- #1
DrKareem
- 101
- 1
Hi. I need to prove the following identity
[tex] \arccos{z} =i \ln { z + (z^2 -1)^\frac{1}{2} } [/tex]
I was given a hint to write
[tex]\cos{A}=z[/tex],
then rewrite
[tex]\cos{A}[/tex]
in terms of the exponential.
[tex]\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z[/tex]
I took the log on both sides and got stuck at that point.
[tex] \ln{\exp{iA} + \exp{-iA}}=\ln{z^2} [/tex]
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 [tex]\arccos{z}[/tex] right?).
[tex] \arccos{z} =i \ln { z + (z^2 -1)^\frac{1}{2} } [/tex]
I was given a hint to write
[tex]\cos{A}=z[/tex],
then rewrite
[tex]\cos{A}[/tex]
in terms of the exponential.
[tex]\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z[/tex]
I took the log on both sides and got stuck at that point.
[tex] \ln{\exp{iA} + \exp{-iA}}=\ln{z^2} [/tex]
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 [tex]\arccos{z}[/tex] right?).
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