&* ± &+) = R3Sℎ&*8QRℎ&+ ± 8QRℎ&*R3Sℎ&+ 
(12.11) 
 
We can figure 8QR(&* ± &+) out following the same 
procedure. Then, 
 
8QR (&* ± &+) = 12 ‚%J(˜™±˜•) + %AJ(˜™±˜•)ƒ 
(12.12) 
 
By the use of (12.3) and (12.4) we can write 
 
8QR (&* ± &+) = 12 [(8QR&* + 3R%S&*)(8QR&+ ± 3R%S&+) +(8QR&* − 3R%S&*)(8QR&+ ∓ 3R%S&+)] 
(12.13) 
 
After algebraic manipulation 
 8QR (&* ± &+) = 8QR&*8QR&+ ∓ R3S&*R3S&+ 
(12.14) 
 
And for 8QRℎ(&* ± &+): 
 
8QRℎ (&* ± &+) = 12 ‚%(˜™±˜•) + %A(˜™±˜•)ƒ 
(12.15) 
 
By the use of (12.8) and (12.9), 
 
8QRℎ (&* ± &+) = 12 [(8QRℎ&* + R%Sℎ&*)(8QRℎ&+ ± R%Sℎ&+) +(8QRℎ&* − R%Sℎ&*)(8QRℎ&+ ∓ R%Sℎ&+)] 
(12.16) 
 
And finally 
 8QRℎ(&* ± &+) = 8QRℎ&*8QRℎ&+ ± R3Sℎ&*R3Sℎ&+ 
(12.17) 
 
 Problem 13
 
Show that the complex function R3S & can vanish only on the 
real axis and, in particular, at points F = SB (S=integer). At 
what points does the function R3Sℎ & vanish? 
Solution: 
The expression for R3S & is given by (12.1). Then, the values 
of & that make null the aforementioned function is given by: 
 %J˜ = %AJ˜ 
(13.1) 
 
 
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Let us set & = F + 3G. Then by the use of (13.1): 
%JW%A = %AJW% 
(13.2) 
So if G = 0, we can write 
%JW = %AJW 
(13.3) 
Hence F = SB, and S is an integer. For R3Sℎ & , the 
expression is given by: 
R3Sℎ & = %˜ − %A˜2 
(13.4) 
The values of &, which make (13.4) null, is given by: 
%˜ = %A˜ 
(13.5) 
The expressionin (13.5) leads to 
%W%J = %AW%AJ 
(13.6) 
So if F = 0, we can write: 
%J = %AJ 
(13.7) 
Hence G = SB, and S is an integer. We can conclude that R3Sℎ & becomes null on imaginary axis. 
 Problem 14
Show that all solutions of equation R3S & = 1000 are given 
approximately by: 
& = hS + *+k B ± 3 ∙ 7,601, 
 
for every even integer S. [Suggestion: Decompose R3S & into 
real and imaginary parts, and solve the resulting equations.] 
Solution: 
Let us set & = F + 3G. And let us use the following sine’s 
expression: 
R3S & = %J˜ − %AJ˜23 
(14.1) 
Then 
R3S & = %A(8QR F + 3 ∙ R3S F) − %(8QR F − 3 ∙ R3S F)23 
(14.2) 
From (14.2), the real and imaginary parts are given by: 
R3S & = R3S F(%A + %)2 −
3 ∙ 8QR F(%A − %)
2 
(14.3) 
 
The imaginary part is null. Then 
 8QR F(%A − %)
2 = 0 ⇒ F =
B
2 + SB or G = 0 
(14.4) 
 
Where S is even integer. And the real part is given by: 
 R3S F(%A + %)
2 = 1000 
(14.5) 
 
If G = 0 then, from (14.5), R3S F = 1000 with F ∈ ℝ. But this 
is impossible. Then, F = <+ + SB and G will be determined by: 
 %A + % = 2000 
(14.6) 
 
Be the variable E given by: 
 % = E 
(14.7) 
 
Then (14.6) becomes 
 E+ − 2000E + 1 = 0 
(14.8) 
 
The solutions of (14.8) are E* ≅ 1999,9995 and E+ = 0,0005. 
Using (14.7) we have: 
 G* = ln 1999,9995 ≅ 7,601 and G+ = ln 0,0005 ≅ −7,601 
(14.9) 
 
