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I have always been wondering why the complex number of the form $\cos\theta + i\sin\theta$ is represented as $e^{i\theta}$. I fully understand how the polar form and euler form works. My only doubt is how the polar is worked out in euler form.

For example, taking two complex numbers in polar form $\cos\theta_1 + i\sin\theta_1$ and $\cos\theta_2 + i\sin\theta_2$.
The division of these two numbers can be evaluated in the euler form. Like $\frac{e^{i\theta_1}}{e^{i\theta_2}}$ is equal to $e^{i(\theta_1 - \theta_2)}$.

So, on the Bottomline my question is : Why can the polar be represented in euler form and also be operated vice versa?.
Hope you get my question intuitively. Please clarify my doubt. Thank you!

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  • $\begingroup$ expand polar form and Euler form using Maclaurin series, they become the same. Does this make sense? $\endgroup$ – qsmy Feb 20 at 2:18
  • $\begingroup$ What is maclaurin series. I have not heard about it sorry $\endgroup$ – Ja jack Feb 20 at 2:37
  • $\begingroup$ MacLaurin series is a Taylor series centered on zero. $\endgroup$ – Andrei Feb 20 at 3:29
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To completely understand why $\cos\theta+i\sin\theta=e^{i\theta}$, you need to understand Maclaurin series. Using the Maclaurin series, we extend the domain of the exponential function to the field of complex numbers $\mathbb C$. But, if you don't know Maclaurin series yet, you can still gain sufficient understanding to feel comfortable using this formula.

For the time being, it is useful to view this formula just as a "convention of notation" - a compact way of writing $\cos\theta+i\sin\theta$.


Why do all the formulas work?

This fact is due to the following formulas: \begin{eqnarray} &(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2) = (\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))& \tag{1}\\ &(\cos\theta+i\sin\theta)^n = (\cos n\theta + i\sin n\theta)& \tag{2} \end{eqnarray} The first formula can be easily proven by just cross-multiplying the LHS and using the additive trigonometric formulas for cos and sin. One can see that the argument of the result is the sum of the arguments of the two inputs, just like when we multiply two complex numbers in Euler form.

The second formula can be easily proven for $n \in\mathbb N$ using mathematical induction, although it is valid for any $n \in\mathbb R$. The formula is called Moivre's formula. Even if you do not know how to prove these, the proofs are easy to find on the internet.

As for why division works, use that $$\frac{e^{i\theta_1}}{e^{i\theta_2}}=e^{i(\theta_1-\theta_2)}$$ is equivalent with $$e^{i\theta_2}e^{i(\theta_1-\theta_2)}=e^{i\theta_1}$$ Changing this into trigonometric form and using formula (1), one can see why this equality holds.

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