An Exponential That Conjugates!😮 | Problem 236

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[@stevemonkey6666]

Would be interesting to see a general form for conversion from a complex number to its conjugate

[@harley_2305]

I don’t wanna be that guy because I agree with you it is interesting, but after working through it I looked back and I saw that you could come up with a general form by looking at the solution to this particular question or any question like it.
The general form would look something like z = [ln(modW) - (argW)i] / [ln(modW) + (argW)i] where W = a + bi. I’m sure you could probably neaten it up a bit but it’s pretty neat

[@Nobodyman181]

Yeeee

[@LucasEccard]

I got the same result but there are infinite roots for z. Just remember that 1-√3 = 2.e^(-π/3 + 2nπ)i.
Thus, the values for z are gonna be
Z = [3ln2 + (6n-1)iπ]/(3ln2 + iπ)

[@scottleung9587]

I got z=6n+5.