Una explicación breve de las principales propiedades de los números complejos

1. COMPLEX
POWERS
AND ROOTS
ESIME Unidad Azcapotzalco
CRUZ CARRERA JOSÉ MANUEL
REYES ORNELAS ADRIAN
TEJEDA GARCÍA ANGEL BALAM
VARELA PEREA GAEL AXEL

2. INTRODUCTION
Let 𝑟 and 𝜃 be the polar
coordinates of the point (𝑥, 𝑦)
that corresponds to a 𝑛𝑜𝑛𝑧𝑒𝑟𝑜
complex number 𝑧 = 𝑥 + 𝑖𝑦.
Since 𝑥 = 𝑟 cos 𝜃 and
𝑦 = 𝑟 sin 𝜃, the number 𝑧 can be
written in polar form as.
𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃)

3. If 𝑧 = 0, the coordinate 𝜃 is undefined; and so, it is
understood that 𝑧 ≠ 0 whenever polar coordinates are
used.
In complex analysis, the real number 𝑟 is not allowed
to be negative and is the length of the radius vector for
𝑧; that is, 𝑟 = 𝑧 . The real number 𝜃 represents the
angle, measured in radians, that 𝑧 makes with the
positive real axis when 𝑧 is interpreted as a radius
vector. As in calculus, 𝜃 has an infinite number of
possible values, including negative ones, that differ by
integral multiples of 2𝜋. Those values can be
determined from the equation tan 𝜃 = 𝑦/𝑥, where the
quadrant containing the point corresponding to 𝑧 must
be specified. Each value of 𝜃 is called an argument of
𝑧, and the set of all such values is denoted by arg 𝑧.
The principal value of arg 𝑧, denoted by 𝐴𝑟𝑔 𝑧, is the
unique value Θ such that −𝜋 < Θ ≤ 𝜋. Evidently,
then,
arg 𝑧 = 𝐴𝑟𝑔 𝑧 + 2𝑛𝜋 (𝑛 = 0, ±1, ±2, … )
Also, when 𝑧 is a negative real number, 𝐴𝑟𝑔 𝑧 has been
the value 𝜋, not −𝜋.

4. EXAMPLE
The complex number −1 − 𝑖, which lies in the third
quadrant, has principal argument −3𝜋/4. That is,
𝐴𝑟𝑔 −1 − 𝑖 = −3𝜋/4
It must be emphasized that because of the restriction
− 𝜋 < Θ ≤ 𝜋 of the principal argument Θ, its not true
that 𝐴𝑟𝑔 −1 − 𝑖 = 5𝜋/4.
arg −1 − 𝑖 = −
3
4
𝜋 + 2𝑛𝜋 (𝑛 = 0, ±1, ±2, … )
Note that the term 𝐴𝑟𝑔 𝑧 on the right-hand side of
equation can be replaced by any particular value of arg 𝑧
and that one can write, for instance,
arg −1 − 𝑖 =
5
4
𝜋 + 2𝑛𝜋 (𝑛 = 0, ±1, ±2, … )

5. EXPONENTIAL
FORM
The symbol 𝑒𝑖𝜋, or exp(𝑖𝜃), is defined by
means of Euler´s formula as
𝑒𝑖𝜃
= cos 𝜃 + 𝑖 sin 𝜃,
Where 𝜃 is to be measured in radians. It
enables us to write the polar form more
compactly in exponential form as
𝑧 = 𝑟𝑒𝑖𝜃
.

6. EXAMPLE
The number −1 − 𝑖 has exponential form
−1 − 𝑖 = 2 exp 𝑖 −
3
4
𝜋
With the agreement that 𝑒−𝑖𝜃
= 𝑒1(−𝜃)
,
this can also be written −1 − 𝑖 =
2𝑒−𝑖3𝜋/4
, this expression is of course,
only one of an infinite number of
possibilities for the exponential form of
− 1 − 𝑖:
−1 − 𝑖 = 2 exp 𝑖 −
3
4
𝜋 + 2𝑛𝜋
𝑛 = 0, ±1, ±2, …

7. Note how expression 𝑧 = 𝑟𝑒𝑖𝜃
with 𝑟 = 1 tell us that the
number 𝑒𝑖𝜃
lie on the circle centered at the origin with
radius unity. Values of 𝑒𝑖𝜃
are, then, immediate from that
figure, without reference to Euler´s formula. It si, for
instance, geometrically obvious that
𝑒𝑖𝜋
= −1, 𝑒−
𝑖𝜋
2 = −𝑖, 𝑒−𝑖4𝜋
= 1
Note too that the equation
𝑧 = 𝑅𝑒𝑖𝜃
(0 ≤ 𝜃 ≤ 2𝜋)
is a parametric representation of the circle 𝑧 = 𝑅,
centered at the origin with radius 𝑅. As the parameter 𝜃
increases from 𝜃 = 0 to 𝜃 = 2𝜋, the point 𝑧 starts from
the positive real axis and traverses the circle once in the
counterclockwise direction.

