Mastering Euler's identity for complex exponential functions is akin to unlocking a door. Once opened, the countless problems that follow become mere pathways, driven by procedure rather than new conceptual revelations.
Isn't most math like that anyway, I mean new revelations occur pretty rarely don't they. Especially right now at a time when math has progressed this far in history.
Here's a solution method that does not make any unwarranted assumptions, so it will work on other complex equations that the method in the video will not. Also, I will not limit the complex exponentiation to the principal value, so all branches of the function will be included (it is easily limited later if desired). i^x = 1 + i exp(lc(i)*x) = 1 + i lc() is the multi-valued complex logarithm function x = a + i*b where a, b are real-valued lc(i) = i*pi/2 + i*m*2*pi for integers m exp( i*pi/2 *(1 + 4*m) * (a + i*b) ) = 1 + i exp( -b * pi/2 *(1 + 4*m) + i*a*pi/2 *(1 + 4*m) ) = 1 + i Definitions u = b*pi/2 *(1 + 4*m) v = a*pi/2 *(1 + 4*m) exp(-u) * exp(i*v) = 1 + i exp(-u) * (cos(v) + i*sin(v)) = 1 + i now set real and imaginary parts equal {1} exp(-u) * cos(v) = 1 real parts {2} exp(-u) * sin(v) = 1 imaginary parts We need to determine u and v that solve these equations, which then determine a and b , the real and imaginary parts of x. We can divide {2} by {1} to get tan(v) = 1 v = arctan(1) = pi/4 + n*2*pi = pi/4 *(1 + 8*n) for integers n Note that we exclude 5*pi/4 + n*2*pi since that will not solve {1} and {2} due to negative values of sin() and cos() not solving the equations. a*pi/2 *(1 + 4*m) = pi/4 *(1 + 8*n) a = 1/2 *(1 + 8*n) / (1 + 4*m) exp(u) = cos(v) = cos( pi/4 *(1 + 8*n) ) = 1/sqrt(2) u = ln(1/sqrt(2)) = -ln(2) / 2 b*pi/2 *(1 + 4*m) = -ln(2) / 2 b = -(ln(2) / pi) / (1 + 4*m) x = 1/2 *(1 + 8*n) / (1 + 4*m) - i*(ln(2) / pi) / (1 + 4*m) If we limit i^x to the principal value of the exponentiation function, then m = 0 x = 1/2 *(1 + 8*n) - i *ln(2) / pi
To solve the equation i^x = 1 + i for x for the principal branch, we need to use the complex logarithm function. The complex logarithm is defined as log(z) = ln|z| + i arg(z), where z is any complex number, |z| is its modulus, and arg(z) is its argument. The argument of a complex number is the angle it makes with the positive real axis in the complex plane, measured in radians. Using this definition, we can rewrite the equation as log(i^x) = log(1 + i). Then, we can use the property of logarithms that log(a^b) = b log(a) to get x log(i) = log(1 + i). Next, we need to find the values of log(i) and log(1 + i) using the complex logarithm formula. We have |i| = 1 and arg(i) = pi/2, so log(i) = ln(1) + i pi/2 = i pi/2. Similarly, we have |1 + i| = sqrt(2) and arg(1 + i) = pi/4, so log(1 + i) = ln(sqrt(2)) + i pi/4. Substituting these values into the equation, we get x i pi/2 = ln(sqrt(2)) + i pi/4. Solving for x, we get x = (ln(sqrt(2)) + i pi/4)/(i pi/2). To simplify this expression, we can multiply both numerator and denominator by -i to get x = (-i ln(sqrt(2)) - pi/4)/(-i i pi/2) = (-i ln(sqrt(2)) - pi/4)/(pi/2). Finally, we can separate the real and imaginary parts of x to get x = -pi/(2pi) - (ln(sqrt(2))/pi)i = -1/2 - (ln(sqrt(2))/pi)i. Therefore, the principal solution to the equation is x = -1/2 - (ln(sqrt(2))/pi)i. You can check this answer by plugging it into the original equation and verifying that both sides are equal. The general solution to the equation i^x = 1 + i for x, for all branches of the complex logarithm, is given by x = -1/2 - (ln(sqrt(2))/pi)i + 2k, where k is any integer. This is because the complex logarithm function is periodic with period 2 pi i, so adding or subtracting any multiple of 2 pi i to the argument does not change the value of the function. Therefore, we can write log(i^x) = log(1 + i) + 2k pi i for any integer k, and then solve for x as before. To verify this solution, we can plug it into the original equation and see that both sides are equal. We have i^x = i^(-1/2 - (ln(sqrt(2))/pi)i + 2k) = i^(-1/2) * i^(-(ln(sqrt(2))/pi)i) * i^(2k) = (-i) * (sqrt(2)/2 - sqrt(2)/2 i) * 1 = 1 + i, where we used the properties of exponentiation and the values of i^n for different values of n.
