I found an interesting case where it seems like an equality sign works wrong.

Let's consider the following construction:

$\frac{1+\Lambda}{2} e^{i\Lambda \phi}$ where $\Lambda = \pm1$, so $\Lambda^2=1$.

Then I apply Euler formula:

$\frac{1+\Lambda}{2} e^{i\Lambda \phi} = \frac{1+\Lambda}{2} (\cos \phi + i\Lambda \sin \phi)= \frac{1}{2} \cos \phi + \frac{\Lambda}{2} \cos \phi + \frac{i\Lambda}{2} \sin \phi + \frac{i}{2} \sin \phi = \frac{1+\Lambda}{2} e^{i\phi}$

where I have used $\sin(\Lambda \phi) = \Lambda \sin \phi$ and $\cos (\Lambda \phi) = \cos \phi$.

However, this is just wrong! $e^{i\Lambda \phi}\neq e^{i \phi}$ even though the equality sign was not broken anywhere in between (at least it doesn't seem to be broken to me). What am I doing wrong?

  • 2
    $\begingroup$ How'd you get $\cos \phi + i\Lambda\sin \phi = e^{i\phi}$? $\endgroup$
    – Brian Tung
    Oct 19, 2016 at 17:10
  • $\begingroup$ Your are forgetting the $\Lambda$ in the second term, hence the LHS does not contain $e^{i\phi}$ $\endgroup$
    – b00n heT
    Oct 19, 2016 at 17:11
  • $\begingroup$ @b00n heT I hope it is more obvious now. $\endgroup$
    – MsTais
    Oct 19, 2016 at 17:13
  • 1
    $\begingroup$ @BrianTung : Don't forget that $\Lambda$ is only $\pm 1$ $\endgroup$
    – MPW
    Oct 19, 2016 at 17:15
  • 1
    $\begingroup$ There are three $\cos$ and a single $\sin$ in your expansion. Fix that. $\endgroup$
    – user65203
    Oct 19, 2016 at 17:17

3 Answers 3


The problem is the division by zero:

The identity $\tfrac{1+\Lambda}{2}e^{i\Lambda\phi}=\tfrac{1+\Lambda}{2}e^{i\phi}$ holds as you have shown.

For $\Lambda=1$, this implies $e^{i\Lambda\phi}=e^{i\phi}$, which is obviously correct. However, for $\Lambda=-1$, this means that $\tfrac{1+\Lambda}{2}=0$ and therefore, the last implication $\tfrac{1+\Lambda}{2}e^{i\Lambda\phi}=\tfrac{1+\Lambda}{2}e^{i\phi}\implies e^{i\Lambda\phi}=e^{i\phi}$ is not correct in this case.


Your third term should be $\frac{i\Lambda}{2}\sin \phi$, not $\frac{i\Lambda}{2}\cos\phi$.


You are correct that $e^{i\Lambda \phi}\neq e^{i \phi}$.

But your computation doesn't show that $e^{i\Lambda \phi}= e^{i \phi}$; it shows that $\frac{1+\Lambda}{2}e^{i\Lambda \phi}= \frac{1+\Lambda}{2}e^{i \phi}$, which is true for $\Lambda=\pm 1$.

  • $\begingroup$ But usually I can just cancel $\frac{1+\Lambda}{2}$ on both sides, which I obviously can't do here. It seems like an analogue of circular equation, which makes me think that I have to use a vector description here, rather than treating it as a double-valued function... $\endgroup$
    – MsTais
    Oct 20, 2016 at 16:46
  • 1
    $\begingroup$ @MsTais: As Lukas has pointed out, you can only cancel a non-zero multiplier on both sides. I really don't see your problem. $\endgroup$
    – TonyK
    Oct 20, 2016 at 16:54

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