der's an easy way-suppose da rows are r1,r2,r3-den r1=r1+r2+r3..take dem common ..you end up with (z+1+w+w2)(a matrix which is not equal to zero)..so for da product to be zero-z+1+w+w2=0 and 1+w+w2=0....so z sud be zero ....dats it rt!!!!!!!!
02:00 should that be positive omega fourth because positive omega squared time positive omega squared should have give positive omega fourth in the last determinant instead of Negative omega fourth in the last determinant ? or is there any other way to do it?
why doesn't z^3 have just 1 root. can you please explain this. since z^n = 1 is nth root of unity. there should be nth root of 0. why is it just 1 ? complex just moved to the next level.
I love it when complex problems simplify so much!
der's an easy way-suppose da rows are r1,r2,r3-den r1=r1+r2+r3..take dem common ..you end up with (z+1+w+w2)(a matrix which is not equal to zero)..so for da product to be zero-z+1+w+w2=0 and 1+w+w2=0....so z sud be zero ....dats it rt!!!!!!!!
Omg
02:00 should that be positive omega fourth because positive omega squared time positive omega squared should have give positive omega fourth in the last determinant instead of Negative omega fourth in the last determinant ? or is there any other way to do it?
Makes me want to do algebra with colored writing utensils, except that it's too much work.
Hi! can you do a video on this problem: what is Arg(Z), if Z = 2 + (sqrt)3 + i ?
pie over 12 sir
why doesn't z^3 have just 1 root. can you please explain this. since z^n = 1 is nth root of unity. there should be nth root of 0. why is it just 1 ? complex just moved to the next level.
Is this problem related to applied calculus?
@FHomeBrew Competition Mathematics is wonderful, it's too bad it's nothing like real math.
oh god... how do they come up with such problems??
This made me feel stupid for not thinking of that on my own...
youler???