# Proof of Complex Numbers Re^{jθ}: R(cosθ + jsinθ)

• lazypast
In summary, the equation Re^{j \theta} = R{(cos \theta + jsin\theta )} can be proven using calculus or Taylor series. It can also be seen as a definition for complex exponents.
lazypast
$$Re^{j \theta} = R{(cos \theta + jsin\theta )}$$

can anyone show me this proof or show me a link please

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What do you know about Taylor series ?

Daniel.

I haven't come across this Taylor series much, using this I take it the proof is harder that i suspected

lazypast said:
I haven't come across this Taylor series much, using this I take it the proof is harder that i suspected
You can also prove this result using calculus. If you have met differentiation, try taking the derivative of f(x) as see where it takes you,

$$f(x):=\frac{R\cos(x)+iR\sin(x)}{Re^{ix}}\hspace{1cm}x\in\Re\;\;\; , \;\;\; R\neq0$$

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Although my consistency with things this level arent good here goes

$$f(x)= \frac {Rcosx + jRsinx} {Re^{jx}} = \frac {ln(Rcosx) + ln(jRsinx)} {ln(Re^{jx}} = \frac {-Rsinx} {Rcosx} + \frac {jRcosx} {jRsinx} /j = \frac {jcotx - tanx} {j}$$

I'm afraid your technique is not quite correct. Have you met thehttps://www.physicsforums.com/showpost.php?p=1140219&postcount=4"yet? Also, I have noticed that you are using $j$ in your questions; I am assuming here you mean the imaginary unit $i=\sqrt{-1}$. It may also perhaps be desirable to remove the $R$ term for the moment to remove the condition that R be non zero. Thus, we have;

$$f(x):=\frac{\cos(x)+i\sin(x)}{e^{ix}}\hspace{1cm} x\in\Re$$

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$$f(x) =\frac {e^{ix}(icosx-sinx) - ie^{ix}(cosx + isinx)} {(e^{ix})^2} f(x) =\frac {ie^{ix}cosx - e^{ix}sinx - ie^{ix}cosx - i^2e^{ix}sinx} {(e^{ix})^2} f(x) =\frac {e^{ix}sinx + e^{ix}sinx} {(e^{ix})^2} f(x) =\frac {2sinx} {e^{ix}}$$

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Your getting closer, I'll show you;

$$\frac{d}{dx}\left (\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$

So in this case $u=\cos(x) + i\sin(x)$ and $v=e^{ix}$, thus;

$$\frac{du}{dx} = -\sin(x) + i\cos(x)\hspace{2cm}\frac{dv}{dx} = i\cdot e^{ix}$$

Thus we obtain;

$$f'(x) = \frac{e^{ix}\cdot (-\sin(x) + i\cos(x)) - (\cos(x) + i\sin(x))\cdot i\cdot e^{ix}}{(e^{ix})^2}$$

$$=\frac{-e^{ix}\cdot\sin(x) + e^{ix}\cdot i\cos(x) - e^{ix}\cdot i \cdot\cos(x) - e^{ix}\cdot i^2\cdot\sin(x)}{(e^{ix})^2}$$

Noting that the two consine terms cancel and that $i^2=(\sqrt{-1})^2 = -1$ ;

$$f'(x) = \frac{-e^{ix}\cdot\sin(x) + e^{ix}\cdot\sin(x)}{(e^{ix})^2}$$

$$= \frac{-\sin(x)+\sin(x)}{e^{ix}} = \frac{0}{e^{ix}} = 0$$

What can we say about a function if it has a derivative of zero everywhere in its domain?

I see now, silly mistakes by me.
When a derivative is zero then the things differentiated must have been a constant

lazypast said:
$$Re^{j \theta} = R{(cos \theta + jsin\theta )}$$

can anyone show me this proof or show me a link please
From a book i have read in complex analysis, and from what i understand about the subject, I believe that we just define this equation. I mean, we define that $$e^{a+j \theta} = e^a{(cos \theta + jsin\theta )}$$ . It's simply the definition for complex exponents. You can also reach this equation by using the Taylor series of e^x=1+x+(x^2)/2!+... , using x=ja but it's not any real proof unless you prove that the series converges for any imaginary number ja. Also, how can you check the above equation while the left part of it cannot be interpreted in a certain way?? So, i see it as a definition and not as an equation that you can prove.

