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**, sine and cosine become involved for some reason. I'm not sure why that happens. Can someone explain this? I finished precalc this year; on to calculus I when school starts in a month. Thanks.**

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- Thread starter GravitatisVis
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In summary, the equation e^z=exp(z) describes how the cosine and sine functions become involved when a whole number is raised to an imaginary non-whole number.

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It's very cute yes. Write e^(pi*i) as cos and i*sin, you'll see why it is true.

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So what you need is a definition of e^z=exp(z) that works for complex numbers z. The best motivation for such is analytic continuation, but that requires to much theory, so one of these is often employed.GravitatisVis said:, sine and cosine become involved for some reason. I'm not sure why that happens. Can someone explain this? I finished precalc this year; on to calculus I when school starts in a month. Thanks.i

1) Infinite series

[tex]exp(z):=\sum_{k=1}^{\infty}\frac{z^k}{k!}[/tex]

2)differential equation

exp'(z)=exp(z) for all complex z

exp(0)=1

3) functional relation

exp(a+b)=exp(a)exp(b) all complex a and b

[tex]\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=1[/tex]

4) unmotivated definition

exp(z)=exp(Re(z))cos(Im(z))+i exp(Re(z))sin(Im(z))

5) various ad hoc definitions that state what properties of exp are wanted and define a unique function.

I like 3) best

consider

z=x+i y

exp(z)=exp(x+i y) for x and y real

then

exp(z)=exp(x)exp(i y)

We already know what to do with exp(x) so we turn attention to exp(i y)

exp((x+y)i)=exp(x i)exp(y i)

now for convienance we will define

exp(i x)=C(x)+i S(x)

exp((x+y)i)=[C(x)C(y)-S(x)S(y)]+i[S(x)C(y)+C(x)S(y)]

C(x+y)=C(x)C(y)-S(x)S(y)

S(x+y)=S(x)C(y)+C(x)S(y)

now if x=x y=-x

1=exp(0)=exp(i(x-x))=exp(i x)exp(-i x)

so exp(-i x)=1/exp(i x)=1/(C(x)+i S(x))

=(C(x)-i S(x))/(C(x)^2+S(x)^2)=exp(-i x)/(C(x)^2+S(x)^2)

hence

1=C(x)^2+S(x)^2 for all real x

Now consider

[tex]\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=1[/tex]

[tex]\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=\lim_{y\rightarrow 0}\frac{C(y)+i S(y)-1}{i y}[/tex]

since we can let z->0 as i y->0

looking at the real part

[tex]\lim_{y\rightarrow 0}\frac{S(y)}{y}=0[/tex]

Now collecting what we know

C(x+y)=C(x)C(y)-S(x)S(y) for all real x and y

S(x+y)=S(x)C(y)+C(x)S(y) for all real x and y

1=C(x)^2+S(x)^2 for all real x

[tex]\lim_{y\rightarrow 0}\frac{S(y)}{y}[/tex]

We recognize this can only be if C(x) and S(x) are our beloved functions

C(x)=cos(x)

S(x)=sin(x)

so

exp(x+i y)=exp(x)(cos(x)+i sin(y))

exp(x+i y)=exp(x)cos(x)+i exp(x)sin(y)

for complex exponetials we define

u^v=exp(v log(u))

where the log used is the natural log so log(e)=1

thus the sin and cos in complex exponentials

In your example

2^2.5i=exp(2.5 i log(2))

2^2.5i=cos(2.5 log(2))+i sin(2.5 log(2))

2^2.5i~-.161363+.986895i

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GravitatisVis said:, sine and cosine become involved for some reason. I'm not sure why that happens. Can someone explain this? I finished precalc this year; on to calculus I when school starts in a month. Thanks.i

son... that is way beyond cal1. It can be proven by series in cal2. But it won't really be decussed until complex analyze.

Proof from MathWorld

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It is a nice precalc/trig type question. The problem is that while many precalc classes cover this material they tend not to be rigorous. Then in introductory calc often due to the large amount of material and the fact that most answers are likely to raise further questions, complex number issues are glossed over. These type of thing tend to be picked up in somewhere along the way and then are delt with in more detail in complex variables or advanced calculus. The idea is simple we want the complex exponential funtion to inherit the sum rule of the real exponential, this along with the rules of complex numbers yeild the famus euler formula.leon1127 said:son... that is way beyond cal1.

