1/04/2023

How can De Moivre's theorem be described? What is the scope of this theorem? Examples for roots and powers.

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. 

According to the De Moivre’s Theorem 

If Z = r(cos θ + isin θ) is a complex number, then Zn = rn[cos(nθ) + isin(nθ)] 

where n is a positive integer. Zn = rn cis(nθ) 

How does it come from? To understand this theorem, we must know the products of complex numbers in polar form and the quotients of complex numbers in polar form first.

Recall that : 

    sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

    sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

    cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

    cos(α – β) = cos(α)cos(β) + sin(α)sin(β)


 


(1+ \sqrt[]{3}i)^3

r= \sqrt[]{ 1^{2}+( \sqrt[]{3})^2 } =2

tan \theta= \sqrt[]{3}  ,  3(\frac{ \pi }{3})= \pi

(1+ \sqrt[]{3}i)^3=2^3(cos\pi +sin \pi)=-8


(\sqrt[]{2}+ \sqrt[]{2}i)^3

r= \sqrt[]{ (\sqrt[]{2})^{2}+( \sqrt[]{2})^2 } =2

tan \theta= 1  ,  \theta= \pi/4

(\sqrt[]{2}+ \sqrt[]{2}i)^3=2^3(cos3\pi/4 +sin3\pi/4)=\sqrt[]{2}(-4+4i)


(1+ \sqrt[]{3}i)^{ \frac{1}{2}}

r= \sqrt[]{(1)^2+ (\sqrt[]{3})^2 } =2

tan \theta= \sqrt[]{3} ,   \theta= \frac{ \pi }{3}

(1+ \sqrt[]{3}i)^{ \frac{1}{2}}   =(2^{ \frac{1}{2} })(cos(\frac{1}{2})(\frac{ \pi }{3})+isin(\frac{1}{2})(\frac{ \pi }{3}))  =  \sqrt[]{2} ( \frac{\sqrt[]{3}}{2} + \frac{1}{2} i)


(\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}

r= \sqrt[]{( \sqrt[]{2})^2+(\sqrt[]{2})^2} =2

tan \theta=1  ,  \theta= \frac{ \pi }{4}

(\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}   = \sqrt[]{2}(cos \frac{ \pi }{8} +isin \frac{ \pi }{8})= \sqrt[]{2}( \sqrt[]{\frac{1+ \frac{\sqrt[]{2}}{2} }{2} } +\sqrt[]{\frac{2-\sqrt[]{2} }{2} })

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The similarity of triangles gives rise to trigonometry.

One of the largest issues in ancient mathematics was accuracy—nobody had calculators that went out ten decimal places, and accuracy generally got worse as the numbers got larger. The famous Eratosthenes experiment, that can be found at https://www.famousscientists.org/eratosthenes/, relied on the fact known to Thales and others that a beam of parallels cut by a transverse straight line determines equal measure for the corresponding angles.  Given two similar triangles, one with small measurements that can be accurately determined, and the other with large measurements, but at least one is known with accuracy, can the other two measurements be deduced? Explain and give an example.

The similarity of triangles gives rise to trigonometry. 

How could we understand that the right triangles of trigonometry with a hypotenuse of measure 1 represent all possible right triangles? Ultimately, the similarity of triangles is the basis for proportions between sides of two triangles, and these proportions allow for the calculations of which we are speaking here. The similarity of triangles is the foundation of trigonometry.



Similar triangles are triangles that have the same shape, but their sizes may vary. If two triangles are similar, then their corresponding angles are congruent and corresponding sides are in equal proportion. Two triangles are similar if they have the same shape but are of different sizes. Thus mathematically, if two triangles are similar, then their corresponding sides are proportional and their corresponding angles are congruent.



Since these two triangles are similar triangles, \frac{h}{y} = \frac{1}{x} ,  h= \frac{y}{x}

For the smaller triangle, tan \theta = \frac{y}{x}

For the larger triangle, tan \theta = \frac{h}{1}











Reference

Abramson, J. (2017). Algebra and trigonometry. OpenStax, TX: Rice University. Retrieved from https://openstax.org/details/books/algebra-and-trigonometry


Admin. (2021, March 10). Similar triangles- Formula, Theorem & Proof of SSS, SAS AAA Similarity. BYJUS. Retrieved December 26, 2022, from https://byjus.com/maths/similar-triangles/ 


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What happens if we graph both f and f^{-1} on the same set of axes, using the x-axis for the input to both f and f^{-1} ?

 What happens if we graph both f and f^{-1} on the same set of axes, using the x-axis for the input to both f and f^{-1} ? 

Suppose f:R \rightarrow R is a function from the set of real numbers to the same set with f(x)=x+1. We write f^{2} to represent f \circ f and f^{n+1}=f^n \circ f. Is it true that f^2 \circ f = f \circ f^2? Why? Is the set {g:R \rightarrow R l g \circ f=f \circ g} infinite? Why? 

Inherently, the range of a function f (x) is the domain of the inverse function f^{-1}. The domain of f  is the range of  (f^{-1} \circ f)(x)=f^{-1}(f(x))=f^{-1}(y)=x . So, what happens if we graph both f and f^{-1} on the same set of axes, using the x-axis for the input to both f and f^{-1} ? Let's try some examples.

If we type y=x^3  \lbrace{-2 < x < 2}\rbrace , y=x^{1/3}  \lbrace{-2 < x < 2}\rbrace , and  y = x   \lbrace{-2 < x < 2}\rbrace  on the graph, all of the three functions will be limited in the same input  \lbrace{-2 < x < 2}\rbrace   and their output will also affected by such a input boundary. The domain of f  become the domain of  (f^{-1} \circ f)(x)  , not its range. 




Let's try another example: C° to F°: Celsius to Fahrenheit Conversion Formula

F = C×( \frac{9}{5} )+32

F^{-1} = (\frac{9}{5})(C-32)





Again, as we can see that these these three lines are more like projections of each other from different angles rather than reflection.

The difficulty of setting the same set of axes for f and f^{-1} is that f^{-1} will become not fully or true inverse function of f  since the output of f^{-1} is limited. Instead of setting the input {-100 < x < 100}, the input of  f^{-1}  should be the range of f 


Suppose f:R \rightarrow R is a function from the set of real numbers to the same set with f(x)=x+1. We write f^{2} to represent f \circ f and f^{n+1}=f^n \circ f. 

When f^{n+1}=f^n \circ f.  and  n \geq1

f^2=f \circ f

If f^{2} represents f \circ f  =f(f(x))  ,  f^2 \circ f =f^2(f(x))  
f^2=f(x+1)=(x+1)+1=x+2

It is similar to an arithmetic progression which the next one is the previous number +1

Recall that  f^2=f \circ f=f(f(x))  
f^2 \circ f=f(f(f(x)))
f \circ f^2=f(f^2(x))=f(f \circ f)=f(f(f(x)))
Therefore,  f^2 \circ f = \( f \circ f^2 = \( f(f(f(x)))

g \circ f =g(f(x)) = f \circ g=f(g(x))
In this case,  f(x)=x+1    g(f(x))=g(x+1)=g(x)+1=f(g(x))
Now, we know that the set has to satisfy the equation  g(x+1)=g(x)+1  
If the domain of  g(x)  is all real numbers, then g(1)=g(0)+1, g(2)=g(1)+1, g(3)=g(2)+1,....... it will be infinite.

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