In my solution ln1 =- 4 (π/2+2nπ)i
Is it correct or what wrong in it.[image][1]
ln1=ln1/1 =ln(1/i)⁴ =-4lni
i is e^i(π/2+2nπ)
So -4lni is ln1 is -4ln(e)^i(π/2+2nπ)
So ln1 is -4(π/2+2nπ)
[
\ln 1 = \ln \frac{1}{1} = \ln \frac{1}{(i)^4} = 4 \ln i^{-1} = 4 \ln i^1 = -4 \ln i
]
[
i = e^{i \left(\frac{\...
Thomas can you please show what is the basic mistake means X ^n why we can't take out in complex how to prove that as like √ab≠√a√b if both are negative.
You haven't shown how you got your result, so I can't tell you what your mistake is. But $0$ is one value of $\ln 1,$ and it is not covered by your formula.
@CyclotomicField The question isn't about $\ln (1^{-4}),$ it's about $\ln( i^{-4}).$ And of course $i^{-4}=1.$ The error is the assumption that $-4\ln z$ is always the value (and the only value) of $\ln(z^4)$ where $z$ is an arbitrary complex number.
Discussion on question by Siddhant Gu…
Discussion on question by Siddhant Gu…