A Shorter Proof of Euler's Formula

Euler's Formula is the very famous equation:

e^ix = cos x + isin x

In a previous blog, I showed how it can be derived using the Taylor Series. In today's blog, I will show how it can be derived in an even simpler way using concepts from calculus.

Theorem: Euler's Formula

e^ix = cos x + isin x

Proof:

(1) For some number x:

let y = cos x + isin x

(2) Taking the first derivative of both sides gives us:

dy/dx = -sin x + icos x

[For details if needed:

(a) dy/dx = d(cos x + isin x)/dx

(b) d(cos x + isin x)/dx = d(cos x)/dx + d(isin x)/dx (see Lemma 3, here)

(c) d(cos x)/dx = -sin x (see Theorem 2, here)

(d) d(isin x)/dx = icos x (see Theorem 1, here) ]

(3) Since i² = -1 by definition, we have

-sin x + icos x = i(isin x + cos x)

(4) Combining step #1, #2, and #3, we get:

dy/dx = iy

[Since:

(a) y = isin x + cos x [from step #1 above]

(b) dy/dx = -sin x + icos x [from step #2 above]

(c) dy/dx = i(isin x + cos x) [from step #3 above]

(d) dy/dx = i(y)

]

(5) If we multiply (dx/y) to each side, we get:

(1/y)dy = (i)dx

(6) Now, if we take the integral of each side we get:

ln y = ix

[For details if needed:

(a) d(ln x)/dy = 1/x [See Lemma 1, here for proof]

(b) So using the Fundamental Theorem of Calculus (see Theorem 2, here), we know that:

∫ (1/x)dx = ln x + C

(c) So, ∫ (1/y)dy = ln y + C

(d) ∫ i(dx) = ix + C [Since d(ix + C)/dx = i, see Lemma 2, here]

(e) So putting this together gives us:

ln y = ix

]

(7) Now putting e the power of both sides, we get:

y = e^ix [Since e^{ln y} = y]

(8) Now combining step #1 with step #7 we get:

e^ix = cos x + isin x

QED

References

• "Proof of Euler's Formula", August 15, 2007, Those Who Can Teach Blog.
• "Proof of Euler's Formula (II)", August 15, 2007, Those Who Can Teach Blog.