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Write $$e^{j\theta}=E(\theta)+jO(\theta)$$,
where $$E(\theta)=\frac{1}{2} (e^{j\theta}+e^{-j\theta})$$
and $$O(\theta)=\frac{1}{2j} (e^{j\theta}-e^{-j\theta})$$,
where the right-hand side is simply the sum of the even and odd parts of the left-hand side. $$j\neq 0$$ has no particular meaning right now, except that it is constant (independent of $$\theta$$).

Then \begin{align*} \frac{d}{d\theta} e^{j\theta} &= \frac{1}{2}\left( j (e^{j\theta}-e^{-j\theta} \right)+ j\frac{1}{2j}\left( j (e^{j\theta}+e^{-j\theta} \right)\\ je^{j\theta} &= j^2O(\theta)+jE(\theta) \end{align*}
where we have noted that
$$\frac{d}{d\theta}E(\theta)=j^2O(\theta)$$ and $$\frac{d}{d\theta}O(\theta)=E(\theta)$$... a coupled set of differential equations.

Continuing on, we find $$\frac{d^2}{d\theta^2}E(\theta)=j^2(E(\theta))$$ and $$\frac{d^2}{d\theta^2}O(\theta)=(j^2O(\theta))$$... two differential equations... or simply, two questions: what functions are proportional to their second derivatives? Let's now formally require that $$j^2=-1$$. So, now our question is: what functions are equal to minus their second derivatives? (With constants of integration, you'll need to consider the initial conditions to completely determine the functions E and O.)

I'll stop here... but you should be able to finish this off.

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We want to show,

$$e^{ix} = \cos x + i \sin x$$

Taylor expanding yields:

$$\sin x = x -\frac{x^3}{3!}+\frac{x^5}{5!} - \ldots$$
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots$$

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \ldots$$

Finish this off. Remember $(i)^2 = -1$

ex) $$(ix)^5 = i^2 i^2 i x^5 = (-1)(-1)i x^5 = ix^5$$

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lazypast said:
I see now, silly mistakes by me.
When a derivative is zero then the things differentiated must have been a constant

Indeed, that is correct. So, try taking our function, let $x=0$ and see where this takes you.

Of course Euler's formula can be derived using Taylor's series (as dextercioby and Frogpad said), you could also prove this using differentials; I was showing this method simply because you said that you hadn't met Taylor series before, but all proofs are equally valid,although I myself find the Taylor series most elegant.

well my original question has been answered now so thank you all.
as for that last this hootenanny do you mean e^ix= or f(x)=

f(x)=(1+0)/1=1 and e^ix: 1=1+0
but i don't understand what either of these prove

lazypast said:
well my original question has been answered now so thank you all.
as for that last this hootenanny do you mean e^ix= or f(x)=

f(x)=(1+0)/1=1 and e^ix: 1=1+0
but i don't understand what either of these prove

We have;

$$f(x):=\frac{\cos(x)+i\sin(x)}{e^{ix}}$$

Since we know that f(x) is a constant function it has the same value everywhere in its domain. Therefore, we can say that;

$$f(x) = f(0) \;\;\; \forall \;\;\;x \Rightarrow \frac{\cos(x)+i\sin(x)}{e^{ix}} = \frac{\cos(0)+i\sin(0)}{e^{i0}} = \frac{1}{1}$$

$$\therefore \frac{\cos(x)+i\sin(x)}{e^{ix}} = 1 \Leftrightarrow \cos(x)+i\sin(x) = e^{ix}$$

$$\Rightarrow R(\cos(x)+i\sin(x)) = R\cdot e^{ix}\hspace{1cm}\text{Q.E.D.}$$

## 1. What is a complex number?

A complex number is a number that contains both a real and an imaginary part. It is typically written in the form a + bi, where a is the real part and bi is the imaginary part (with i representing the imaginary unit).

## 2. What is the significance of the "j" in Re^{jθ}?

The "j" in Re^{jθ} represents the imaginary unit, which is equal to the square root of -1. It is used in complex numbers to distinguish the imaginary part from the real part.

## 3. What does the "θ" in Re^{jθ} represent?

The "θ" in Re^{jθ} represents the angle or phase of the complex number in polar form. It is measured in radians and determines the direction of the vector from the origin to the complex number.

## 4. How do you convert a complex number from rectangular form to polar form?

To convert a complex number from rectangular form (a + bi) to polar form (Re^{jθ}), you can use the following formulas: R = √(a^2 + b^2) and θ = tan^-1(b/a). This will give you the magnitude and angle of the complex number.

## 5. What is the significance of "e" in Re^{jθ}?

The "e" in Re^{jθ} represents Euler's number, which is a mathematical constant equal to approximately 2.71828. It is used in complex numbers to convert from rectangular form to polar form, and vice versa.

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