interesting problem, indeed. But I thing the final answer is wrong. See below... Firstly, I'd like to give you a few comments if you don't mind: 1. at 6:55 you say "... we don't have to worry about all of the branches, because x will take care of that...". This is not really correct. In fact, if x is not integer, and theta = angle + 2*pi*n (n - integer), then x * theta = x*angle + 2*pi*x*n will affect the "cyclic symmetry" of the complex number on the complex plane, so I wouldn't count on that statement. It kinda works anyway, because x in the end nicely goes out of the exponent when you log it, but I think it's good to be a bit rigorous with such conclusions anyway ;). 2. there is no big issue with log of the complex variable, but a few things need to be taken into consideration! when there's an equation z1 = z2, and say z1 = r1*exp(i*theta_1+2*pi*n), z2 = r2*exp(i*theta_2+2*pi*m), theta's are main (!!!) arguments here, then ln(z1) = ln(z2) will finally bring to this equation: ln(r1/r2) = -i *(theta_1-theta2 + 2*pi*k), where k is also integer (k = n-m). The thing here is that we have a real value on the LHS and purely imaginary value on RHS. The only option for this to hold true is when LHS=RHS=0, from where the equatin splits into two parts: r1 = r2 and theta_1 = theta_2 + 2*pi*k. Any kind of equation with complex numbers always imply 2 equations: for real and imaginary part correspondingly, or for modulus and argument correspondingly. You can't really reduce it all to only one eqution! In this case, for moduli we have |sqrt(5)^x| = 5, then assuming x = u + i*v, |sqrt(5)^(u+i*v)| = |sqrt(5)^u|*|e^(i*1/2*ln(5)*v)| = |sqrt(5)^u| * 1 = sqrt(5)^u, and from equation we get sqrt(5)^u = 5 and then u = Re(x) = 2. For arguments: for LHS we have multi-branch expression: x*theta + 2*pi*n*x, and for RHS: 2*theta + 2*pi*k. If we put x = u + i*v as before, then we can spot that there are no terms proportional to i, (like, i*... ) on the RHS, which makes v be 0 in order for this equality to hold. And also, x= 2 will make the argument equation hold for all of the branches, if we establish the correspondance between "n" branches and "k" branches, so that for each "n" there will be a corresponding "k" like using 2n = k equality. So, actually x = 2 makes both the moduli equal and arguments equal, and there are no any other complex value that would satisfy both these conditions.
And the only solution (no more angles added to 2π multiples as his method produces) because the magnitude (or he calls radius r = ✓5 for (2 - i) and r = 5 for (3 - 4i)! For √5 having only a power of 2 to equal 5 means x = 2 as the only solution in this problem.
well we could have figured this out in this way also , (considering x in R):\\ we set z=(2-i)^{x} we have |z|=5^{x/2} and also |3-4i|=5 now since z=3-4i we must have |z|=|3-4i| or 5^{x/2}=5 implies x/2=1 or x=2
But his ∆ + 2nπi complex plane angle should have been explained better. He moves after finding x = 2 only solution to this problem for magnitude in vector mathematics (radius in his complex polar plane general mathematics) only matches up at x = 2 only solution. This only solution has a ∆ only solution also that is not the angle found of 3 - 4i in the fourth polar complex quadrant. It is the ∆ = 4π - tan -1(4/3) only angle, despite 5 at ∆ = 2π - tan-1(4/3) starting assumption of n=0 is false but n=1 is true and the only one in tan-1(4/3) + 2nπi. This is true because if we ever took the ✓ (3 - 4i) one of the roots came from 4π - tan-1(4/3) forming 2 - i and the other root came from 2π - tan-1(4/3) forming -2 + i ... Too many people are forgetting domain and range analysis when applying √ in the complex plane (cube roots, fourth roots, etc. because they haven't figured out what domain and range to functions were all about mathematics)!
I approached this by setting w^x = z, so x = (ln z) / (ln w), which I defined as (a + bi) / (c + di). Multiply top and bottom by the complex conjugate of the denominator to get x = (ac - bd) / (c^2 + d^2) + [(bc - ad) / (c^2 + d^2)] i. Define that as e + fi and hope that e = 2 and f = 0 for the principle branch :) . Do a bunch of math to get the following: a = ln 5 b = arctan (-4/3) + 2n(pi) c = (1/2) ln 5 d = arctan (-1/2) + 2m(pi) I think that's right.
