Proof that $e^{i\theta}/e^ {i\phi} = e^{i(\theta - \phi)}$

Question

Could anyone please help me with proving that

\begin{equation} \frac{e^{i\theta}}{e^{i\phi}}= e^{i(\theta-\phi)} \end{equation}

Is $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ useful in this proof?

Thank you

Answer 1

You can rationalise the denominator with the complex conjugate of $e^{i\phi}$ which will then give you: $$\frac{e^{i\theta}}{e^{i\phi}}=\frac{\left(\cos\theta+i\sin\theta\right)\left(\cos\phi-i\sin\phi\right)}{\left(\cos\phi+i\sin\phi\right)\left(\cos\phi-i\sin\phi\right)}\\ =\frac{\left(\cos\theta\cos\phi+\sin\theta\sin\phi\right)+i\left(\sin\theta\cos\phi-\cos\theta\sin\phi\right)}{\cos^2\phi-i^2\sin^2\phi}\\ =e^{i\left(\theta-\phi\right)}$$

Answer 2

Hint: The identity you provide is useful, the other key idea is :

$$\frac{a+b}{c+d} = \frac{a+b}{c+d} \frac{c-d}{c-d} = \frac{(a+b)(c-d)}{c^2-d^2} $$