n = √3 + i ← this is a complex number The modulus of n is: m = √[(√3)² + (1)²] = √(3 + 1) = 2 The argument of n is β such as: tan(β) = 1/√3 = (√3)/3 → β = π/6 N = 2 + 2i√3 ← this is a new complex number The modulus of N is: M = √[(2)² + (2√3)²] = √(4 + 12) = √16 = 4 The argument of N is ω such as: tan(ω) = (2√3)/2 = √3 → ω = π/3 You want to get: n^(z) = N For the modulus: m^(z) = M 2^(z) = 4 2^(z) = 2^(2) z = 2 For the argument: z * β = ω z * (π/6) = π/3 z = (π/3) * (6/π) z = (6π)/(3π) z = 2 You can conclude that: z = 2
You can slightly simplify the solution by writing the imaginary part of the numerator as i (pi + 6pi.k + 12pi.n - 12pi.n) A whole load of stuff cancels giving z = 2 + i [(12pi.n - 6pi.k)/(6ln2 + i(pi + 12pi.n))] This also shows us that when k = 2n, z = 2.
n = √3 + i ← this is a complex number
The modulus of n is: m = √[(√3)² + (1)²] = √(3 + 1) = 2
The argument of n is β such as: tan(β) = 1/√3 = (√3)/3 → β = π/6
N = 2 + 2i√3 ← this is a new complex number
The modulus of N is: M = √[(2)² + (2√3)²] = √(4 + 12) = √16 = 4
The argument of N is ω such as: tan(ω) = (2√3)/2 = √3 → ω = π/3
You want to get: n^(z) = N
For the modulus: m^(z) = M
2^(z) = 4
2^(z) = 2^(2)
z = 2
For the argument: z * β = ω
z * (π/6) = π/3
z = (π/3) * (6/π)
z = (6π)/(3π)
z = 2
You can conclude that: z = 2
When the obvious guess and check really isn't enough.
Nice!
Thanks!
(√3+i)^z=2+2√3i
2^(z-2)·e^(i(π/6+2nπ)z)=e^(i(π/3+2kπ))
∴(1/6+2n)z=(1/3+2k), 2^(z-2)=1
→k=2n ∴z=2
You can slightly simplify the solution by writing the imaginary part of the numerator as
i (pi + 6pi.k + 12pi.n - 12pi.n)
A whole load of stuff cancels giving
z = 2 + i [(12pi.n - 6pi.k)/(6ln2 + i(pi + 12pi.n))]
This also shows us that when k = 2n, z = 2.
Nice