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Complex Integration — Howell & Matthews
Mathematics — Complex Analysis · inverse-trig-hyperbolic-identities
Problem Establish the following identities.
(a) arccos z = − i log [ z + i ( 1 − z 2 ) 1 2 ] \arccos z = -i\log\left[z + i(1-z^2)^{\frac{1}{2}}\right] arccos z = − i log [ z + i ( 1 − z 2 ) 2 1 ] d d z arccos z = − 1 ( 1 − z 2 ) 1 2 \dfrac{d}{dz}\arccos z = \dfrac{-1}{(1-z^2)^{\frac{1}{2}}} d z d arccos z = ( 1 − z 2 ) 2 1 − 1 arctan z = i 2 log ( i + z i − z ) \arctan z = \dfrac{i}{2}\log\left(\dfrac{i+z}{i-z}\right) arctan z = 2 i log ( i − z i + z ) d d z arctan z = 1 1 + z 2 \dfrac{d}{dz}\arctan z = \dfrac{1}{1+z^2} d z d arctan z = 1 + z 2 1 arcsin z + arccos z = π 2 + 2 n π \arcsin z + \arccos z = \dfrac{\pi}{2} + 2n\pi arcsin z + arccos z = 2 π + 2 nπ n n n d d z arctanh z = 1 1 − z 2 \dfrac{d}{dz}\operatorname{arctanh} z = \dfrac{1}{1-z^2} d z d arctanh z = 1 − z 2 1 arcsinh z = log [ z + ( z 2 + 1 ) 1 2 ] \operatorname{arcsinh} z = \log\left[z + (z^2+1)^{\frac{1}{2}}\right] arcsinh z = log [ z + ( z 2 + 1 ) 2 1 ] d d z arcsinh z = 1 ( z 2 + 1 ) 1 / 2 \dfrac{d}{dz}\operatorname{arcsinh} z = \dfrac{1}{(z^2+1)^{1/2}} d z d arcsinh z = ( z 2 + 1 ) 1/2 1 arccosh z = log [ z + ( z 2 − 1 ) 1 2 ] \operatorname{arccosh} z = \log\left[z + (z^2-1)^{\frac{1}{2}}\right] arccosh z = log [ z + ( z 2 − 1 ) 2 1 ] d d z arccosh z = 1 ( z 2 − 1 ) 1 2 \dfrac{d}{dz}\operatorname{arccosh} z = \dfrac{1}{(z^2-1)^{\frac{1}{2}}} d z d arccosh z = ( z 2 − 1 ) 2 1 1 Browse the problem library