Alexander Bass
This note is one of many taken during 2025

Proof by Multiplicative Inverse

We have functions f(x),g(x),h(x)f(x), g(x), h(x).

Given h(x)f(x)=1h(x) f(x) = 1. (they are multiplicative inverses)

We would like to determine whether f(x)=g(x)f(x) = g(x).

Naturally, h(x)g(x)=1   ⟹   f(x)=g(x)h(x)g(x) = 1\ \implies\ f(x) = g(x)

Application

I discovered this trick when searching for a proof to Euler’s Formula:

eit=cos⁡t+isin⁡t e^{it} = \cos{t} + i\sin{t}

We define

f(x)=eitg(x)=(cos⁡t+isin⁡t) \begin{aligned} f(x) &= e^{it}\\ g(x) &= \left(\cos{t} + i\sin{t}\right) \end{aligned}

Naturally, eith(x)=1e^{it}h(x) = 1, so

h(x)=e−it h(x) = e^{-it}

So, if we can prove that h(x)g(x)=1h(x)g(x) = 1, then it is clear that f(x)=g(x)f(x)=g(x). Introduce j(x)j(x) for brevity.

h(x)g(x)=j(x)=e−it(cos⁡t+isin⁡t) h(x)g(x) = j(x) = e^{-it}\left(\cos{t} + i\sin{t}\right)

Differentiate j(x)j(x)

j′(x)=e−it(−sin⁡t+icos⁡t)−ie−it(cos⁡t+isin⁡t)=e−it(−sin⁡t+icos⁡t)+e−it(−icos⁡t+sin⁡t)=e−it(icos⁡t−sin⁡t+sin⁡t−icos⁡t)=0 \begin {aligned} j'(x) &= e^{-it}\left(-\sin{t} + i\cos{t}\right) - i e^{-it}\left(\cos{t} + i\sin{t}\right)\\ &= e^{-it}\left(-\sin{t} + i\cos{t}\right) + e^{-it}\left(-i\cos{t} + \sin{t}\right)\\ &= e^{-it}\left(i\cos{t} -\sin{t} + \sin{t} -i\cos{t} \right)\\ &= 0 \end{aligned}

That is, j′(x)=0j'(x) =0 for all xx, which means that j(x)j(x) is constant for all xx.

Find value of j(x)j(x)

j(0)=e0(cos⁡0+isin⁡0)=1 j(0) = e^{0}\left(\cos{0} + i\sin{0}\right) = 1

So, j(x)=1j(x) = 1 for all xx, which shows that g(x)g(x) is the multiplicative inverse of h(x)h(x). Given that g(x)g(x) is also the multiplicative inverse of h(x)h(x):

f(x)=g(x) f(x) = g(x)

Source: https://math.stackexchange.com/a/8612

This note is one of many taken during 2025