8. More generally, the circle 𝑧 − 𝑧0 = 𝑅,
whose center is 𝑧0 and whose radius is
𝑅, has the parametric representation
𝑧 = 𝑧0 + 𝑅𝑒𝑖𝜃
(0 ≤ 𝜃 ≤ 2𝜋)
This can be seen vectorially by nothing
that a point 𝑧 traversing the circle
𝑧 − 𝑧0 = 𝑅 once in the
counterclockwise direction corresponds
to the sum of the fixed vector 𝑧0 and a
vector of length 𝑅 whose angle of
inclination 𝜃 varies from 𝜃 = 0 to 𝜃 =
2𝜋

9. PRODUCTS OF
COMPLEX
NUMBERS
Simple trigonometry tells us about that 𝑒𝑖𝜃
has
the familiar additive property of the exponential
function in calculus:
𝑒𝑖𝜃1𝑒𝑖𝜃2
= (cos 𝜃1 + 𝑖 sin 𝜃1)(cos 𝜃2 + 𝑖 sin 𝜃2)
= cos 𝜃1 cos 𝜃2 − sin 𝜃1 sin 𝜃2
+ 𝑖 sin 𝜃1 cos 𝜃2 + cos 𝜃1 sin 𝜃2
= cos(𝜃1 + 𝜃2) + 𝑖 sin(𝜃1 + 𝜃2) = 𝑒𝑖(𝜃1+𝜃2)
Thus, if 𝑧1 = 𝑟1𝑒𝑖𝜃1 and 𝑧2 = 𝑟2𝑒𝑖𝜃2, the
product 𝑧1𝑧2 has the exponencial form
𝑧1𝑧2 = 𝑟1𝑒𝑖𝜃1𝑟2𝑒𝑖𝜃2 = 𝑟1𝑟2𝑒𝑖𝜃1𝑒𝑖𝜃2
= (𝑟1𝑟2)𝑒𝑖(𝜃1+𝜃2)

10. PRODUCTS AND
POWERS

11. 3.1 - METHOD
• The exponential form of a complex number is z=reiΩ, where r
represents the distance from the origin to the complex number and Ω
represents the angle of the complex number.
• If we have a complex number z=a+bi, we can find its radius with the
formula:
•r2 = a2+b2
• And we find its angle with the formula:
•tan(θ)= b÷a

12. • Then, we apply any power n to the complex number in it is
polar form:
•zn=rniinΩ
• The expression can be simplified and written in the form
a+bi using Euler's formula:
•eiΩ=cos(θ)+ i sin(θ)

13. 3.2 – FOR EXAMPLE
Simplify the expression i58
• We can simplify this by remembering that i4=1. Then, we obtain factors that have
exponents divisible by 4 and simplify:
i58=i56×i2
=(i4)14×i2
=114×i2
=i2
I58=−1

14. 3.3 – APPLICATIONS IN MECHANICAL
ENGINEERING
In mechanical engineering, complex numbers are used to represent
material and to put the behavior of fluids into numbers.

15. For dynamic analysis of structures and for numerical control of
machine tool actions through numbers.

16. Complex numbers are used in modeling; Among these processes is the
analysis of electric current and electronic signals.

17. SQUARE ROOT
OF COMPLEX
NUMBERS
•The square root of a complex
number z is another complex
number w such that w^2=z.
•Just like the square root of a real
number can be either positive or
negative, a complex number has
two square roots, which are
complex conjugates of each other.

18. METHOD
• To find the square root of a complex number z
in rectangular form (z = a + bi), we can use the
following formula:
• √(a + bi) = ± (√{[√(a^2 + b^2) + a]/2} + ib/|b|
√{[√(a^2 + b^2) - a]/2})
• where ± indicates that either sign is possible.