Use polar form on both side and using (r*cos(a)+i*sin(a))^n=r^n*(cos(na)+i*sin(na)) to express left hand side. Finally comparing both side real part and imaginary part to get the answer. 😉😉😉😉😉😉
I wonder if the answer isn't just examining the magnitude of each side (a different solution than your's by magnitude manipulations). Since the left side has magnitude √5 = 5^(1/2) and right side = magnitude 5^(1) forming 5^[(1/2)x] = 5^[(1)] or 1/2x = 2 has the one and only solution x=2 verified by (√5)^2 = 5 ... So there are no i(n)(2π∆) because magnitude passes 5^1 and the magnitudes don't equal despite phases or complex plane angle matches.
Also a unique magnitude of 5 value at the unique angle of that going to the fourth quadrant of 2π to 4π (or the n=1 choice in infinite possible angles ... without domain and range analysis on functions in squaring variables as much as square root functions of variables in a range). The unique angle is therefore magnitude 5 but it's angle at 4π - tan-1(4/3) is clear to be that angle specifically because taking the square root of 5 at angle 4π - tan-1(4/3) lands inside the 0 to 2π new domain at 2π - tan-1(1/2) specific angle of magnitude √5 (positive radius in complex plane equivalent). The √5 at π - tan-1(1/2) is from the 0 to 2π domain equivalent 5 angle 2π - tan-1(4/3) to form as -2 + i other root from the square root applied to a complex variable.
⎷5 = 5^½, so you can move the ½ to the front. Not sure if it would help anything. My teacher said, "Whenever possible, reduce the argument of a Log to be >0 and
interesting problem, indeed. But I thing the final answer is wrong. See below...
Firstly, I'd like to give you a few comments if you don't mind:
1. at 6:55 you say "... we don't have to worry about all of the branches, because x will take care of that...". This is not really correct. In fact, if x is not integer, and theta = angle + 2*pi*n (n - integer), then x * theta = x*angle + 2*pi*x*n will affect the "cyclic symmetry" of the complex number on the complex plane, so I wouldn't count on that statement. It kinda works anyway, because x in the end nicely goes out of the exponent when you log it, but I think it's good to be a bit rigorous with such conclusions anyway ;).
2. there is no big issue with log of the complex variable, but a few things need to be taken into consideration! when there's an equation z1 = z2, and say z1 = r1*exp(i*theta_1+2*pi*n), z2 = r2*exp(i*theta_2+2*pi*m), theta's are main (!!!) arguments here, then ln(z1) = ln(z2) will finally bring to this equation: ln(r1/r2) = -i *(theta_1-theta2 + 2*pi*k), where k is also integer (k = n-m). The thing here is that we have a real value on the LHS and purely imaginary value on RHS. The only option for this to hold true is when LHS=RHS=0, from where the equatin splits into two parts: r1 = r2 and theta_1 = theta_2 + 2*pi*k. Any kind of equation with complex numbers always imply 2 equations: for real and imaginary part correspondingly, or for modulus and argument correspondingly. You can't really reduce it all to only one eqution!
In this case, for moduli we have
|sqrt(5)^x| = 5,
then assuming x = u + i*v, |sqrt(5)^(u+i*v)| = |sqrt(5)^u|*|e^(i*1/2*ln(5)*v)| = |sqrt(5)^u| * 1 = sqrt(5)^u, and from equation we get sqrt(5)^u = 5 and then u = Re(x) = 2.
For arguments:
for LHS we have multi-branch expression:
x*theta + 2*pi*n*x,
and for RHS:
2*theta + 2*pi*k.
If we put x = u + i*v as before, then we can spot that there are no terms proportional to i, (like, i*... ) on the RHS, which makes v be 0 in order for this equality to hold. And also, x= 2 will make the argument equation hold for all of the branches, if we establish the correspondance between "n" branches and "k" branches, so that for each "n" there will be a corresponding "k" like using 2n = k equality.
So, actually x = 2 makes both the moduli equal and arguments equal, and there are no any other complex value that would satisfy both these conditions.
