You've been teaching me well. I saw the link to the video, read the problem, and then solved it in my head in less than 30 seconds. That is entirely because I've been learning from your videos. I had completely forgotten the whole DeMoivre/Euler thing until I started watching your videos a few months ago. Thanks! By the way, the way I did i was by seeing that 1+√3i = 2(cos(π/3) + i sin(π/3)), and 2⁶=64, and 6π/3=2π, so the answer is 6 because cos(2π)=1 and sin(2π)=0.
Unfortunately, you are missing a lot of solutions. Whenever a number that is not a positive real-value is raised to an exponent that is not an integer, the result is multi-valued. This problem has an explicitly complex number raised to an exponent, so the most reasonable assumption is that we must do a full complex analysis. Here is the correct way to solve this problem. z^x = 64 z = 1 + i * sqrt(3) = 2*exp(i * pi/3) = r*exp(i*t) r = 2 t = pi/3 x = a + i * b z^x = exp(x * lc(z)) lc() multi-valued complex logarithm = exp( (a + i*b) * ( ln(r) + i*t + i*k*2*pi ) ) k integer index = exp( a*ln(r) - b*t - k*b*2*pi + i*( b*ln(r) + a*t + k*a*2*pi ) ) = exp( c + i*d ) c = a*ln(r) - b*(t + k*2*pi) d = b*ln(r) + a*(t + k*2*pi) z^x = 64 = exp(c) * ( cos(d) + i * sin(d) ) which separates into two equations: {1} exp(c) * cos(d) = 64 {2} exp(c) * sin(d) = 0 Since exp(c) > 0 , {2} requires that d = n*pi for integer index n , however, {1} can only be satisfied by even values of n, or equivalently d = n*2*pi which makes cos(d) = 1 and {1} becomes exp(c) = 64 c = ln(64) = 6 * ln(2) and plugging the values into the defnitions for c, d : 6*ln(2) = a*ln(r) - b*(t + k*2*pi) r and t are known, a and b are unknowns n*2*pi = b*ln(r) + a*(t + k*2*pi) and these equations can be solved for a, b : a = 6*(n*pi^2 * (6*k + 1) + 9*ln(2)^2 ) / D b = 18*pi*ln(2) * (n - (6*k + 1)) / D D = pi^2 * (6*k + 1)^2 + 9*ln(2)^2 x = a + i * b which simplifies to x = ( n*6*pi - i*18*ln(2) ) / ( (6*k + 1)*pi - i*3*ln(2) ) Note that if n = 6*k + 1, this simplifies to x = 6 which is the only real-valued solution
I am unfamiliar with complex numbers so plz help me out. Since e^πi is -1 and -1^1/3 is -1 then would that not make 1+sqrt3×i a real number or that operation can't be done?
At age 80 don't have much time left, so quickly cast into the Euler form re^(iθ) = r(cosθ + i sinθ) where r = 2 and θ = π/3. (2e^(iπ/3))^6 = 2^6 (e^(iπ/3))^6 = 64 e^(i2π) = 64, so x = 6.
Let z = (1 + i sqrt(3)). Note that z = 2 exp(i pi/3). z^x = 2^x * exp(xi pi/3). This is the standard polar form for complex numbers. Obvious solution at x = 6.
Taking the result for x and expressing it as a complex number, we get x = 6 + i [(pi)(n-1)(ln8)]/[(ln8)^2 + (pi)^2]. So the only real solution corresponds to n =1, with the integer answer x = 6.
Why didn't you consider the periodicity and therefore e^(π/3 +kπ)i?
2:36 Don’t we have to add 2n*pi the argument here too?
You've been teaching me well. I saw the link to the video, read the problem, and then solved it in my head in less than 30 seconds. That is entirely because I've been learning from your videos. I had completely forgotten the whole DeMoivre/Euler thing until I started watching your videos a few months ago. Thanks!
