Q. Euler’s formula \(e^{ix} = \cos(x) + i \sin(x)\).

Q: What is a quick proof of Euler's formula?

Expand e^{ix}=\sum_{n=0}^\infty \frac{(ix)^n}{n!} and compare real/imaginary parts with the Taylor series for \cos x and \sin x; they match, giving e^{ix}=\cos x+i\sin x.

Q: What is Euler's identity and why is it famous?

Substitute x=\pi to get e^{i\pi}+1=0. It elegantly links e,\;i,\;\pi,\;1,\;0 in one equation, often cited for aesthetic unity in mathematics.

Q: How do I get \cos and \sin from exponentials?

Use the identities \cos x = \frac{e^{ix} + e^{-ix}}{2} and \sin x = \frac{e^{ix} - e^{-ix}}{2i}, obtained by solving e^{ix} and e^{-ix} for real and imaginary parts.

Q: What is the geometric interpretation of e^{ix}?

e^{ix} represents a point on the unit circle at angle x from the positive real axis: coordinates (\cos x,\sin x). Multiplication by e^{i\theta} rotates complex numbers by \theta.

Q: How does Euler's formula give trig addition formulas?

Use e^{i(a+b)}=e^{ia}e^{ib}. Equate real and imaginary parts to derive \cos(a+b)=\cos a\cos b-\sin a\sin b and \sin(a+b)=\sin a\cos b+\cos a\sin b.

Q: What is De Moivre's formula and its relation?

De Moivre: (\cos x+i\sin x)^n=\cos(nx)+i\sin(nx). It follows directly from Euler: e^{inx}=(e^{ix})^n.

Q: How do derivatives and integrals work with e^{ix}?

Differentiate: \frac{d}{dx}e^{ix} = ie^{ix}, so \frac{d}{dx}\cos x = -\sin x and \frac{d}{dx}\sin x = \cos x. Integrals follow by inverting differentiation or using the exponential form.

Answer

Start from the power series definitions:

\[e^z = \sum_{n=0}^\infty \frac{z^n}{n!}, \quad \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}, \quad \sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}.\]

Put \(z = ix\):

\[e^{ix} = \sum_{n=0}^\infty \frac{(ix)^n}{n!} = \sum_{n=0}^\infty \frac{i^{2n} x^{2n}}{(2n)!} + \sum_{n=0}^\infty \frac{i^{2n+1} x^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} + i\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}.\]

Hence

\[e^{ix} = \cos x + i \sin x.\]

Detailed Explanation

Derivation of Euler’s formula: \(e^{ix} = \cos x + i \sin x\)

Start with the Taylor series definitions (valid for all complex numbers).

Recall the power series expansions centered at 0 for the exponential, cosine, and sine functions.

\[e^{z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\]

\[\cos x=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}\]

\[\sin x=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}\]

These series converge absolutely for every complex number \(z\), so substitution and regrouping of terms are justified.
Substitute \(z = ix\) into the exponential series.

Replace \(z\) by \(ix\) in the expansion of \(e^{z}\) to obtain the series for \(e^{ix}\):

\[e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^{n}}{n!}\]
Expand \((ix)^{n}\) using powers of \(i\).

Compute the factor \((ix)^{n}\) as \(i^{n} x^{n}\), and recall the cycle of powers of \(i\):

\[i^{0}=1,\quad i^{1}=i,\quad i^{2}=-1,\quad i^{3}=-i,\quad i^{4}=1,\ \text{and so on.}\]

Thus the series becomes

\[e^{ix}=\sum_{n=0}^{\infty}\frac{i^{n}x^{n}}{n!}\]
Separate the series into even and odd terms.

Break the index \(n\) into even \(n=2k\) and odd \(n=2k+1\) and write two sums.

\[e^{ix}=\sum_{k=0}^{\infty}\frac{i^{2k}x^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{i^{2k+1}x^{2k+1}}{(2k+1)!}\]

Evaluate \(i^{2k}\) and \(i^{2k+1}\) using \(i^{2}=-1\):

\[i^{2k}=(i^{2})^{k}=(-1)^{k},\qquad i^{2k+1}=i\cdot i^{2k}=i(-1)^{k}\]

Substitute these into the two sums:

\[e^{ix}=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}+i\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}\]
Recognize the cosine and sine series.

Compare the two sums with the known series for cosine and sine given in step 1. The first sum equals \(\cos x\) and the second sum equals \(\sin x\). Therefore

\[e^{ix}=\cos x + i\,\sin x\]
Conclusion and justification.

The equality holds for every real \(x\) because the power series for \(e^{z}\), \(\cos x\), and \(\sin x\) converge absolutely and uniformly on compact sets of the complex plane, which permits the substitution \(z=ix\) and the rearrangement into even and odd parts. This yields Euler’s formula:

\[e^{ix}=\cos x + i\sin x\]

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FAQs

What is a quick proof of Euler's formula?

Expand \(e^{ix}=\sum_{n=0}^\infty \frac{(ix)^n}{n!}\) and compare real/imaginary parts with the Taylor series for \(\cos x\) and \(\sin x\); they match, giving \(e^{ix}=\cos x+i\sin x\).

What is Euler's identity and why is it famous?

Substitute \(x=\pi\) to get \(e^{i\pi}+1=0\). It elegantly links \(e,\;i,\;\pi,\;1,\;0\) in one equation, often cited for aesthetic unity in mathematics.

How do I get \(\cos\) and \(\sin\) from exponentials?

Use the identities \(\cos x = \frac{e^{ix} + e^{-ix}}{2}\) and \(\sin x = \frac{e^{ix} - e^{-ix}}{2i}\), obtained by solving \(e^{ix}\) and \(e^{-ix}\) for real and imaginary parts.

What is the geometric interpretation of \(e^{ix}\)?

\(e^{ix}\) represents a point on the unit circle at angle \(x\) from the positive real axis: coordinates \((\cos x,\sin x)\). Multiplication by \(e^{i\theta}\) rotates complex numbers by \(\theta\).

How does Euler's formula give trig addition formulas?

Use \(e^{i(a+b)}=e^{ia}e^{ib}\). Equate real and imaginary parts to derive \(\cos(a+b)=\cos a\cos b-\sin a\sin b\) and \(\sin(a+b)=\sin a\cos b+\cos a\sin b\).

How does this help compute powers and roots of complex numbers?

What is De Moivre's formula and its relation?

De Moivre: \((\cos x+i\sin x)^n=\cos(nx)+i\sin(nx)\). It follows directly from Euler: \(e^{inx}=(e^{ix})^n\).

How do derivatives and integrals work with \(e^{ix}\)?

Differentiate: \(\frac{d}{dx}e^{ix} = ie^{ix}\), so \(\frac{d}{dx}\cos x = -\sin x\) and \(\frac{d}{dx}\sin x = \cos x\). Integrals follow by inverting differentiation or using the exponential form.

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