e^(ik2 π N/N) = 1, 1 is the identify operator of multiplication, N is positive integer, k is zero or positive integer (up to N-1). -1 = e^(i π) N = 4 (-1)^(1/4) = (-1(1))^(1/4) = (e^(i π) e^(ik2 π 4/4))^(1/4) = (e^(i π/4) e^(ik2 π/4)), k = 0, 1 , 2, 3 (k= 4 is a repetition of k =0) = e^(i π/4, e^(i 3π/4), e^(i 5π/4), e^(i 7π/4)
Before watching: 4root(-1) sqrt(sqrt(-1)) sqrt(i) -> i=sqrt(-1) suppose sqrt(i)=a+bi s.t. a and b are real, non-complex numbers i=(a+bi)^2 i=a^2+2abi-b^2 2abi=i; a^2-b^2=0 since there are no ways f(a,b) can be imaginary 2ab=1; (a+b)(a-b)=0 2ab=1; a=b or a=-b 2a^2=1 or -2a^2=1 but in case 2, a is not real. a=sqrt2/2, b=sqrt2/2 Therefore, sqrt2/2(1+i)
You lost 3 solutions. The first two at 1:20, the third at 8:25. ⁴√(-1)=k Let k=a+bi {a,b∈R} (a+bi)⁴=-1 a⁴+4a³bi-6a²b²-4ab³i+b⁴=-1+0i Comparing the real and imaginary parts of both sides of the equation a⁴-6a²b²+b⁴=-1 ① and 4a³b-4ab³=0 => 4ab(a²-b²)=0 => a=0 (rejected: from ① b⁴=-1, but b∈R) or b=0 (rejected: from ① a⁴=-1, but a∈R) or a²-b²=0 => a²=b² => (a=b or a=-b) ② Considering ② in ① (no matter which with an even power) we get -4b⁴=-1 => b⁴=1/4 => b=±⁴√(1/4)=±1/√2 ③ From ② and ③: for b=1/√2 => a=1/√2 or a=-1/√2 for b=-1/√2 => a=1/√2 or a=-1/√2 So k=1/√2+i/√2 or k=-1/√2+i/√2 or k=-1/√2-i/√2 or k=1/√2-i/√2
e^(ik2 π N/N) = 1, 1 is the identify operator of multiplication, N is positive integer, k is zero or positive integer (up to N-1).
-1 = e^(i π)
N = 4
(-1)^(1/4) = (-1(1))^(1/4) = (e^(i π) e^(ik2 π 4/4))^(1/4)
= (e^(i π/4) e^(ik2 π/4)), k = 0, 1 , 2, 3 (k= 4 is a repetition of k =0)
= e^(i π/4, e^(i 3π/4), e^(i 5π/4), e^(i 7π/4)
Bro forgot Oxford is for english
I know oxford is for English
And this video is also for English 😊
And its approaching finely.. don't be worried 🙂
Another method
⁴√(-1) = (-1)^(1/4) = (e^((1+2n)πi))^(1/4) = e^((1+2n)πi/4) when n∈Z
for n=0: ⁴√(-1) = e^(πi/4) = cos(π/4)+i∙sin(π/4) = 1/√2+i/√2
for n=1: ⁴√(-1) = e^(3πi/4) = cos(3π/4)+i∙sin(3π/4) = -1/√2+i/√2
for n=2: ⁴√(-1) = e^(5πi/4) = cos(5π/4)+i∙sin(5π/4) = -1/√2-i/√2
for n=3: ⁴√(-1) = e^(7πi/4) = cos(7π/4)+i∙sin(7π/4) = 1/√2-i/√2
Great approach 🙂
Before watching:
4root(-1)
sqrt(sqrt(-1))
sqrt(i) -> i=sqrt(-1)
suppose sqrt(i)=a+bi s.t. a and b are real, non-complex numbers
i=(a+bi)^2
i=a^2+2abi-b^2
2abi=i; a^2-b^2=0 since there are no ways f(a,b) can be imaginary
2ab=1; (a+b)(a-b)=0
2ab=1; a=b or a=-b
2a^2=1 or -2a^2=1 but in case 2, a is not real.
a=sqrt2/2, b=sqrt2/2
Therefore, sqrt2/2(1+i)
incorrect - you only get one of the four solutions.
Nice effort 😊
(➖ 1)^4= ➖ 1*➖ 1*➖ 1* ➖ 1 (x ➖ 1ix+1).
You lost 3 solutions. The first two at 1:20, the third at 8:25.
⁴√(-1)=k
Let k=a+bi {a,b∈R}
(a+bi)⁴=-1
a⁴+4a³bi-6a²b²-4ab³i+b⁴=-1+0i
Comparing the real and imaginary parts of both sides of the equation
a⁴-6a²b²+b⁴=-1 ①
and
4a³b-4ab³=0 => 4ab(a²-b²)=0 =>
a=0 (rejected: from ① b⁴=-1, but b∈R) or
b=0 (rejected: from ① a⁴=-1, but a∈R) or
a²-b²=0 => a²=b² => (a=b or a=-b) ②
Considering ② in ① (no matter which with an even power) we get
-4b⁴=-1 => b⁴=1/4 => b=±⁴√(1/4)=±1/√2 ③
From ② and ③:
for b=1/√2 => a=1/√2 or a=-1/√2
for b=-1/√2 => a=1/√2 or a=-1/√2
So
k=1/√2+i/√2 or k=-1/√2+i/√2 or k=-1/√2-i/√2 or k=1/√2-i/√2
Superb approach 🙂
Good
Thanks 😊