If you can see how that works, and you are sure you don't need it, then by all means, don't generalize, but if you don't know or you're unsure, then you should use +2mπ on the x side and solve regardless of m, and then see how the answers change based on different values for m
These problems are of the complex Euler angles (instead of i multiplied by real number angles). This one is of the class of (i)^(i) situations where [(i)^(i)]^(-i/2) multiplied with [(i)^(i)]^ (-0.2206356) defining (i)^x
Mastering Euler's identity for complex exponential functions is akin to unlocking a door. Once opened, the countless problems that follow become mere pathways, driven by procedure rather than new conceptual revelations.
Check my channel for simplified maths videos like this
Isn't most math like that anyway, I mean new revelations occur pretty rarely don't they. Especially right now at a time when math has progressed this far in history.
Here's a solution method that does not make any unwarranted assumptions, so it
will work on other complex equations that the method in the video will not. Also,
I will not limit the complex exponentiation to the principal value, so all branches
of the function will be included (it is easily limited later if desired).
i^x = 1 + i
exp(lc(i)*x) = 1 + i lc() is the multi-valued complex logarithm function
x = a + i*b where a, b are real-valued
lc(i) = i*pi/2 + i*m*2*pi for integers m
exp( i*pi/2 *(1 + 4*m) * (a + i*b) ) = 1 + i
exp( -b * pi/2 *(1 + 4*m) + i*a*pi/2 *(1 + 4*m) ) = 1 + i
Definitions
u = b*pi/2 *(1 + 4*m)
v = a*pi/2 *(1 + 4*m)
exp(-u) * exp(i*v) = 1 + i
exp(-u) * (cos(v) + i*sin(v)) = 1 + i now set real and imaginary parts equal
{1} exp(-u) * cos(v) = 1 real parts
{2} exp(-u) * sin(v) = 1 imaginary parts
We need to determine u and v that solve these equations, which then determine
a and b , the real and imaginary parts of x.
We can divide {2} by {1} to get
tan(v) = 1
v = arctan(1) = pi/4 + n*2*pi = pi/4 *(1 + 8*n) for integers n
Note that we exclude 5*pi/4 + n*2*pi since that will not solve {1} and {2} due
to negative values of sin() and cos() not solving the equations.
a*pi/2 *(1 + 4*m) = pi/4 *(1 + 8*n)
a = 1/2 *(1 + 8*n) / (1 + 4*m)
exp(u) = cos(v) = cos( pi/4 *(1 + 8*n) ) = 1/sqrt(2)
u = ln(1/sqrt(2)) = -ln(2) / 2
b*pi/2 *(1 + 4*m) = -ln(2) / 2
b = -(ln(2) / pi) / (1 + 4*m)
x = 1/2 *(1 + 8*n) / (1 + 4*m) - i*(ln(2) / pi) / (1 + 4*m)
If we limit i^x to the principal value of the exponentiation function, then m = 0
x = 1/2 *(1 + 8*n) - i *ln(2) / pi
To solve the equation i^x = 1 + i for x for the principal branch, we need to use the complex logarithm function. The complex logarithm is defined as log(z) = ln|z| + i arg(z), where z is any complex number, |z| is its modulus, and arg(z) is its argument. The argument of a complex number is the angle it makes with the positive real axis in the complex plane, measured in radians.