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lurflurf said:It is a nice precalc/trig type question. The problem is that while many precalc classes cover this material they tend not to be rigorous. Then in introductory calc often due to the large amount of material and the fact that most answers are likely to raise further questions, complex number issues are glossed over. These type of thing tend to be picked up in somewhere along the way and then are delt with in more detail in complex variables or advanced calculus. The idea is simple we want the complex exponential funtion to inherit the sum rule of the real exponential, this along with the rules of complex numbers yeild the famus euler formula.

yeah, i agree what you have said. But i don't see it can be proven with pre-cal level math technique. That is why i said it is way beyond pre-cal level. You used limit technique to proof it, but i don't think limit and series is taugh in pre-cal or was taugh in that deep (you were using complex limit and power series, precal barely touch geometric series and real limit).

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I type with one arm and a baby in hand! So I keep it short and suggestive.

--SR

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Take the McLaurin or Taylor of e^(x*i), i*sin(x) and i*cos(x) and all will be revealed

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SteveRives said:

I type with one arm and a baby in hand! So I keep it short and suggestive.

--SR

That's a good question. I find it very difficult to visualize a number raised to the i power. My knowledge of complex numbers is slim to nil to say the least; i = root(-1). I’m just going to make a guess here, but does multiplying a number on the complex plane flip it ninety degrees? 4i and 4 form a 90 degrees angle at the origin.

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You asked a simple enough question to answer. I am surprised nobody has given you a simple answer yet.

GravitatisVis said:On a similar note, when I raise a whole number to an imaginary non whole number, say 2^2.5, sine and cosine become involved for some reason. I'm not sure why that happens.i

It happens because of this definition:

[tex]e^{ix} \equiv \cos x + i\sin x[/tex]

Using the definition we have:

[tex]e^{i \pi} = \cos \pi + i \sin \pi = -1 + i \times 0 = -1[/tex]

and therefore

[tex]e^{i \pi} + 1 = 0[/tex]

If you want something like

[tex]2^{i \theta}[/tex]

then you need to know that

[tex]2^x = e^{x \ln 2}[/tex].

Does that help?

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Gauss is reported to have commented that if this formula was not immediately obvious, the reader would never be a first-class mathematician

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James R said:GravitatisVis:

You asked a simple enough question to answer. I am surprised nobody has given you a simple answer yet.

Sombody has given a simple answer

It happens because of this definition:

[tex]e^{ix} \equiv \cos x + i\sin x[/tex]

This is one possible definition. It is a special case of my #4. I call it unmotivated because it is not obvious why anyone would think to make that definition. It would also be helpful to know what propertys of the real exponential this preserves and which it does not. Also why is the definition better than another say

exp(i x)=cos(2x)+i sin(2x)

Using the definition we have:

[tex]e^{i \pi} = \cos \pi + i \sin \pi = -1 + i \times 0 = -1[/tex]

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Yes this is a way to do it.GravitatisVis said:That's a good question. I find it very difficult to visualize a number raised to the i power. My knowledge of complex numbers is slim to nil to say the least; i = root(-1). I’m just going to make a guess here, but does multiplying a number on the complex plane flip it ninety degrees? 4i and 4 form a 90 degrees angle at the origin.

consider the unit circle in the complex plain

numbers for which |z|=1

or x^2+y^2=1 (if z=x+i y for x and y real)

the "obvious" way of parametrizing this is

f(t)=cos(t)+i sin(t)

f(s)f(t)=(cos(s)+i sin(s))(cos(t)+i sin(t))

=[cos(s)cos(t)-sin(s)sin(t)]+i[sin(s)cos(t)+cos(s)sin(t)]

=cos(s+t)+i sin(s+t)

=f(s+t)

so

f(s)f(t)=f(s+t)

this looks like an exponential function but it is hard to fit all the pieces together.

