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

$$$$\frac{e^{i\theta}}{e^{i\phi}}= e^{i(\theta-\phi)}$$$$

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

Thank you

• Refer to the quotient rule of exponents: $a^x/a^y=a^{x-y}$ Feb 17, 2022 at 9:26
• @Luthier It is not obvious that the quotient rule for exponents is valid for complex numbers. It would need a little more explanation to be a full solution. Feb 17, 2022 at 9:27
• They are still subjected to the quotient rule, so $a^{ix}/a^{iy}=a^{ix-iy}=a^{i(x-y)}$ Feb 17, 2022 at 9:30
• @Luthier As Arthur said, this need more explanation since exponents rules are not always valid for complex numbers Feb 17, 2022 at 9:34

$$\frac{a+b}{c+d} = \frac{a+b}{c+d} \frac{c-d}{c-d} = \frac{(a+b)(c-d)}{c^2-d^2}$$
You can rationalise the denominator with the complex conjugate of $$e^{i\phi}$$ which will then give you: $$\frac{e^{i\theta}}{e^{i\phi}}=\frac{\left(\cos\theta+i\sin\theta\right)\left(\cos\phi-i\sin\phi\right)}{\left(\cos\phi+i\sin\phi\right)\left(\cos\phi-i\sin\phi\right)}\\ =\frac{\left(\cos\theta\cos\phi+\sin\theta\sin\phi\right)+i\left(\sin\theta\cos\phi-\cos\theta\sin\phi\right)}{\cos^2\phi-i^2\sin^2\phi}\\ =e^{i\left(\theta-\phi\right)}$$
• @MathNoob I just substituted the formula of $\cos\left(\theta-\phi\right)$ and $\sin\left(\theta-\phi\right)$ and the denominator becomes $\sin^2\phi+\cos^2\phi=1$ Feb 18, 2022 at 7:20