Euler made this problem easy by noting that 1=e^{2ijπ}, where e is Euler's number, π is the ratio of circle circumference to diameter, i=√(-1), and j is any integer. For x^n=1=e^{2ijπ}, n any positive integer, the n roots would be given by x_j=e^{2ijπ/n}=cos(2jπ/n)+i*sin(2jπ/n), j=0,1,...,n-1 For x^5=1=e^{2ijπ} we have x_j=e^{2ijπ/5}=cos(2jπ/5)+i*sin(2jπ/5), j=0,1,...,4, which are easily evaluated using a pocket calculator.
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Euler made this problem easy by noting that 1=e^{2ijπ}, where e is Euler's number, π is the ratio of circle circumference to diameter, i=√(-1), and j is any integer.
For x^n=1=e^{2ijπ}, n any positive integer, the n roots would be given by x_j=e^{2ijπ/n}=cos(2jπ/n)+i*sin(2jπ/n), j=0,1,...,n-1
For x^5=1=e^{2ijπ} we have x_j=e^{2ijπ/5}=cos(2jπ/5)+i*sin(2jπ/5), j=0,1,...,4, which are easily evaluated using a pocket calculator.
Very nice! ❤
1:57 - не по правилам! (x+1)(x³+x+1)=0. x=-1 ∨ x³+x+1=0.
Very nice! ❤
My stupid ass immiedietely thought of 1
It’s one of the answers though, right?
I thought of the same answer just after seeing the thumbnail