Stanford University Admission Interview Tricks | Calculators NOT Allowed.✍️🖋️📘💙
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in polar form -1=e^i(2n+1)π so x=(-1)^(1/4) =
e^(iπ(2n+1)/4)
x=e^(iπ1/4) , e^(iπ3/4), e^(iπ5/4), e^(iπ7/4)
This corresponds to (±1±i)/√2 ie four complex roots.
Viewed on the Argand diagram, the roots form an ×
x^2= +i , -i, x^4=i^2 = -1
in polar form
x^2= e^(iπ1/2) , e^(iπ3/2), e^(iπ5/2) = e^(iπ1/2), e^(iπ7/2)=e^(iπ3/2)
x^4= e^(iπ)=-1
You should come up with four answers to this problem because you are taking 4th root of negative 1. What you found is only the Principle Value of 4th root of negative 1. There are three more answers. I enjoy watching your channel.
The Principal Root is what is called for when you use the radical symbol.
We are not solving for all values of X such that X⁴ = -1, we are just evaluating/simplifying the expression that is read as Principal (because the radical symbol was used) 4th root of -1
Thanks for sharing your perspective and for watching! 💪I appreciate your feedback! 👍
Totally agree.
It is better to use:
-1 = e^(i(π + 2nπ)) for n ∈ ℤ.
And then take the fourth root of -1 and divide by 4 the exponent of e.
Great stuff Professor! Thank you so much for the detailed walk through of your examples - extremely helpful.
You're very welcome!Glad you enjoyed it!✅💯🎉
-1 = e^(iπ); (-1)^(1/4) = e^(iπ/4)=cos(π/4)+isin(π/4)=1/sqrt(2)+i*1/sqrt(2)
❤❤❤❤
Let a+bi=⁴√(-1) {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 (complex roots rejected, b/c b∈R)
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
⁴√(-1) = 1/√2+i/√2 (pincipal root) or
⁴√(-1) = -1/√2+i/√2 or
⁴√(-1) = -1/√2-i/√2 or
⁴√(-1) = 1/√2-i/√2
Another method
⁴√(-1) = (-1)^(1/4) = (e^((1+2k)πi))^(1/4) = e^((1+2k)πi/4) where k∈Z
for k=0: ⁴√(-1) = e^(πi/4) = cos(π/4)+i∙sin(π/4) = 1/√2+i/√2 (pincipal root)
for k=1: ⁴√(-1) = e^(3πi/4) = cos(3π/4)+i∙sin(3π/4) = -1/√2+i/√2
for k=2: ⁴√(-1) = e^(5πi/4) = cos(5π/4)+i∙sin(5π/4) = -1/√2-i/√2
for k=3: ⁴√(-1) = e^(7πi/4) = cos(7π/4)+i∙sin(7π/4) = 1/√2-i/√2
For the previous and next k the solutions are repeated.
U should be math major...
Thanks for incisive solution 💯🎉💪✅🙏
As already suggested,polar form is far more appropriate. Also using cosθ+isinθ in de Moivre’s theorem is a better alternative.
Both these methods avoid losing roots. At university entrance that would probably be an automatic disqualifier.
U can't leave sqrt to nominator - sqrt(-1) = sqrt(2)/2+sqrt(2)i/2
Lost 2 roots
No, lost 3 roots.
0:50 i²=-1 but (-i)²=-1 (lost 2 roots)
6:57 (a^4)^(1/4)=±(1/4)^(1/4) (lost 1 root)
Didn't lose any roots, given that the radical symbol calls for only the Principal Root.
Why you nake simlpe steps?
I fell asleep....