Hence: 
& = B2 + SB ± 37,601 
(14.10) 
 
 Problem 15
 
If & = F + 3G, show that: 
 
E6Sℎ & = R3Sℎ 2F + 3R3S 2G8QRℎ 2F + 8QR 2G 
 
Solution: 
The tangent’s expression is given by: 
 
E6Sℎ & = R3Sℎ &8QRℎ & =
R3Sℎ (F + 3G)
8QRℎ(F + 3G) 
(15.1) 
 
 
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For additions, 
R3Sℎ (F + 3G) = R3Sℎ F ∙ 8QRℎ 3G + 8QRℎ F ∙ R3Sℎ 3G 8QRℎ(F + 3G) = 8QRℎ F ∙ 8QRℎ 3G + R3Sℎ F ∙ R3Sℎ 3G 
(15.2) 
By the use of properties: R3Sℎ3G = 3 ∙ R3SG and 8QRℎ3G =8QRG, (15.1) becomes 
E6Sℎ & = R3SℎF ∙ 8QRG + 38QRℎF ∙ R3S G8QRℎF ∙ 8QRG + 3R3SℎF ∙ R3S G 
(15.3) 
Let us multiply (15.3) for 8QRℎF ∙ 8QRG − 3R3SℎF ∙ R3S G, and 
using 8QRℎ+F − R3Sℎ+F = 1: 
E6Sℎ & = R3SℎF ∙ 8QRℎF + 38QRG ∙ R3S G8QRℎ+F ∙ 8QR+G + R3Sℎ+F ∙ R3S+G 
(15.4) 
We have the following properties: 
R3Sℎ2F = 2R3SℎF ∙ 8QRℎF; 8QRℎ2F = 8QRℎ+F + R3Sℎ+F = 28QRℎ+F − 1 = 1 + 2R3Sℎ+F; R3S2G = 2 R3S G ∙ 8QR G; 8QR2G = 8QR+G − R3S+G = 1 − 2R3S+G = 28QR+G − 1 
(15.5) 
By the help of (15.5), the expression in (15.4) becomes: 
E6Sℎ & = R3Sℎ2F 2† + 3
R3S2G 2†(8QRℎ2F + 1)8QR+G2 + (8QRℎ2F − 1)R3S
+G2
 
(15.6) 
Hence: 
E6Sℎ & = R3Sℎ2F − 3R3S2G8QRℎ2F + 8QR2G 
(15.7) 
 Problem 16
There is no difficulty to figure the following additions out: 
a) ∑ %XW£X¤¥ ; 
b) ∑ %JXW£X¤¥ , 
(F = 0%6¦), because both of them represent a geometric 
progression. Use the Euler’s formula to deduce: 
§ 8QR SF
£
X¤¥
= R3S 
12 (¨ + 1)F
R3S 12 F
∙ 8QR ¨F2 (F = 0%6¦) 
What is the expression regarded to R3S F? 
Solution: 
Let us write the expression “b” as: 
© = § &X
£
X¤¥
= 1 + & + &+ + ⋯ + &£ 
(16.1) 
 
Where & = %JW . Then 
 © − &© = 1 − &£Y* 
∴ © = 1 − &£Y*1 − & 
(16.2) 
 
By the help of Euler’s formula, we can write 
 & = %JW = 8QR F + 3R3S F 
(16.3) 
 
By the use of (16.3) into (16.2) 
 
§ 8QR SF + 3R3S SF
£
X¤¥
= 1 − 8QR(¨ + 1)F − 3R3S (¨ + 1)F1 − 8QR F − 3R3S F 
(16.4) 
 
After some algebraic manipulations 
 
§ 8QR SF + 3R3S SF
£
X¤¥
= 
= 1 − 8QR(¨ + 1)F − 3R3S (¨ + 1)F1 − 8QR F − 3R3S F ∙
(1 − 8QR F + 3R3S F)
(1 − 8QR F + 3R3S F) 
 
= (1 − 8QR(¨ + 1)F)(1 − 8QR F) + R3S (¨ + 1)FR3S F2(1 − 8QR F) + 
 3[R3S F(1 − cos(¨ + 1)F) − R3S (¨ + 1)F(1 − 8QR F)]
2(1 − 8QR F) 
(16.5) 
 
