i = e^(i(2n+½)pi) i^x = e^(i(2n+½)pi*x) e = e*e^(i2mpi) = e^(i2mpi +1) n,m € Z i^x = e e^(i(2n+½)pi*x) = e^(i2mpi + 1) i(2n+½)pi*x = i2mpi + 1 Divide by i*pi (2n+½)x = 2m - i/pi x = (2m - i/pi)/(2n+½) n,m € Z For n = 0 you get x = 4m - 2i/pi, which is your solution.
If you back substitute the general solution the original equation and considering i = e^i(4k+1)PI/2?and x=2(1+2nPIi)/PIi i^x = e^((4k+1)(1+2nPIi) For this to satisfy the original equation k=n=0 That means there is only one solution x =-2i/PI No general solution.
I thought a little further why the general solutions didn’t work. On both RHS and LHS you are evaluating the function (which is e^(y) in this case) and take the inverse of it ( ln(e^y)in this case) which should bring back the original variable y. In RHS y=1 but you get y=1+2iPIn instead. For these two values to be equal n must be zero. ‘This is because n=0 covers the entire unit circle. When you are evaluating the kth root it is fine because you are inverting the kth root of the function not the function itself. In this case to complete the unit circle n=0,…..k Makes sense?
The imaginary unit i in polar form is e^(iπ/2). So [e^(iπ/2)]^x=e. By the property (a^b)^c=a^bc: e^(iπx/2)=e Taking the natural logarithm ln of both sides: iπx/2=1 Multiplying both sides by 2 and dividing by iπ yields us with x=2/iπ. To simplify this, multiply both the numerator and denominator by i to get x=2i/-π=-2i/π. So x=-2i/π for x ∈ ℤ
You persist in the mistake of calling some of the solutions you found as the "principal value". Again, that is incorrect. All of the solutions you found are for the principal branch of the complex exponentiation. Real-valued constants do NOT have branch cuts. As I have repeated several times now, when you raise a number that is not positive real-valued to a non-integer exponent, the result is multi-valued. That is what creates branch cuts. Not arbitrarily choosing a real constant and multiplying it by exp(i*n*2*pi). That doing so will magically produce all solutions is just an unwarranted assumption that will fail to produce correct results in general. If you want the full solution set for this problem, it is: x = (2/pi) * (n*2*pi - i) / (4*k + 1) k, n are integers where the PRINCIPAL BRANCH of the complex exponentiation is taken by choosing k = 0 and any integer n.
I have made system of eqn, please solve this for real solutions : (a+b+c+d)/4 = 4th root of (abcd).........(1) ln(ln(ln(a^b^c^2))) = 2........(2) That's my challenge for you.
If you write the equation as e^(x*i*π/2) = e1,it follows immediately that x= -2 i/π .
i = e^(i(2n+½)pi)
i^x = e^(i(2n+½)pi*x)
e = e*e^(i2mpi) = e^(i2mpi +1)
n,m € Z
i^x = e
e^(i(2n+½)pi*x) = e^(i2mpi + 1)
i(2n+½)pi*x = i2mpi + 1
Divide by i*pi
(2n+½)x = 2m - i/pi
x = (2m - i/pi)/(2n+½) n,m € Z
For n = 0 you get x = 4m - 2i/pi, which is your solution.
Very nice👍
Our brave syber
If you back substitute the general solution the original equation and considering i = e^i(4k+1)PI/2?and x=2(1+2nPIi)/PIi
i^x = e^((4k+1)(1+2nPIi)
For this to satisfy the original equation k=n=0
That means there is only one solution x =-2i/PI
No general solution.
I tried adding the 2mPI to the LHS to generate a general solution. When resubmit to verify the answer it works only if m=0
I thought a little further why the general solutions didn’t work. On both RHS and LHS you are evaluating the function (which is e^(y) in this case) and take the inverse of it ( ln(e^y)in this case) which should bring back the original variable y. In RHS y=1 but you get y=1+2iPIn instead. For these two values to be equal n must be zero. ‘This is because n=0 covers the entire unit circle. When you are evaluating the kth root it is fine because you are inverting the kth root of the function not the function itself. In this case to complete the unit circle n=0,…..k
Makes sense?
The imaginary unit i in polar form is e^(iπ/2).
So [e^(iπ/2)]^x=e.
By the property (a^b)^c=a^bc:
e^(iπx/2)=e
Taking the natural logarithm ln of both sides:
iπx/2=1
Multiplying both sides by 2 and dividing by iπ yields us with x=2/iπ. To simplify this, multiply both the numerator and denominator by i to get x=2i/-π=-2i/π.
So x=-2i/π for x ∈ ℤ
対数の底が複素数まで拡張されているなら、x＝log i e
で間違いではない。つまり、正しい。
Nice. Thanks ❤
I chose the third method and since I don't have any problem at all seeing i in the denominator I let it x=2/πi
You can’t have imaginary numbers in the denominator
I like these simple equations :)
You persist in the mistake of calling some of the solutions you found as the "principal value". Again, that is incorrect. All of the solutions you found are for the principal branch of the complex exponentiation. Real-valued constants do NOT have branch cuts.
As I have repeated several times now, when you raise a number that is not positive real-valued to a non-integer exponent, the result is multi-valued. That is what creates branch cuts. Not arbitrarily choosing a real constant and multiplying it by exp(i*n*2*pi). That doing so will magically produce all solutions is just an unwarranted assumption that will fail to produce correct results in general.
If you want the full solution set for this problem, it is:
x = (2/pi) * (n*2*pi - i) / (4*k + 1) k, n are integers
where the PRINCIPAL BRANCH of the complex exponentiation is taken by choosing k = 0 and any integer n.
Does this mean 1^x=e has a solution?
Interesting. I haven't taken complex analysis yet, but I naively took 4th power on each side which led to xln(1)=4 which obviously doesn't work
Yeah, sometimes squaring or raising to fourth power will produce something that's obviously not true
Multiplication of exponents is not necessarily true in complex numbers
General solution is
x=(4nπ-2i)/((4k+1)π) n,k integers
I got the principal value, but I find the general solution tricky.
Why € Z? Sorry.
Nice❤
I have made system of eqn, please solve this for real solutions :
(a+b+c+d)/4 = 4th root of (abcd).........(1)
ln(ln(ln(a^b^c^2))) = 2........(2)
That's my challenge for you.
x=-2i/pi
x=-2i/(1+4n)π
Good Morning Sir e tt i(丌÷2)=i ⇒ i tt -i(丌÷2)=e