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I would like to proof that: $$z = |z|\big(\cos(\phi) + i \sin(\phi)\big)$$ of course the second part can be shown using Euler's formula. That's why I would like to prove that: $$z = |z|e^{i \phi}$$ I can use only the series definition of exp and some basic properties like $e^{a+b} = e^ae^b$.
I think it must be quite trivial but I really don't know how to do it properly.

I think that my problem is a bit different to the one in the suggested post. I do know how to prove that $e^{i\phi} = \cos \phi + i\sin \phi$ but I do not know how to prove that $$\forall_{z \in \mathbb{C}} z = |z|\exp(ix)$$ for some $x \in \mathbb{R}$.

Hendrra
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3 Answers3

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I don't know if this is what you want, but you can alwyas prove Euler's formula:\begin{align}e^{x+yi}&=e^xe^{yi}\\&=e^x\left(1+yi+\frac{(yi)^2}{2!}+\frac{(yi)^3}{3!}+\cdots\right)\\&=e^x\left(\left(1-\frac{y^2}{2!}+\frac{y^4}{4!}-\cdots\right)+\left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots\right)i\right)\\&=e^x\bigl(\cos(y)+\sin(y)i\bigr).\end{align}

José Carlos Santos
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Every $z\in\mathbb{C}$ can be written as $$z=x+iy$$ with $x,y\in\mathbb{R}$. By definition $|z|:=\sqrt{x^2+y^2}$. If $\theta$ is the angle the vector $(x,y)$ makes with the horizontal axis (the real axis) in the positive direction then $x=|z|\cos\theta$ and $y=|z|\sin\theta$. Therefore $$z=x+iy=|z|\cos\theta+i|z|\sin\theta=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}$$ where the last equality follows from Euler's identity $$e^{i\theta}=\cos\theta+i\sin\theta$$

Arian
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If

$z \ne 0, \tag 1$

then

$\left \vert \dfrac{z}{\vert z \vert} \right \vert = \dfrac{\vert z \vert}{\vert z \vert} = 1; \tag 2$

thus

$\dfrac{z}{\vert z \vert} \in S^1, \tag 3$

the unit circle in $\Bbb C$; therefore, we may always find some $\phi \in \Bbb R$ with

$\dfrac{z}{\vert z \vert} = \cos \phi + i \sin \phi; \tag 4$

it remains to show that

$\cos \phi + i \sin \phi = e^{i \phi}; \tag 5$

but this may easily be done by expanding $e^{i \phi}$ in a power series

$e^{i\phi} = \displaystyle \sum_0^\infty \dfrac{(i\phi)^n}{n!}, \tag 6$

and separating out the real and imaginary parts, as is shown in this wikipedia article, as well as by Jose Carlos Santos in his answer. Then (4) becomes

$\dfrac{z}{\vert z \vert} = \cos \phi + i \sin \phi = e^{i\phi}, \tag 7$

or

$z = \vert z \vert e^{i\phi}. \tag 8$

Note Added in Edit, Tuesday 3 April 2018 10:34 AM PST: This in response to the comment to this answer made by our OP Hendrra. The easiest way to see that

$z \in S^1 \Longrightarrow \exists \phi \in \Bbb R, \; z = \cos \phi + i \sin \phi, \tag 9$

is via simple geometry and trigonometry. Since $z$ is a point on the unit circle, there is a radial line segment 'twixt the origin $O$ and $z$, $\overline{Oz}$; the length of this segment is $1$, since $S^1$ is the "unit circle". Then let $\phi$ be the angle 'twixt the positive $x$-axis and the segment $\overline{Oz}$; the $x$-coordinate of the point $z$ is then the real part of $z$ considered as a complex number:

$\Re(z) = \vert \overline{Oz} \vert \cos \phi = \cos \phi, \tag{10}$

since $\vert \overline{Oz} \vert = 1$; likewise the $y$-coordinate is

$\Im(z) = \sin \phi; \tag{11}$

thus

$z = \Re(z) + i\Im(z) = \cos \phi + i \sin \phi. \tag{12}$

End of Note.

Robert Lewis
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