# Taking out the exponential by lowering its exponent

In a book I am studying, while solving something, it does the following thing:

From here $$e^{-iωτ}=\frac{1}{2}-iω$$

it goes here

$$τ=\frac{-arg(1/2-iω)}{ω}$$

However I don't get how this happened. I looked up properties to do it my self, but I can't find it. Can anybody tell me what properties where used to do this?

P.S. I hope this question is valid for this Q&A. I have spent 2 hours trying to solve this, something that I think is really easy.

• We have $e^{i(-\omega \tau)} = \frac 1 2 - i\omega$. By definition of $\mathrm{arg}$, we have $-\omega \tau = \mathrm{arg}(1/2 - i\omega)$. Then you just solve for $\tau$. – Trevor Norton Mar 26 '18 at 19:30
• @TrevorNorton now I see what you mean. I was obvious but I still couldn't see it. Thanks – Dimitris Pantelis Mar 27 '18 at 17:00

The property used is ( based on the definitions of $arg()$ and $e^{θi}$ ): $$arg(e^{θi})=θ$$
So by applying $arg()$ on both sides, we can get the exponential out.