f(n t)=f(t)^n

is obvious when n is an integer

with a litle thought it can be seen

f(x t)=f(t)^x

where x is a real number

the trouble is we do not know how to do

f(z t)=f(t)^z

when z is a complex number

this is why since we can define

u^v=exp(v log(u))

for real numbers we take this definition for complex numbers

we wil need to define exp(z) for z complex

for this we can adopt one of the five commonly used (equivelent) definitions of

exp(z) for z complex

at some point we are led to conclude that

exp(a+b)=exp(a)exp(b)

holds when a and b are complex

hence

exp(x+i y)=exp(x)exp(i y)

it turns out we want

exp(i y)=f(y)=cos(y)+i sin(y)

because what we like about exp is

exp(a+b)=exp(a)exp(b)

and

exp(x)~1+x if x is a small number

f(t) is a good canidate for exp(i y) since

f(a+b)=f(a)f(b)

and

f(y)~1+i y if y is a small number

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Again the main trouble is of the five commonly used definitions this one gives no reason why it should be used. The other four are more clear, but are difficult to explain on the precalc. level

And the initial poster clearly said that he had only just finished precalc.

As a mathematician, it's good to know that you want everything justified and proven from first principles. But somebody who knows nothing about complex numbers is probably better off being presented with some basic results as a first step. They can worry about detailed proofs and motivations later.

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GravitatisVis said:That's a good question. I find it very difficult to visualize a number raised to the i power. My knowledge of complex numbers is slim to nil to say the least; i = root(-1). I’m just going to make a guess here, but does multiplying a number on the complex plane flip it ninety degrees? 4i and 4 form a 90 degrees angle at the origin.

Let's back up again and think about what it means to take a thing to a power.

Normally, we might think of repeated multiplication when we think of taking something to a power. But there is another way we can think of this. Taking something to a power it a mapping function.

If we only think of taking powers as repeated multiplication, then of course taking to the i power will confuse us! But, if we think of powers as transforming, or mapping, or moving from one set to another, then we might have a better go at it.

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I can't find anything in a quick google, but there should be something on Joost Burgi either on the net or in a library that provides a geometric interpretation of logarithms. Burgi actually came up with the idea of logarithms before Napier, even though Napier's work was published earlier.DaveC426913 said:

Burgi did work on "prosthaphaeresis" - a method where trigonometry tables were used as an aid in multiplication problems. He expanded on this to develop a geometric approach to logarithms, something that hopefully would be simpler to use for multiplication than trig tables.

Napier also did work in prosthaphaeresis, and developed an algebraic approach to logarithms.

Neither came up with the logarithms used today, but the principles were the same, even if the base was different.

One thing I noticed as a side note was that you could develop a slide rule based on prosthaphaeresis that could easily be accurate up to four significant digits if skillfully made. That's kind of interesting since the motivation for developing logarithms was to find a shortcut to long multiplication problems. With a mechanical device to do the calculations, the motivation to progress to logarithms would have been much less. Napier probably wouldn't have invented his "Napier's bones", an invention that extended the ability to perform multiplication to the average business person. Euler probably wouldn't have had Napier's logarithms to read while analyzing infinite series and who knows when the natural log as we know it today would have been invented.

As things turned out, http://www.cee.hw.ac.uk/~greg/calculators/napier/about.html [Broken] revolutionized arithmetic for the average person - except now, instead of getting to actually use sticks for multiplication, elementary school students are expected to memorize them. Brigg's converted Napier's logs to base 10 and called them common logs. The first sophisticated calculating device wound up being based on Brigg's common logs by Edmund Gunter, even though it was kind of a pain to use. William Oughtred turned Gunter's static scales into sliding scales, inventing the first slide rule - a calculating device that was expanded to include trig functions and natural log scales (log log scales) and served as a powerful handful calculator for 350 years until the introduction of the electronic calculator (I doubt the development of natural logs would have been delayed for 350 years, though).

