An Exponent That Negates Euler's Number | Problem 387

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This problem was suggested by ‪@viniaz2997‬!
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[@elmer6123]

e^{iπ}=-1 and e^1=e. Thus, e^{iπ}*e^1=e^{1+iπ}=(-1)*e=-e, so z=1+iπ

[@joelproko]

My first thought seeing the problem was basically the same, except -1=e^{i(π+2nπ)}

[@DavidIsFrenchTemp]

​@@joelprokoyeah I totally understand what's going on here

[@orangenostril]

I think it's more intuitive if you divide both sides by e at the beginning
e^(z-1)=-1, and e^(i(pi+2pi*n))=-1, so z-1=i(pi+2pi*n)

[@viniaz2997]

Thank you! I thought it would be easy and fun, and it is!

[@SweetSorrow777]

This one looks easy. The rhs can be rewritten as -1 * e. Then -1 can be e^(i*(PI + 2PI))*n.

[@joelproko]

The *n needs to be inside the brackets right next to 2PI ;-)

[@seitanarchist]

“And since the bases are equal, the exponents are equal…”
This is incorrect. You nearly have a correct argument, but this line is off. The complex exponential function is not one-to-one. Instead, this line is where you could introduce the additional 2pi*n terms: e^z = e^(1+i*pi) iff z = 1+i*(pi + 2pi*n).

[@mikecaetano]

Also, what about ln(-1) where i^2 is substituted for -1?

[@scottleung9587]

Nice!

[@TazwaarAhmed09]

e^z = -e
ln(e^z) = ln(-1 x e)
z = ln(-1) +ln(e)
z = ln(-1) + 1
hows this?
edit: i was halfway in the vid at time of comment

[@yogesh193001]

z= ipi+1+2npi

[@trojanleo123]

z = 1 + i(2n+1)π
Where n ∈ ℤ