Another Exponent That Negates | Problem 403

[@tedczotter5297]

Huh? I saw by inspection that z = 1/3 works (i is a cube root of -i). For sake of brevity, define q = i*pi/2, then -i is e^(q(3 + 4k)) and i is e^(q(1 + 4n)). This reveals the general formula is z = (1 + 4n) / (3 + 4k). Is that right or no?

[@scottleung9587]

Nice job!

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@Blaqjaqshellaq]

z=(4*n+3)/(4*m+1) where n and m are integers. (The denominator is never 0.)

[@Don-Ensley]

z = (4M+1)/[4(K+N)+3], K,N,M integers

[@RyanLewis-Johnson-wq6xs]

(-i)^z=i z=4k+3 k=z z as in any integer z=-1

[@stevemonkey6666]

An American says CONgruent but a Brit or Australian says conGRUent

[@viniaz2997]

Funny thing - Russians stress the third syllable in that word. 😅

[@TypoKnig]

American here - I also stress the second syllable. Prof. Michael Penn does also.

[@adamrussell658]

Ive heard it both ways, but I think I would stress 2nd syllable. Maybe CONgruent is an east coast thing? 🙂

[@seanfraser3125]

z = 4n+3

[@ionlyemergeafterdark]

( -i )( -i )( -i ) = ( i )( i )( -i ) = - ( -i) = i. So I conclude that z = 3.