In 1971, you could buy a Post Versalog bamboo slide rule with leather case for about $30 ($146 in today's dollars) - a device that could do multiplication, division, squares, cubes, and square roots, solve quadratic equations, had 8 "log-log" scales to give natural logs for numbers from about .0001 to 20000, could perform vector problems using the trig functions ... or, you could pay $345 ($1685 in today's dollars) for a Canon pocketronic calculator that could add, subtract, multiply, or divide. In fact, an architect named Paul Hemphill had to finance the first four-function calculator he bought in 1970 - he paid $200 down and paid installments on the remaining $1300 (that's about $7700 in today's dollars). He was still paying off his four-function calculator when Hewlett Packard put out the HP-35 scientific calculator for only $395 - the first hand-held calculator as powerful as a slide rule. Now a days, calculators are so cheap that people can graduate from high school without having to learn anything so complex as 17000*80.

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SteveRives said:Let's back up again and think about what it means to take a thing to a power.

Normally, we might think of repeated multiplication when we think of taking something to a power. But there is another way we can think of this. Taking something to a power it a mapping function.

If we only think of taking powers as repeated multiplication, then of course taking to the i power will confuse us! But, if we think of powers as transforming, or mapping, or moving from one set to another, then we might have a better go at it.

Mr. Rives, could you explain that in a little more detail? The most we've delved into exponents is, as you said, "repeated multiplication." That's why I find it so difficult to wrap my head around something like a term raised to a non whole number. I've never heard of an exponent "mapping" numbers from one set to another; and to be honest, I have no idea what that even means. Thanks, I appreciate all the answers so far guys.

- Christopher

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Let's start with a simple example. Car squared is not a car that is exponentially bigger. If we apply the idea of squaring to non integers (in this case a car), what do we get? We get gibberish, that's what we get!GravitatisVis said:Mr. Rives, could you explain that in a little more detail? The most we've delved into exponents is, as you said, "repeated multiplication." That's why I find it so difficult to wrap my head around something like a term raised to a non whole number. I've never heard of an exponent "mapping" numbers from one set to another; and to be honest, I have no idea what that even means. Thanks, I appreciate all the answers so far guys.

- Christopher

Likewise, when we take something to a fractional power, like x to the 1.345 power, we are not talking about a geometric object. Nor are we talking about repeated multiplication. What we are talking about is a

Think of cryptography. If you take my English sentence and do a mathematical transform on each letter, the end result will be a sentence that is encoded. The end result is that you map the plain text to code. And that is quite useful. The transformation of each letter is a mapping, and might not correspond to some geometric shape.

Extending the example, taking something to a power is like encrypting plain text, and the log is like the reverse (decrypting the cipher). It is a mapping thing. You are transforming one set to another. It's not something you picture in terms of a graph.

My goal in this short reply is to hint at a different way to think of the exponent idea (so that you are free from thinking only in terms of geometric objects or whole number math). I won't go deeper than this for now, but will wait and see if this made any sense.

Cheers,

Steve Rives

The equation e^(pi*i) + 1 = 0 is an example of Euler's identity, which is a special case of the Euler's formula. This formula relates the exponential function, the trigonometric functions, and the imaginary unit i. In this case, the complex number e^(pi*i) has a magnitude of 1 and an angle of pi radians, resulting in a value of 1 + 0i, which equals 1. Adding 1 to this complex number results in a magnitude of 1 and an angle of 0 radians, or 0 + 0i, which equals 0.

The unit circle is a circle with a radius of 1, centered at the origin on a Cartesian plane. The equation e^(pi*i) + 1 = 0 can be represented on the unit circle as the point (1,0), which is the point at which the unit circle intersects with the x-axis. This point represents the complex number 1 + 0i, which is the result of e^(pi*i) + 1.

This equation is considered significant in mathematics as it connects various fundamental concepts such as the exponential function, trigonometric functions, complex numbers, and the unit circle. It is also used in many mathematical proofs and has applications in fields such as physics and engineering.

No, this equation cannot be simplified or solved for a specific value. The value of e^(pi*i) is an irrational number, meaning it cannot be expressed as a finite decimal or fraction. The value of e^(pi*i) + 1 is also an irrational number, making it impossible to solve for a specific value.

This equation is an example of a complex number, which is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit. The equation e^(pi*i) + 1 = 0 is a complex number with a real part of 1 and an imaginary part of 0, making it a real number. It is a special case of a complex number as it has a magnitude of 1 and an angle of pi radians, resulting in a simple form of 1 + 0i, or just 1.

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