19. EXAMPLE
• if we want to find the square root of the
complex number z = 9 + 40i, we can use the
following steps:
• Calculate the modulus of z: |z| = √(9^2 +
40^2) = 41
• Calculate the square root of the modulus of z:
√(|z|) = √(41) = √(41)
• Calculate the two square roots of z:
• ± (√{[√(9^2 + 40^2) + 9]/2} + i40/|40| √{[√(9^2
+ 40^2) - 9]/2})
• = ± (√{[41 + 9]/2} + i40/40 √{[41 - 9]/2})
• = ± (5 + 4i)
• Therefore, the two square roots of the
complex number z = 9 + 40i are 5 + 4i and -5 -
4i.

20. METHOD
Square root of complex numbers in
polar form
• The square root of a complex number z in
polar form (z = r(cos θ + i sin θ)) can be found
using the following formula:
• √z = √r (cos θ/2 + i sin θ/2)

21. EXAMPLE
• if we want to find the square root of the
complex number z = 41(cos π/4 + i sin π/4), we
can use the following steps:
• Calculate the square root of the modulus of z:
√r = √(41) = √(41)
• Calculate the square root of z:
• √z = √r (cos θ/2 + i sin θ/2)
• = √(41) (cos π/8 + i sin π/8)
• = 5(cos π/8 + i sin π/8)
• Therefore, the square root of the complex
number z = 41(cos π/4 + i sin π/4) is 5(cos π/8
+ i sin π/8).

22. APPLICATIONS IN
MECHANICAL
ENGENIREENG

23. APPLICATIONS
• When designing a new car. To determine the
natural frequencies of the car, the engineer
must calculate the roots of the car's
characteristic polynomial. These natural
frequencies will determine how the car will
vibrate when subjected to external forces,
such as road bumps.
• When designing a control system for a
production plant. To determine whether the
system will be stable, the engineer must
calculate the roots of the system's transfer
polynomial. If any of the roots have a positive
real part, the system will be unstable and
could enter into oscillation.

24. POWER EXERCISES OF COMPLEX NUMBERS
Power and Root 1-.:
- (2 - i)^3
- (2 - i)^3 = (2 - i)^2 times
(2 - i)
- = (4 - 4i + i^2) times (2 -
i)
Remember that (i^2 = -1)
- = (-3 - 4i) times (2 - i)
- = 6 + 11i - 4i^2
- = 10 + 11i
- Square root of (3 + 4i)^2
- (3 + 4i)^2 = 9 + 24i - 16
- = -7 + 24i 2.
Power and Root:2-.
- (1 - 3i)^2
- (1 - 3i)^2 = (1 - 3i)(1 -
3i)
- = 1 - 6i + 9i^2
- = -8 - 6i
- Cube root of (4 +
2i)^3
- (4 + 2i)^3 = 64 + 48i +
12i^2
- = -40 + 48i 3.
Power and Root:3-.
- (2i)^4 )
- (2i)^4 = (2i)^2 times
(2i)^2
- = -4 times -4
- =16
- Fourth root of (-1 - i)^4
- (-1 - i)^4 = 1 + 4i + 6i^2 +
4i^3 + i^4
- = 1 - 4 - 6i - 4i + 1
- = -6i - 6

26. BIBLIOGRAPHIC REFERENCES
• Garben, C. (s/f). Unidad 1: Numeros Complejos. Blogspot.com. Recuperado el 9 de noviembre de 2023, de
http://mecanica-unefa.blogspot.com/2011/06/unidad-1-numeros-complejos.html
• Complejos, N. (s/f). Leonhard Paul Euler. Wordpress.com. Recuperado el 9 de noviembre de 2023, de
https://matematicasiesoja.files.wordpress.com/2022/02/aplicioones-de-los-numeros-complejos.pdf
• Potencia de numeros complejos. (s/f). Material Didáctico - Superprof. Recuperado el 9 de noviembre de
2023, de https://www.superprof.es/apuntes/escolar/matematicas/aritmetica/complejos/potencia-de-
numeros-complejos.html
• Ayres, F. (2006). Análisis matemático (4.ª ed.). Madrid: Addison-Wesley.
• Stewart, J. (2015). Cálculo multivariable (8.ª ed.). Ciudad de México: Cengage Learning.
• Zill, D. G., & Shanahan, D. A. (2016). Cálculo con geometría analítica (6.ª ed.). Ciudad de México: Pearson.