For real part: 
 
§ 8QR SF
£
X¤¥
= 
(1 − 8QR(¨ + 1)F)(1 − 8QR F) + R3S (¨ + 1)F ∙ R3S F
2(1 − 8QR F) 
(16.6) 
 
After some manipulations 
 
§ 8QR SF
£
X¤¥
= 1 − 8QR(¨ + 1)F − 8QR F + 8QR ¨F2(1 − 8QR F) 
(16.7) 
 
Some properties were used to reach the expression (16.7): 
 
8QR(¨ + 1)F ∙ 8QR F = 12 (8QR ¨F + 8QR(¨ + 2)F) 
R3S(¨ + 1)F ∙ R3SF = 12 (8QR ¨F − 8QR(¨ + 2)F) 
(16.8) 
 
www.profafguimaraes.net 
9 
Now, let us use the following properties into (16.7): 
8QR ¨F − 8QR(¨ + 1)F = 2R3S 12 (2¨ + 1)F ∙ R3S
F
2 
1 − 8QR F = 2R3S+ F2 
(16.9) 
Then: 
§ 8QR SF
£
X¤¥
= 2R3S
12 (2¨ + 1)F ∙ R3S F2 + 2R3S+ F2
4R3S+ F2
 
(16.10) 
By the help of the following expression 
R3S 12 (2¨ + 1)F + R3S
F
2 = 2R3S
1
2 (¨ + 1)F ∙ 8QR
¨F
2 
(16.11) 
The expression (16.10) becomes 
§ 8QR SF
£
X¤¥
= R3S 
12 (¨ + 1)F
R3S 12 F
∙ 8QR ¨F2 
(16.12) 
Let us use the imaginary part for sine’s expression 
§ R3S SF
£
X¤¥
= 
[R3S F(1 − cos(¨ + 1)F) − R3S (¨ + 1)F(1 − 8QR F)]
2(1 − 8QR F) 
(16.13) 
By the help of the following expressions: 
R3S (¨ + 1)F ∙ 8QR F = 12 [R3S ¨F + R3S (¨ + 2)F] 
R3S F ∙ 8QR(¨ + 1)F = 12 [R3S (−¨F) + R3S (¨ + 2)F] 
(16.14) 
And R3S (−¨F) = −R3S ¨F, we can write 
§ R3S SF
£
X¤¥
= R3S F + R3S ¨F − R3S (¨ + 1)F2(1 − 8QR F) 
(16.15) 
Let us use the expressions: 
R3S ¨F − R3S (¨ + 1)F = 2 8QR 12 (2¨ + 1) ∙ R3S h
−F
2 k 
R3S F = 2 ∙ R3S F2 ∙ 8QR
F
2 
(16.16) 
Then 
§ R3S SF
£
X¤¥
= 8QR
F2 − 8QR h¨ + 12k F
2R3S F2
 
(16.17) 
 
By the help of the following expression: 
 
8QR F2 − 8QR \¨ +
1
2^ F = 2R3S
1
2 (¨ + 1)F ∙ R3S 
¨F
2 
(16.18) 
 
(16.17) becomes 
 
§ R3S SF
£
X¤¥
= R3S
F2 (¨ + 1) ∙ R3S ¨F2
R3S F2
 
(16.19) 
 
 Problem 17
 
Apply the same idea of the problem 16 to calculate: 
 
§ 6X 8QR SF, § 6XR3S SF
«
X¤¥
«
X¤¥
 
 
What are the values of a so that these result are valid? 
Solution: 
 
Let us set & = 6(8QR F + 3R3S F) and the expression Erro! 
Fonte de referência não encontrada., then 
 
§ 6X(8QR SF + 3R3S SF)
£
X¤¥
= 
= 1 − 6£Y*(8QR(¨ + 1)F + 3R3S (¨ + 1)F)1 − 6 ∙ 8QR F − 36 ∙ R3S F 
(17.1) 
 
