(1+i)^Sqrt[2]=Sqrt[2]^Sqrt[2]Cos[0.25Sqrt[2]π]+Sqrt[2]^Sqrt[2]Sin[0.25Sqrt[2]π]i Input (1 + i)^sqrt(2) = sqrt(2)^sqrt(2) cos(0.25 sqrt(2) π) + sqrt(2)^sqrt(2) sin(0.25 sqrt(2) π) i Result True It’s in my head
Why only take the principal log. This expression will actually have infinitely many distinct representations. 1+i is not a positive number so the convention of treating an exponential as a function does not apply here.
So here's a question I've been pondering without success for the last few days: When does sin z = sinh z? Obviously z=0 is a solution, but are there other complex solutions? Can we calculate the other solutions? (Or prove that there are none?) I think I've figured out how to solve cos z = cosh z (and it's a rather nice answer) but I can't figure out how to do it for sin and sinh
Why not apply Euler's formula first and exponentiate to get the same result without any logarithms? (1 +i)^sqrt(2) = (sqrt(2) * e^(i*pi/4))^sqrt(2) = sqrt(2)^sqrt(2) * e^(i*pi*sqrt(2)/4).
Why do you never truncate sqrt(2)/2 = 1/sqrt(2) ? The un-shortened form makes it more fuzzy, and less elegant. Also moe elegant: 4π/sqrt(2) = 2 sqrt(2) π
It's an option but it's generally considered to be good practice to keep denominators as simple as possible as they are generally easier to work with in this form. I would normally prefer ½√2 over 1/√2.
@TheMathManProfundities With all respect, I dispute the claimed generality of keeping denominators simple at the cost of a more bloated complete fraction. E.g., where I studied keeping sqrt(2)/2 in an end result would had cost me a few points. Apart from that it's just the visual familarity of 1/sqrt(2). To me, it's just a simple old friend. Taking your argument into account, and knowing that sqrt(2)/2 = 1/sqrt(2) = sin(π/4) = cos(π/4), you should opt for the trigonometric expressions because they provide the simplest denominator.
problem ( 1 + i ) ᵛ²̅ = ? Call this complex number z. The polar form of the base is 1 + i = √2 e^(i π (1/4+2k)), k ∈ ℤ This is seen in the Argand plane with 1 unit along the real axis and 1 unit along the imaginary axis for a modulus of √2 and general angle θ of π(1/4 +2k), k ∈ ℤ. This just goes to show that there is no better oiler than Euler when it comes to rotations in the complex plane! Replacing from the above discussion, z = (√2 ᵛ²̅) e^[ i π √2(1/4 +2k) ] , where √2 exponent has been distributed. Now take natural logarithms on each side. ln z = √2 ln (√2) + i π √2(1/4 +2k) ln z = √2 ln (2)/2 + i π √2(1/4 +2k) Take e to each side. z = 2ᵛ²̅ᐟ² e^(i π√2[1/4+2k]) Euler's formula renders this as = 2ᵛ²̅ᐟ² [ cos π√2(1/4+2k) + i sin π√2(1/4+2 k) ] k ∈ ℤ Principal value is at k = 0. answer ( 1 + i ) ᵛ²̅ = = 2ᵛ²̅ᐟ² [ cos π√2(1/4+2k) + i sin π√2(1/4+2 k) ] k ∈ ℤ
❤❤❤❤
(1+i)^Sqrt[2]=Sqrt[2]^Sqrt[2]Cos[(π/4)Sqrt[2]]+Sqrt[2]^Sqrt[2]Sin[(π/4)Sqrt[2]]
Nice!
I’m glad you like it! 😊
(1+i)^Sqrt[2]=Sqrt[2]^Sqrt[2]Cos[0.25Sqrt[2]π]+Sqrt[2]^Sqrt[2]Sin[0.25Sqrt[2]π]i Input
(1 + i)^sqrt(2) = sqrt(2)^sqrt(2) cos(0.25 sqrt(2) π) + sqrt(2)^sqrt(2) sin(0.25 sqrt(2) π) i
Result
True It’s in my head
You solved (1+I)^2. Why not just square root that to get your answer? Or write (1+I) in polar form then raise to sqrt 2
Square root of (1+i)^2 is +-(1+i)
@ haha, oops, I wrote that late at night.
Yes, polar form is simpler- ( sqrt2 * e^i*PI/4 ) ^ sqrt2 = sqrt2^sqrt2 * e^i*sqrt2*PI/4.
Why only take the principal log. This expression will actually have infinitely many distinct representations. 1+i is not a positive number so the convention of treating an exponential as a function does not apply here.
So here's a question I've been pondering without success for the last few days: When does sin z = sinh z?
Obviously z=0 is a solution, but are there other complex solutions? Can we calculate the other solutions? (Or prove that there are none?)
I think I've figured out how to solve cos z = cosh z (and it's a rather nice answer) but I can't figure out how to do it for sin and sinh
e^((√2/2)In2)＝2^(√2/2 ) ともう少し簡単な式に出来ます😅
Why not apply Euler's formula first and exponentiate to get the same result without any logarithms?
(1 +i)^sqrt(2) = (sqrt(2) * e^(i*pi/4))^sqrt(2) = sqrt(2)^sqrt(2) * e^(i*pi*sqrt(2)/4).
Why do you never truncate sqrt(2)/2 = 1/sqrt(2) ?
The un-shortened form makes it more fuzzy, and less elegant.
Also moe elegant: 4π/sqrt(2) = 2 sqrt(2) π
It's an option but it's generally considered to be good practice to keep denominators as simple as possible as they are generally easier to work with in this form. I would normally prefer ½√2 over 1/√2.
@TheMathManProfundities With all respect, I dispute the claimed generality of keeping denominators simple at the cost of a more bloated complete fraction.
E.g., where I studied keeping sqrt(2)/2 in an end result would had cost me a few points.
Apart from that it's just the visual familarity of 1/sqrt(2). To me, it's just a simple old friend.
Taking your argument into account, and knowing that sqrt(2)/2 = 1/sqrt(2) = sin(π/4) = cos(π/4), you should opt for the trigonometric expressions because they provide the simplest denominator.
base / bass joke 👍
problem
( 1 + i ) ᵛ²̅ = ?
Call this complex number z.
The polar form of the base is
1 + i = √2 e^(i π (1/4+2k)),
k ∈ ℤ
This is seen in the Argand plane with 1 unit along the real axis and 1 unit along the imaginary axis for a modulus of √2 and general angle θ of π(1/4 +2k), k ∈ ℤ.
This just goes to show that there is no better oiler than Euler when it comes to rotations in the complex plane!
Replacing from the above discussion,
z = (√2 ᵛ²̅) e^[ i π √2(1/4 +2k) ]
, where √2 exponent has been distributed.
Now take natural logarithms on each side.
ln z = √2 ln (√2) + i π √2(1/4 +2k)
ln z = √2 ln (2)/2 + i π √2(1/4 +2k)
Take e to each side.
z =
2ᵛ²̅ᐟ² e^(i π√2[1/4+2k])
Euler's formula renders this as
= 2ᵛ²̅ᐟ² [ cos π√2(1/4+2k) +
i sin π√2(1/4+2 k) ]
k ∈ ℤ
Principal value is at k = 0.
answer
( 1 + i ) ᵛ²̅ =
= 2ᵛ²̅ᐟ² [ cos π√2(1/4+2k) +
i sin π√2(1/4+2 k) ]
k ∈ ℤ