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Question Number 58153 by behi83417@gmail.com last updated on 18/Apr/19

arctan((√2)−i)=?          [i=(√(−1))]

arctan(2i)=?[i=1]

Answered by MJS last updated on 18/Apr/19

found these:  tan (a+bi) =((2e^(2b) sin 2a)/(2e^(2b) cos 2a +e^(4b) +1))+((e^(4b) −1)/(2e^(2b) cos 2a +e^(4b) +1))  arctan (a+bi) =(1/2)(πsign a −arctan ((b+1)/a) +arctan ((b−1)/a))+(i/4)ln ((a^2 +(b+1)^2 )/(a^2 +(b−1)^2 ))  ⇒ arctan ((√2)−i) =(π/4)+(1/2)arctan ((√2)/2) −(i/4)ln 3 ≈1.09314−.274653i

foundthese:tan(a+bi)=2e2bsin2a2e2bcos2a+e4b+1+e4b12e2bcos2a+e4b+1arctan(a+bi)=12(πsignaarctanb+1a+arctanb1a)+i4lna2+(b+1)2a2+(b1)2arctan(2i)=π4+12arctan22i4ln31.09314.274653i

Commented by behi83417@gmail.com last updated on 18/Apr/19

dear MJS sir! it is a great knowledge!

dearMJSsir!itisagreatknowledge!

Commented by MJS last updated on 18/Apr/19

sin (a+bi) =sin a cosh b +i cos a sinh b  cos (a+bi) =cos a cosh b −i sin a sinh b  arcsin (a+bi) =arcsin (((√(a^2 +b^2 +2a+1))−(√(a^2 +b^2 −2a+1)))/2) +i sign b ln ((1/2)((√(a^2 +b^2 −2a+1))+(√(a^2 +b^2 +2a+1))+(√(2(a^2 +b^2 −1+(√((a^2 +b^2 −2a+1)(a^2 +b^2 +2a+1)))))))  arccos (a+bi) =arccos (((√(a^2 +b^2 +2a+1))−(√(a^2 +b^2 −2a+1)))/2) −i sign b ln ((1/2)((√(a^2 +b^2 −2a+1))+(√(a^2 +b^2 +2a+1))+(√(2(a^2 +b^2 −1+(√((a^2 +b^2 −2a+1)(a^2 +b^2 +2a+1)))))))

sin(a+bi)=sinacoshb+icosasinhbcos(a+bi)=cosacoshbisinasinhbarcsin(a+bi)=arcsina2+b2+2a+1a2+b22a+12+isignbln(12(a2+b22a+1+a2+b2+2a+1+2(a2+b21+(a2+b22a+1)(a2+b2+2a+1)))arccos(a+bi)=arccosa2+b2+2a+1a2+b22a+12isignbln(12(a2+b22a+1+a2+b2+2a+1+2(a2+b21+(a2+b22a+1)(a2+b2+2a+1)))

Commented by behi83417@gmail.com last updated on 18/Apr/19

thanks again sir.

thanksagainsir.

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