0
$\begingroup$

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

$\endgroup$
7
  • 1
    $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ Feb 17, 2022 at 9:21
  • 3
    $\begingroup$ Refer to the quotient rule of exponents: $a^x/a^y=a^{x-y}$ $\endgroup$ Feb 17, 2022 at 9:26
  • 4
    $\begingroup$ @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. $\endgroup$
    – Arthur
    Feb 17, 2022 at 9:27
  • $\begingroup$ They are still subjected to the quotient rule, so $a^{ix}/a^{iy}=a^{ix-iy}=a^{i(x-y)}$ $\endgroup$ Feb 17, 2022 at 9:30
  • $\begingroup$ @Luthier As Arthur said, this need more explanation since exponents rules are not always valid for complex numbers $\endgroup$
    – jjagmath
    Feb 17, 2022 at 9:34

2 Answers 2

6
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ Thank you, that's exactly how I tried to solve this problem but I got stuck at the end. $\endgroup$
    – MathNoob
    Feb 17, 2022 at 14:54
0
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ Thank you, this is what I was looking for. I didn't know how to solve the second to last step, even now I am a bit confused just by looking at it. Could you please elaborate on that? $\endgroup$
    – MathNoob
    Feb 17, 2022 at 14:58
  • $\begingroup$ @MathNoob I just substituted the formula of $\cos\left(\theta-\phi\right)$ and $\sin\left(\theta-\phi\right)$ and the denominator becomes $\sin^2\phi+\cos^2\phi=1$ $\endgroup$
    – DeBARtha
    Feb 18, 2022 at 7:20
  • $\begingroup$ I didn't know that formula, now it all makes sense. Thank you very much for your help. $\endgroup$
    – MathNoob
    Feb 18, 2022 at 9:04

Not the answer you're looking for? Browse other questions tagged .