The answer is indeed surprising ,but very easy to check.
And the only solution (no more angles added to 2π multiples as his method produces) because the magnitude (or he calls radius r = ✓5 for (2 - i) and r = 5 for (3 - 4i)! For √5 having only a power of 2 to equal 5 means x = 2 as the only solution in this problem.
well we could have figured this out in this way also , (considering x in R):\\
we set z=(2-i)^{x} we have |z|=5^{x/2} and also |3-4i|=5 now since z=3-4i we must have |z|=|3-4i| or 5^{x/2}=5 implies x/2=1 or x=2
But his ∆ + 2nπi complex plane angle should have been explained better. He moves after finding x = 2 only solution to this problem for magnitude in vector mathematics (radius in his complex polar plane general mathematics) only matches up at x = 2 only solution. This only solution has a ∆ only solution also that is not the angle found of 3 - 4i in the fourth polar complex quadrant. It is the ∆ = 4π - tan -1(4/3) only angle, despite 5 at ∆ = 2π - tan-1(4/3) starting assumption of n=0 is false but n=1 is true and the only one in tan-1(4/3) + 2nπi. This is true because if we ever took the ✓ (3 - 4i) one of the roots came from 4π - tan-1(4/3) forming 2 - i and the other root came from 2π - tan-1(4/3) forming -2 + i ... Too many people are forgetting domain and range analysis when applying √ in the complex plane (cube roots, fourth roots, etc. because they haven't figured out what domain and range to functions were all about mathematics)!
@@lawrencejelsma8118 well yes
I approached this by setting w^x = z, so x = (ln z) / (ln w), which I defined as (a + bi) / (c + di). Multiply top and bottom by the complex conjugate of the denominator to get x = (ac - bd) / (c^2 + d^2) + [(bc - ad) / (c^2 + d^2)] i. Define that as e + fi and hope that e = 2 and f = 0 for the principle branch :) .
Do a bunch of math to get the following:
a = ln 5
b = arctan (-4/3) + 2n(pi)
c = (1/2) ln 5
d = arctan (-1/2) + 2m(pi)
I think that's right.
Excellent
I think that ln(e ^ (i * theta)) is actually i * (theta + 2 * m * pi), where m is a natural number, so there are even more solutions.
The thing with this one, if you know even a little about Gaussian primes this is as obvious as 3^x = 9. If you do not, then it's a bit of toughie.
Use polar form on both side and using (r*cos(a)+i*sin(a))^n=r^n*(cos(na)+i*sin(na)) to express left hand side. Finally comparing both side real part and imaginary part to get the answer. 😉😉😉😉😉😉
I wonder if the answer isn't just examining the magnitude of each side (a different solution than your's by magnitude manipulations). Since the left side has magnitude √5 = 5^(1/2) and right side = magnitude 5^(1) forming 5^[(1/2)x] = 5^[(1)] or 1/2x = 2 has the one and only solution x=2 verified by (√5)^2 = 5 ... So there are no i(n)(2π∆) because magnitude passes 5^1 and the magnitudes don't equal despite phases or complex plane angle matches.
Also a unique magnitude of 5 value at the unique angle of that going to the fourth quadrant of 2π to 4π (or the n=1 choice in infinite possible angles ... without domain and range analysis on functions in squaring variables as much as square root functions of variables in a range). The unique angle is therefore magnitude 5 but it's angle at 4π - tan-1(4/3) is clear to be that angle specifically because taking the square root of 5 at angle 4π - tan-1(4/3) lands inside the 0 to 2π new domain at 2π - tan-1(1/2) specific angle of magnitude √5 (positive radius in complex plane equivalent). The √5 at π - tan-1(1/2) is from the 0 to 2π domain equivalent 5 angle 2π - tan-1(4/3) to form as -2 + i other root from the square root applied to a complex variable.
In the denominator couldn't you write ln(sqrt(5)+ i(theta +2m(pi)? Then x=2 when m=n (m and n need not be equal)
We quickly get x=2 by comparing the moduli of each side. The only real solution.
But yes, an infinite number of complex solutions 👍
⎷5 = 5^½, so you can move the ½ to the front. Not sure if it would help anything.
My teacher said, "Whenever possible, reduce the argument of a Log to be >0 and
x=2