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