# Thread: Prove e^z = cos (z) + i sin (z)

1. ## Prove e^{iz} = cos (z) + i sin (z)

Euler's formula states that $\displaystyle e^{\mathbf{i}\theta}=cos(\theta)+\mathbf{i}sin(\th eta), \ \ \forall\theta\in\mathbb{R}$.

Prove $\displaystyle e^{\mathbf{i}z}=cos(z)+\mathbf{i}sin(z), \ \ \forall z\in\mathbb{C}$

This seems trivial but I am going to ask for verification anyways.

Let $\displaystyle \ z\in\mathbb{C}, \ z=x+\mathbf{i}y, \ x,y\in\mathbb{R}$

$\displaystyle e^{\mathbf{i}z}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\ mathbf{i}x-y}=e^{-y}(cos(x)+\mathbf{i}sin(x))=e^{-y}e^{x\mathbf{i}}$

$\displaystyle =e^{x\mathbf{i}-y}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\mathbf{i}z}$

Is it just this simple?

2. Your title and the problem do not match. Which is it?

$\displaystyle \cos(z)=\cos(x)\cosh(y)-i~\sin(x)\sinh(y)$

$\displaystyle i~\sin(z)=-\cos(x)\sinh(y)+i~\sin(x)\cosh(y)$

3. I forgot the i in e^{iz} in the title.

4. Originally Posted by dwsmith
Euler's formula states that $\displaystyle e^{\mathbf{i}\theta}=cos(\theta)+\mathbf{i}sin(\th eta), \ \ \forall\theta\in\mathbb{R}$.

Prove $\displaystyle e^{\mathbf{i}z}=cos(z)+\mathbf{i}sin(z), \ \ \forall z\in\mathbb{C}$

This seems trivial but I am going to ask for verification anyways.

Let $\displaystyle \ z\in\mathbb{C}, \ z=x+\mathbf{i}y, \ x,y\in\mathbb{R}$

$\displaystyle e^{\mathbf{i}z}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\ mathbf{i}x-y}=e^{-y}(cos(x)+\mathbf{i}sin(x))=e^{-y}e^{x\mathbf{i}}$

$\displaystyle =e^{x\mathbf{i}-y}=e^{\mathbf{i}(x+\mathbf{i}y)}=e^{\mathbf{i}z}$

Is it just this simple?
Your proof seems circular to me. The usual approach is to use power series.

5. $\displaystyle \cos z+i\sin z=\dfrac{e^{iz}+e^{-iz}}{2}+i\;\dfrac{e^{iz}-e^{-iz}}{2i}=e^{iz}$

Fernando Revilla

6. Originally Posted by FernandoRevilla
$\displaystyle \cos z+i\sin z=\dfrac{e^{iz}+e^{-iz}}{2}+i\;\dfrac{e^{iz}-e^{-iz}}{2i}=e^{iz}$

Fernando Revilla
Circular proof.

The usual way of proving the formulas you have used is to use power series ....

7. Originally Posted by mr fantastic
Your proof seems circular to me. The usual approach is to use power series.
I am little confused about using the power series but here is what I have thought of so far.

$\displaystyle \displaystyle e^{\mathbf{i}}e^{z}\Rightarrow e^{\mathbf{i}}e^{z}=e^{\mathbf{i}}\sum_{n=0}^{\inf ty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3! }+\cdots +$

Now, how can I use this to show the e^{i} sum is cos(z)+ i sin(z)??

8. Originally Posted by mr fantastic
Circular proof.
No, it is not.

(i) You can define $\displaystyle e^z=e^x(\cos y+i\sin y),\quad z=x+iy\;(x,y\in \mathbb{R})$ where $\displaystyle \cos y,\;\sin y$ are the standard $\displaystyle \cos$ and $\displaystyle \sin$ on $\displaystyle \mathbb{R}$

(ii) You can define $\displaystyle \sin z=\dfrac{e^{iz}-e^{-iz}}{2i}$ and $\displaystyle \cos z=\dfrac{e^{iz}+e^{-iz}}{2}$ in strict terms of $\displaystyle e^{iz}$ and $\displaystyle e^{-iz}$ which have been already defined

The usual way of proving the formulas you have used is to use power series ....
It is a way, but not the only one.

Fernando Revilla

9. Originally Posted by dwsmith
I am little confused about using the power series but here is what I have thought of so far.

$\displaystyle \displaystyle e^{\mathbf{i}}e^{z}\Rightarrow e^{\mathbf{i}}e^{z}=e^{\mathbf{i}}\sum_{n=0}^{\inf ty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3! }+\cdots +$

Now, how can I use this to show the e^{i} sum is cos(z)+ i sin(z)??
$\displaystyle e^{iz} \neq e^{i} e^z$

10. Originally Posted by mr fantastic
$\displaystyle e^{iz} \neq e^{i} e^z$
I don't know what I was thinking. I am going to work on this later.

11. You can choose:

First proof:

$\displaystyle \cos z+i\sin z=\dfrac{e^{iz}+e^{-iz}}{2}+i\;\dfrac{e^{iz}-e^{-iz}}{2i}=e^{iz}$

Second proof:

$\displaystyle \cos z+i\sin z=$
$\displaystyle 1-\dfrac{z^2}{2!}+\dfrac{z^4}{4!}-\ldots$
$\displaystyle +i\;\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}-\right)=$
$\displaystyle 1+\dfrac{iz}{1!}+\dfrac{(iz)^2}{2!}+\dfrac{(iz)^3} {3!}+\dfrac{(iz)^4}{4!}+\dfrac{(iz)^5}{5!}+\ldots= e^{iz}$

Fernando Revilla

12. You can also doing this, if you know the theorems:

A) By appealing to certain facts about analytic continuation (namely, you can prove that your function is true on a certain line [namely the imaginary one] and go from there)

B) Use (although for this one I'm less sure that it works) particular theorems about the uniqueness to complex valued initial condition equations.

13. Originally Posted by Drexel28
A) By appealing to certain facts about analytic continuation (namely, you can prove that your function is true on a certain line [namely the imaginary one] and go from there)
Right. From

$\displaystyle e^{ix}=\cos x +i\sin x\;(\forall x\in\mathbb{R})$

and using the Milne-Thompson method, we inmediately obtain:

$\displaystyle e^{iz}=\cos z +i\sin z\;(\forall z\in\mathbb{C})$

Fernando Revilla

14. I liked this question even though I didn't know how to approach it originally.