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?
@scottleung95873 күн бұрын
Nice job!
@RyanLewis-Johnson-wq6xs3 күн бұрын
It’s in my head.
@Blaqjaqshellaq3 күн бұрын
z=(4*n+3)/(4*m+1) where n and m are integers. (The denominator is never 0.)
@Don-Ensley2 күн бұрын
z = (4M+1)/[4(K+N)+3], K,N,M integers
@RyanLewis-Johnson-wq6xs3 күн бұрын
(-i)^z=i z=4k+3 k=z z as in any integer z=-1
@stevemonkey66663 күн бұрын
An American says CONgruent but a Brit or Australian says conGRUent
@viniaz29973 күн бұрын
Funny thing - Russians stress the third syllable in that word. 😅
@TypoKnig3 күн бұрын
American here - I also stress the second syllable. Prof. Michael Penn does also.
@adamrussell6582 күн бұрын
Ive heard it both ways, but I think I would stress 2nd syllable. Maybe CONgruent is an east coast thing? 🙂
@seanfraser31253 күн бұрын
z = 4n+3
@ionlyemergeafterdark3 күн бұрын
( -i )( -i )( -i ) = ( i )( i )( -i ) = - ( -i) = i. So I conclude that z = 3.