Using this definition, we can rewrite the equation as log(i^x) = log(1 + i). Then, we can use the property of logarithms that log(a^b) = b log(a) to get x log(i) = log(1 + i). Next, we need to find the values of log(i) and log(1 + i) using the complex logarithm formula.
We have |i| = 1 and arg(i) = pi/2, so log(i) = ln(1) + i pi/2 = i pi/2. Similarly, we have |1 + i| = sqrt(2) and arg(1 + i) = pi/4, so log(1 + i) = ln(sqrt(2)) + i pi/4. Substituting these values into the equation, we get x i pi/2 = ln(sqrt(2)) + i pi/4. Solving for x, we get x = (ln(sqrt(2)) + i pi/4)/(i pi/2).
To simplify this expression, we can multiply both numerator and denominator by -i to get x = (-i ln(sqrt(2)) - pi/4)/(-i i pi/2) = (-i ln(sqrt(2)) - pi/4)/(pi/2). Finally, we can separate the real and imaginary parts of x to get x = -pi/(2pi) - (ln(sqrt(2))/pi)i = -1/2 - (ln(sqrt(2))/pi)i.
Therefore, the principal solution to the equation is x = -1/2 - (ln(sqrt(2))/pi)i. You can check this answer by plugging it into the original equation and verifying that both sides are equal.
The general solution to the equation i^x = 1 + i for x, for all branches of the complex logarithm, is given by x = -1/2 - (ln(sqrt(2))/pi)i + 2k, where k is any integer. This is because the complex logarithm function is periodic with period 2 pi i, so adding or subtracting any multiple of 2 pi i to the argument does not change the value of the function. Therefore, we can write log(i^x) = log(1 + i) + 2k pi i for any integer k, and then solve for x as before.
To verify this solution, we can plug it into the original equation and see that both sides are equal. We have i^x = i^(-1/2 - (ln(sqrt(2))/pi)i + 2k) = i^(-1/2) * i^(-(ln(sqrt(2))/pi)i) * i^(2k) = (-i) * (sqrt(2)/2 - sqrt(2)/2 i) * 1 = 1 + i, where we used the properties of exponentiation and the values of i^n for different values of n.
Shouldnt the general solution be
((8n+1)*pi-i*2*ln2)/((8*k+2)*pi), k,n integers?
I’m wondering the same thing.
So before I do the problem or watch the video, are we supposed to generalize the side with x or not?
You should generalize it, and if you do, you'll see at the end that the answers will be a subset of what was given by the base value
If you can see how that works, and you are sure you don't need it, then by all means, don't generalize, but if you don't know or you're unsure, then you should use +2mπ on the x side and solve regardless of m, and then see how the answers change based on different values for m
I was so lost on this one, I don't have enough breadcrums to get back home.... 😅😂
These problems are of the complex Euler angles (instead of i multiplied by real number angles). This one is of the class of (i)^(i) situations where [(i)^(i)]^(-i/2) multiplied with [(i)^(i)]^ (-0.2206356) defining (i)^x
Hallo everytwo!
😁
x=1/2-i ln2/pi
Polar form formula is good to use:
{r[cos(a)+i*sin(a)]}^n= r^n[cos(n*a)+i*sin(n*a)]😋😋😋😋😋😋
Maybe I am wrong, but the final answer is not for x but ix.
Sample text
Nice ? I keep this word only for girls, sorry
I'm assuming you're down bad
@@DeJay7 or just honest, this video has nothing to deserve that we can use this word. Sorry.