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

arctan((√2)−i)=?          [i=(√(−1))]

$$\boldsymbol{\mathrm{arctan}}\left(\sqrt{\mathrm{2}}−\boldsymbol{\mathrm{i}}\right)=?\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{i}}=\sqrt{−\mathrm{1}}\right] \\ $$

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

$$\mathrm{found}\:\mathrm{these}: \\ $$$$\mathrm{tan}\:\left({a}+{b}\mathrm{i}\right)\:=\frac{\mathrm{2e}^{\mathrm{2}{b}} \mathrm{sin}\:\mathrm{2}{a}}{\mathrm{2e}^{\mathrm{2}{b}} \mathrm{cos}\:\mathrm{2}{a}\:+\mathrm{e}^{\mathrm{4}{b}} +\mathrm{1}}+\frac{\mathrm{e}^{\mathrm{4}{b}} −\mathrm{1}}{\mathrm{2e}^{\mathrm{2}{b}} \mathrm{cos}\:\mathrm{2}{a}\:+\mathrm{e}^{\mathrm{4}{b}} +\mathrm{1}} \\ $$$$\mathrm{arctan}\:\left({a}+{b}\mathrm{i}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi\mathrm{sign}\:{a}\:−\mathrm{arctan}\:\frac{{b}+\mathrm{1}}{{a}}\:+\mathrm{arctan}\:\frac{{b}−\mathrm{1}}{{a}}\right)+\frac{\mathrm{i}}{\mathrm{4}}\mathrm{ln}\:\frac{{a}^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}−\mathrm{i}\right)\:=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\frac{\mathrm{i}}{\mathrm{4}}\mathrm{ln}\:\mathrm{3}\:\approx\mathrm{1}.\mathrm{09314}−.\mathrm{274653i} \\ $$

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

dear MJS sir! it is a great knowledge!

$${dear}\:{MJS}\:{sir}!\:{it}\:{is}\:{a}\:{great}\:{knowledge}! \\ $$

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)))))))

$$\mathrm{sin}\:\left({a}+{b}\mathrm{i}\right)\:=\mathrm{sin}\:{a}\:\mathrm{cosh}\:{b}\:+\mathrm{i}\:\mathrm{cos}\:{a}\:\mathrm{sinh}\:{b} \\ $$$$\mathrm{cos}\:\left({a}+{b}\mathrm{i}\right)\:=\mathrm{cos}\:{a}\:\mathrm{cosh}\:{b}\:−\mathrm{i}\:\mathrm{sin}\:{a}\:\mathrm{sinh}\:{b} \\ $$$$\mathrm{arcsin}\:\left({a}+{b}\mathrm{i}\right)\:=\mathrm{arcsin}\:\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}}}{\mathrm{2}}\:+\mathrm{i}\:\mathrm{sign}\:{b}\:\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}}+\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{1}+\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}\right)}\right.}\right)\right) \\ $$$$\mathrm{arccos}\:\left({a}+{b}\mathrm{i}\right)\:=\mathrm{arccos}\:\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}}}{\mathrm{2}}\:−\mathrm{i}\:\mathrm{sign}\:{b}\:\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}}+\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{1}+\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}\right)}\right.}\right)\right) \\ $$

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

thanks again sir.

$${thanks}\:{again}\:{sir}. \\ $$

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