Writing the expression when ¨ → ∞ 
 
§ 6X(8QR SF + 3R3S SF)
«
X¤¥
= 
lim£→«
1 − 6£Y*(8QR(¨ + 1)F + 3R3S (¨ + 1)F)
1 − 6 ∙ 8QR F − 36 ∙ R3S F 
(17.2) 
 
This converges if |6| < 1. For instance, it can be verified by 
means of root’s test. Then 
 
§ 6X(8QR SF + 3R3S SF)
«
X¤¥
= 11 − 6 ∙ 8QR F − 36 ∙ R3S F 
(17.3) 
 
After algebraic manipulations 
 
§ 6X(8QR SF + 3R3S SF)
«
X¤¥
= 1 − 6 ∙ 8QR F + 36 ∙ R3S F1 − 268QR F + 6+ 
(17.4) 
 
www.profafguimaraes.net 
10 
A 
B 
L 
R 
C i(t) 
The real part becomes 
§ 6X 8QR SF = 1 − 6 ∙ 8QR F1 − 268QR F + 6+
«
X¤¥
 
(17.5) 
And the imaginary one: 
§ 6X R3S SF = 6 ∙ R3S F1 − 268QR F + 6+
«
X¤¥
 
(17.6) 
 Problem 18
Figure 18.1 represents a portion of an alternating current 
circuit with the potential difference (electromotive force – 
EMF) between points A and B given by: 
®(E) = °¯ − ²¯ = ¥¯ 8QR ³E 
 
 
 
 
 
 
 
 
 
Figure 18.1 
Show that 3(E) (as shown) must satisfy the differential 
equation: 
– 9+39E+ + $
93
9E +
3
´ =
9®
9E 
 
Show how the method of complex exponential can be used 
to find a complex solution of 3(E) which varies harmonically 
in time. Perform the transition to real variables, show that: 
3(E) = ,¥ 8QR(³E + µ), 
and evaluate ,¥ and µ in terms of –, $, ´ and ¥¯. 
Solution: 
According to the first Kirchhoff’s law of electric circuits: 
– 93(E)9E + $3(E) +
¶(E)
´ = ®(E) 
(18.1) 
Where – ·J·K , $3 and ¸¹ are respectively the potential 
differences across the inductance, resistance and 
capacitance. Differentiating (18.1) with to the time, 
– 9+39E+ + $
93
9E +
3
´ =
9®
9E 
(18.2) 
With 3(E) = ·¸·K . Let us write the expression for ®(E) in polarform and consider only the real part. Then 
 ®(E) = $%( ¥¯%AJºK) 
(18.3) 
 
Let the current strength be 
 3(E) = $%(,¥%AJºK) 
(18.4) 
 
With a complex constant (,¥). We can use (18.2). Then 
 
−–³+,¥%AJºK − 3$³,¥%AJºK + ,¥%
AJºK
´ = −3³ ¥¯%AJºK 
,¥ = ¥¯»$ − 3 h³– − 1³´k¼
 
(18.5) 
 
The last expression in (18.5) has a complex denominator, 
which is the circuit’s impedance (½). Impedance has a real 
part, the resistance ($) and imaginary one, which is known 
as reactance (¾). The reactance can be formed by the 
inductive reactance (¾¿ = ³–) and the capacitive reactance 
h¾¹ = *º¹k. The reactance is complex because there is 
difference of phase between ®(E) and 3(E). Let us write the 
impedance in a polar form 
 ½ = $ − 3(¾¿ − ¾¹) ½ = T$+ + (¾¿ − ¾¹)+%JÀ 
(18.6) 
 
With 
 
µ = E6SA* ¾¿ − ¾¹$ 
(18.7) 
 
Let us apply (18.6) into (18.5), 
 
3(E) = ¥¯T$+ + (¾¿ − ¾¹)+%JÀ =
¥¯%AJÀ
T$+ + (¾¿ − ¾¹)+ 
(18.8) 
 
Now, we can substitute in (18.4), so 
 
3(E) = $% Á ¥¯%AJ(ºKYÀ)T$+ + (¾¿ − ¾¹)+ 
(18.9) 
 
With real part given by: 
 3(E) = ,¥ 8QR(³E + µ) 
(18.10) 
 
And, 
 
,¥ = ¥¯T$+ + (¾¿ − ¾¹)+ 
(18.11)