$e^{i\theta}$ versus $\cos\theta+i\sin\theta$

Question

I am teaching an basic university maths course, and have been thinking about the complex numbers part. Specifically, I was wondering why I should include Euler's formula in my course.

This led me to the following "big list" question, which I thought interesting.

What can you do with $e^{i\theta}$ that is much harder/impossible with $\cos\theta+i\sin\theta$?

For example, $e^{i\theta}$ makes it easy to see that $\operatorname{arg}(uv)=\operatorname{arg}(u)+\operatorname{arg}(v)$.

Answer 1

It makes De Moivre's Formula much more obvious. Also n-th roots.

Are these engineering students? It's very convenient for dealing with alternating current.

There's bound to be more, but these spring to mind first.

Also this

Answer 2

It becomes trivial to derive the multiple-angle or half-angle formulae u\sing Euler's identity. The geometric derivation is long-winded and painful to visualise, but it's easy if you use the fact that $\frac{e^{i\theta}+e^{-i\theta}}{2}=\cos(\theta)$ and $\frac{e^{i\theta}-e^{-i\theta}}{2}=i \sin(\theta)$, and just apply Euler's identity and the rules for exponents. For example:

$$\begin{align}i \sin(\alpha + \beta) & =\frac{e^{i(\alpha + \beta)}-e^{-i(\alpha + \beta)}}{2} \\[1ex] &=\frac{e^{i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}}{2} \\[1ex] &=\frac{(\cos(\alpha) + i \sin(\alpha))(\cos(\beta) + i \sin(\beta))-(\cos(-\alpha) + i \sin(-\alpha))(\cos(-\beta) + i \sin(-\beta))}{2} \\[1ex] &=\frac{(\cos(\alpha) + i \sin(\alpha))(\cos(\beta) + i \sin(\beta))-(\cos(\alpha) - i \sin(\alpha))(\cos(\beta) - i \sin(\beta))}{2} \\[1ex] &=\frac{(\cos(\alpha)\cos(\beta)+i \sin(\alpha)\cos(\beta)+i\cos(\alpha)\sin(\beta)-\sin(\alpha)\sin(\beta))-(\cos(\alpha)\cos(\beta)-i \sin(\alpha)\cos(\beta)-i\cos(\alpha)\sin(\beta)-\sin(\alpha)\sin(\beta))}{2} \\[1ex] &=\frac{2i \sin(\alpha)\cos(\beta) + 2i\cos(\alpha)\sin(\beta)}{2} \\[1ex] &=i(\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)) \end{align}$$

Hence,

$$\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$$

Answer 3

• Calculating integrals like $$\int_0^\infty\frac{\cos(x)}{1+x^2}dx$$ is difficult without Euler's identity.
• The physics of waves and optics is greatly simplified.
• Fourier transforms are easier to compute and work with, and have a more intuitive meaning, I think.
• There are a lot of applications in quantum mechanics.
• Almost all trigonometric identities are easily derived from this.

Answer 4

Complex Analysis uses a lot the logarithm. As you said with $e^{i\theta}$ you can talk more easily for the argument. The important thing is that a branch of the argument exists iff exist a branch of the logarithm.

Therefore, exponential goes with logarithm. Also of the expressions $z^a$ are exponential forms...

Answer 5

Developing products of several terms is painful in the $\text{cis}$ notation as the number of terms grows exponentially.

Compare

$$e^{ia}e^{ib}e^{ic}=e^{i(a+b+c)}$$

to

$$(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))(\cos(c)+i\sin(c))=\\ \cos(a)\cos(b)\cos(c)-\sin(a)\sin(b)\cos(c)-\sin(a)\sin(c)\cos(b)-\sin(b)\sin(c)\cos(a)+\\i\sin(a)\cos(b)\cos(c)+i\sin(b)\cos(a)\cos(c)+i\sin(c)\cos(a)\cos(b)-i\sin(a)\sin(b)\sin(c)=\\ \cos(a+b+c)+i\sin(a+b+c).$$

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Now try by hand

$$\frac{e^{ia}e^{ib}}{e^{ic}e^{id}}=e^{i(a+b-c-d)}$$and let me know.