By the way, the way I did i was by seeing that 1+√3i = 2(cos(π/3) + i sin(π/3)), and 2⁶=64, and 6π/3=2π, so the answer is 6 because cos(2π)=1 and sin(2π)=0.
Excellent! Np. Glad to hear that 🤩
Nice!
Thanks!
Great one...💯💯💯.
Thanks 💯
Unfortunately, you are missing a lot of solutions. Whenever a number that is not a positive real-value is raised to an exponent that is not an integer, the result is multi-valued. This problem has an explicitly complex number raised to an exponent, so the most reasonable assumption is that we must do a full complex analysis. Here is the correct way to solve this problem.
z^x = 64
z = 1 + i * sqrt(3) = 2*exp(i * pi/3) = r*exp(i*t)
r = 2
t = pi/3
x = a + i * b
z^x = exp(x * lc(z)) lc() multi-valued complex logarithm
= exp( (a + i*b) * ( ln(r) + i*t + i*k*2*pi ) ) k integer index
= exp( a*ln(r) - b*t - k*b*2*pi + i*( b*ln(r) + a*t + k*a*2*pi ) )
= exp( c + i*d )
c = a*ln(r) - b*(t + k*2*pi)
d = b*ln(r) + a*(t + k*2*pi)
z^x = 64 = exp(c) * ( cos(d) + i * sin(d) ) which separates into two equations:
{1} exp(c) * cos(d) = 64
{2} exp(c) * sin(d) = 0
Since exp(c) > 0 , {2} requires that
d = n*pi for integer index n , however, {1} can only be satisfied by even values
of n, or equivalently
d = n*2*pi which makes cos(d) = 1 and {1} becomes
exp(c) = 64
c = ln(64) = 6 * ln(2) and plugging the values into the defnitions for c, d :
6*ln(2) = a*ln(r) - b*(t + k*2*pi) r and t are known, a and b are unknowns
n*2*pi = b*ln(r) + a*(t + k*2*pi) and these equations can be solved for a, b :
a = 6*(n*pi^2 * (6*k + 1) + 9*ln(2)^2 ) / D
b = 18*pi*ln(2) * (n - (6*k + 1)) / D
D = pi^2 * (6*k + 1)^2 + 9*ln(2)^2
x = a + i * b which simplifies to
x = ( n*6*pi - i*18*ln(2) ) / ( (6*k + 1)*pi - i*3*ln(2) )
Note that if n = 6*k + 1, this simplifies to
x = 6 which is the only real-valued solution
x = 6 , Am I right ?
It's just an one solution
I am unfamiliar with complex numbers so plz help me out. Since e^πi is -1 and -1^1/3 is -1 then would that not make 1+sqrt3×i a real number or that operation can't be done?
No. z^3 is not a 1-1 function so it does not have a unique inverse. (-1)^(1/3) has three choices, only one of which is real.
@@rickdesper ok
At age 80 don't have much time left, so quickly cast into the Euler form
re^(iθ) = r(cosθ + i sinθ) where r = 2 and θ = π/3.
(2e^(iπ/3))^6 = 2^6 (e^(iπ/3))^6 = 64 e^(i2π) = 64, so x = 6.
Thank
Let z = (1 + i sqrt(3)). Note that z = 2 exp(i pi/3). z^x = 2^x * exp(xi pi/3). This is the standard polar form for complex numbers. Obvious solution at x = 6.
Taking the result for x and expressing it as a complex number, we get x = 6 + i [(pi)(n-1)(ln8)]/[(ln8)^2 + (pi)^2]. So the only real solution corresponds to n =1, with the integer answer x = 6.
For the real solution, taking logs: ln( 1 + i √3)ˣ = ln64
xln( 1 + i √3) = ln(2^6)
x(ln2 + iπ/3) = 6ln2
x = 6, comparing real and imaginary parts
Be careful when you say "taking logs" with a complex number. It's not that simple.