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Why does $4e^{i\pi}=-4$?

It means that $e^{i\pi}=-1$ why is that?

mrf
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gbox
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    $e^{ix}$ is defined as $\cos x + i\sin x$ for $x \in \mathbb{R}$. See for example: http://math.stackexchange.com/questions/155068 – mrf Oct 20 '15 at 15:50
  • Technically, you can define $e^z$ in general, and prove it. For example, if you define $e^z$ in terms of the power series, or $\lim_{n\to\infty} \left(1+z/n\right)^n$. @mrf – Thomas Andrews Oct 20 '15 at 15:57
  • I'm well aware that there are many different ways to define $e^z$, but going via Euler's formulas seem to be the most common one. – mrf Oct 20 '15 at 15:58

1 Answers1

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Notice,

Using Euler's formula $e^{i\theta}=\cos \theta+i\sin \theta$ as follows $$4e^{i\pi}=4(\cos \pi+i\sin \pi)=4(-1+i(0))=-4$$

Harish Chandra Rajpoot
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