DrKareem
- 101
- 1
Hi. I need to prove the following identity
\arccos{z} =i \ln { z + (z^2 -1)^\frac{1}{2} }
I was given a hint to write
\cos{A}=z,
then rewrite
\cos{A}
in terms of the exponential.
\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z
I took the log on both sides and got stuck at that point.
\ln{\exp{iA} + \exp{-iA}}=\ln{z^2}
I know it's a correct method becaues the right hand is starting to take form. But i just couldn't solve for A (which will be \arccos{z} right?).
\arccos{z} =i \ln { z + (z^2 -1)^\frac{1}{2} }
I was given a hint to write
\cos{A}=z,
then rewrite
\cos{A}
in terms of the exponential.
\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z
I took the log on both sides and got stuck at that point.
\ln{\exp{iA} + \exp{-iA}}=\ln{z^2}
I know it's a correct method becaues the right hand is starting to take form. But i just couldn't solve for A (which will be \arccos{z} right?).
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