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Комментарии : 23   
@SALogics
@SALogics День назад
Very nice problem with a nice solution! ❤❤
@GillesF31
@GillesF31 День назад
Or (maybe shorter): √i = i^(1/2) || || set i on exponential form: || || z = 0 + 1i || || |z| = √(0² + 1²) = √1 = 1 || || θ = π/2 || || z = e^(i·(π/2)) || = [e^(i·(π/2))]^(1/2) = e^(i·(π/2)·(1/2)) = e^(i·(π/4)) = (cos(π/4) + i·sin(π/4)) = √2/2 + i·√2/2 note: (√2/2)·(√2/√2) = 2/(2·√2) = 1/√2 = 1/√2 = i/√2 /// final result: ■ √i = 1/√2 + i/√2 🙂
@payoo_2674
@payoo_2674 6 часов назад
but we need all results √i = i^(1/2) = (e^((π/2 + 2kπ)*i))^(1/2) = e^((π/4 + kπ)*i) k ∈ Z for k=0: √i = e^(πi/4) = cos(π/4) + i*sin(π/4) = √2/2 + i*√2/2 (or 1/√2 + i*1/√2) for k=1: √i = e^(5πi/4) = cos(5π/4) + i*sin(5π/4) = -√2/2 - i*√2/2 (or -1/√2 - i*1/√2) for the next k the result will be repeated
@payoo_2674
@payoo_2674 3 часа назад
√i = i^(1/2) = (e^((π/2 + 2kπ)*i))^(1/2) = e^((π/4 + kπ)*i) k ∈ Z for k=0: √i = e^(πi/4) = cos(π/4) + i*sin(π/4) = √2/2 + i*√2/2 (or 1/√2 + i*1/√2) for k=1: √i = e^(5πi/4) = cos(5π/4) + i*sin(5π/4) = -√2/2 - i*√2/2 (or -1/√2 - i*1/√2) for the next k the result will be repeated
@payoo_2674
@payoo_2674 3 часа назад
from 4:00 a²-b²=0 => a²=b² => (a=b ˅ a=-b) ① 2ab=1 => sign(a)=sign(b) ② ①˄② => a=b ③ 2a²=1 a²=1/2 a=1/√2, b=1/√2 ˅ a=-1/√2, b=-1/√2 so √i = 1/√2 + i*1/√2 ˅ √i = -1/√2 - i*1/√2
@is7728
@is7728 День назад
Love this problem that needs our creativity
@thesagefoxbat
@thesagefoxbat День назад
e to the i pi power -1 equals 0.
@edsonluisvedovatto9174
@edsonluisvedovatto9174 8 часов назад
Falta substituir o "i" e fazer a prova.
@lornacy
@lornacy День назад
Is this a total of four answers?
@elmer6123
@elmer6123 2 дня назад
I asked Euler and he said √i=(√2/2)(1+i). Is that true? [(√2/2)(1+i)]^2=(1/2)(1+2i-1)=(1/2)(2i)=i, so it is. Actually he said, i=e^{iπ/2} and √i=[e^{iπ/2}]^{1/2}=e^{iπ/4}=cos(π/4)+i*sin(π/4)=(√2/2)(1+i)
@payoo_2674
@payoo_2674 6 часов назад
Euler didn't mention the second solution? √i = i^(1/2) = (e^((π/2 + 2kπ)*i))^(1/2) = e^((π/4 + kπ)*i) k ∈ Z for k=0: √i = e^(πi/4) = cos(π/4) + i*sin(π/4) = √2/2 + i*√2/2 (or 1/√2 + i*1/√2) for k=1: √i = e^(5πi/4) = cos(5π/4) + i*sin(5π/4) = -√2/2 - i*√2/2 (or -1/√2 - i*1/√2) for the next k the result will be repeated
@walterwen2975
@walterwen2975 2 дня назад
i = (2i)/2 = (1 + 2i - 1)/2 = (1 + 2i + i²)/2 = [(1 + i)²]/2 = [ (1 + i)/√2]² √i = ± (1 + i)/√2 = ± (√2 + i√2)/2
@is7728
@is7728 День назад
From your expression, it should be +- ( +- √2 + √2 i ) / 2
@Kanadgaming-q3b
@Kanadgaming-q3b День назад
Root of root of -1
@reginadevakis6899
@reginadevakis6899 2 дня назад
Nice. Bye bye
@learncommunolizer
@learncommunolizer День назад
Have a nice day
@johnnychen6205
@johnnychen6205 2 дня назад
Gooooooood❤❤❤❤❤
@learncommunolizer
@learncommunolizer День назад
Thanks you very much ❤️🙏❤️
@jonathanr520
@jonathanr520 День назад
(-1)^¼
@jonathanr520
@jonathanr520 День назад
±(-1)^¼
@thesagefoxbat
@thesagefoxbat День назад
So what? Will this knowledge provide us with incorruptible government?
@irenehartlmayr8369
@irenehartlmayr8369 День назад
Mathematics is not about government !!! And mathematics is about solving problems in Mathematics and mathematical sciences....! Ever been to school...?.