Please note that the theorem "If a, b, c is real, then for each complex root of az²+bz+c=0 its conjugate is also a root" is true regardless of the sign of the discriminant b²-4ac (it was not used in the proof at all), but only for a negative discriminant we can say "The two complex roots are conjugates of each other". For discriminant 0 we have just one (real) root, and for positive discriminant we'll have two real roots, which are not conjugates of each other (but each is its own conjugate)..
Wouldn't it be easier to just show ( conj(z·w) = conj(z)·conj(w) ) and use that to take the conjugate of the polynomial apart, instead of proving it separately for real factors and squares?
Great 👍🏻 I inspired by you and started to teach on UA-cam ❤️
Please note that the theorem "If a, b, c is real, then for each complex root of az²+bz+c=0 its conjugate is also a root" is true regardless of the sign of the discriminant b²-4ac (it was not used in the proof at all), but only for a negative discriminant we can say "The two complex roots are conjugates of each other". For discriminant 0 we have just one (real) root, and for positive discriminant we'll have two real roots, which are not conjugates of each other (but each is its own conjugate)..
Magneficient, I also have a channel for maths.. You can check it out
I like how simple these conjugate properties are. I was a bit surprised about the squaring one.
It's even more general: conj(z·w) = conj(z)·conj(w), the square and the real factor are just special cases of this.
Magneficient, I also have a channel for maths.. You can check it out
Watching this on Pi day.
Happy Pi day
3.14
Big fan sir from India
Same here bro
Magneficient, I also have a channel for maths.. You can check it out
✊💕 do u have a video teaching functions
Wouldn't it be easier to just show ( conj(z·w) = conj(z)·conj(w) ) and use that to take the conjugate of the polynomial apart, instead of proving it separately for real factors and squares?
I’d rather watch you than my teacher 😒